Sample Tests for a Population Mean

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Course: MA121: Introduction to Statistics
Book: Sample Tests for a Population Mean
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Date: Wednesday, May 22, 2024, 7:34 AM

Description

This section talks about using the central limit theorem to test a population mean when the sample size is large. It also addresses how to interpret the test results in the application background. Then, it discusses testing a population mean when the sample size is small, outlines a five-step testing procedure, and illustrates the procedure with an example. Study the example carefully and complete the relevant exercises and applications. Finally, it talks about large sample tests for a population proportion. The critical value and p-value approach are introduced based on a standardized test statistic.

Large Sample Tests for a Population Mean

In this section we describe and demonstrate the procedure for conducting a test of hypotheses about the mean of a population in the case that the sample size n is at least 30. The Central Limit Theorem states that \bar{X} is approximately normally distributed, and has mean \mu_{\bar{X}}=\mu and standard deviation \sigma_{\bar{X}}=\sigma / \sqrt{n}, where \mu and \sigma are the mean and the standard deviation of the population. This implies that the statistic

\frac{\bar{x}-\mu}{\sigma / \sqrt{n}}

has the standard normal distribution, which means that probabilities related to it are given in Figure 12.2 "Cumulative Normal Probability" and the last line in Figure 12.3 "Critical Values of".

If we know \sigma then the statistic in the display is our test statistic. If, as is typically the case, we do not know \sigma, then we replace it by the sample standard deviation s. Since the sample is large the resulting test statistic still has a distribution that is approximately standard normal.

Standardized Test Statistics for Large Sample Hypothesis Tests Concerning a Single Population Mean

If \sigma is known: \quad Z=\frac{\bar{x}-\mu_{0}}{\sigma / \sqrt{n}}

If \sigma is unknown: \quad Z=\frac{\bar{x}-\mu_{0}}{s / \sqrt{n}}

The test statistic has the standard normal distribution.

The distribution of the standardized test statistic and the corresponding rejection region for each form of the alternative hypothesis (left-tailed, right-tailed, or two-tailed), is shown in Figure 8.4 "Distribution of the Standardized Test Statistic and the Rejection Region".

Figure 8.4 Distribution of the Standardized Test Statistic and the Rejection Region



EXAMPLE 4

It is hoped that a newly developed pain reliever will more quickly produce perceptible reduction in pain to patients after minor surgeries than a standard pain reliever. The standard pain reliever is known to bring relief in an average of 3.5 minutes with standard deviation 2.1 minutes. To test whether the new pain reliever works more quickly than the standard one, 50 patients with minor surgeries were given the new pain reliever and their times to relief were recorded. The experiment yielded sample mean \bar{x}=3.1 minutes and sample standard deviation s=1.5 minutes. Is there sufficient evidence in the sample to indicate, at the 5% level of significance, that the newly developed pain reliever does deliver perceptible relief more quickly?


Solution:

We perform the test of hypotheses using the five-step procedure given at the end of Section 8.1 "The Elements of Hypothesis Testing".

  • Step 1. The natural assumption is that the new drug is no better than the old one, but must be proved to be better. Thus if \mu denotes the average time until all patients who are given the new drug experience pain relief, the hypothesis test is

\begin{aligned} H_{0}: \mu &=3.5 \\ \text { vs. } H_{a}: \mu & < 3.5 @ \alpha=0.05 \end{aligned}

  • Step 2. The sample is large, but the population standard deviation is unknown (the 2.1 minutes pertains to the old drug, not the new one). Thus the test statistic is

Z=\frac{\bar{x}-\mu_{0}}{s / \sqrt{n}}

and has the standard normal distribution.

  • Step 3. Inserting the data into the formula for the test statistic gives

Z=\frac{\bar{x}-\mu_{0}}{s / \sqrt{n}}=\frac{3.1-3.5}{1.5 / \sqrt{50}}=-1.886

  • Step 4. Since the symbol in H_{a} is "<" this is a left-tailed test, so there is a single critical value, -z_{\alpha}=-z_{0.05}, which from the last line in Figure 12.3 "Critical Values of " we read off as -1.645. The rejection region is (-\infty,-1.645].
  • Step 5. As shown in Figure 8.5 "Rejection Region and Test Statistic for " the test statistic falls in the rejection region. The decision is to reject H_{0}. In the context of the problem our conclusion is:

The data provide sufficient evidence, at the 5% level of significance, to conclude that the average time until patients experience perceptible relief from pain using the new pain reliever is smaller than the average time for the standard pain reliever.

Figure 8.5

Rejection Region and Test Statistic for Note 8.27 "Example 4"



EXAMPLE 5

A cosmetics company fills its best-selling 8-ounce jars of facial cream by an automatic dispensing machine. The machine is set to dispense a mean of 8.1 ounces per jar. Uncontrollable factors in the process can shift the mean away from 8.1 and cause either underfill or overfill, both of which are undesirable. In such a case the dispensing machine is stopped and recalibrated. Regardless of the mean amount dispensed, the standard deviation of the amount dispensed always has value 0.22 ounce. A quality control engineer routinely selects 30 jars from the assembly line to check the amounts filled. On one occasion, the sample mean is \bar{x}=8.2 ounces and the sample standard deviation is s= 0.25 ounce. Determine if there is sufficient evidence in the sample to indicate, at the 1% level of significance, that the machine should be recalibrated.


Solution:

  • Step 1. The natural assumption is that the machine is working properly. Thus if \mu denotes the mean amount of facial cream being dispensed, the hypothesis test is

\begin{aligned} H_{0}: \mu &=8.1 \\ \text { vs. } H_{a}: \mu & \neq 8.1 @ \alpha=0.01 \end{aligned}

  • Step 2. The sample is large and the population standard deviation is known. Thus the test statistic is

Z=\frac{\bar{x}-\mu_{0}}{\sigma / \sqrt{n}}

and has the standard normal distribution.

  • Step 3. Inserting the data into the formula for the test statistic gives

Z=\frac{\bar{x}-\mu_{0}}{\sigma / \sqrt{n}}=\frac{8.2-8.1}{0.22 / \sqrt{30}}=2.490

  • Step 4. Since the symbol in H_{a} is " \neq " this is a two-tailed test, so there are two critical values, \pm z_{\alpha / 2}=\pm z_{0.005}, which from the last line in Figure 12.3 "Critical Values of " we read off as \pm 2.576. The rejection region is (-\infty,-2.576] \cup[2.576, \infty).

  • Step 5. As shown in Figure 8.6 "Rejection Region and Test Statistic for " the test statistic does not fall in the rejection region. The decision is not to reject H_{0}. In the context of the problem our conclusion is:

The data do not provide sufficient evidence, at the 1% level of significance, to conclude that the average amount of product dispensed is different from 8.1 ounce. We conclude that the machine does not need to be recalibrated.

Figure 8.6

Rejection Region and Test Statistic for Note 8.28 "Example 5"




Creative Commons License This text was adapted by Saylor Academy under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License without attribution as requested by the work's original creator or licensor.

KEY TAKEAWAYS

  • There are two formulas for the test statistic in testing hypotheses about a population mean with large samples. Both test statistics follow the standard normal distribution.
  • The population standard deviation is used if it is known, otherwise the sample standard deviation is used.
  • The same five-step procedure is used with either test statistic.

EXERCISES

BASIC

1. Find the rejection region (for the standardized test statistic) for each hypothesis test.

  1. H_{0}: \mu=27 vs. H_{a}: \mu.
  2. H_{0}: \mu=52 vs. H_{a}: \mu \neq 52 @ \alpha=0.05.
  3. H_{0}: \mu=-105 vs. H_{a}: \mu>-105 @ \alpha=0.10.
  4. H_{0}: \mu=78.8 vs. H_{a}: \mu \neq 78.8 @ \alpha=0.10.

3. Find the rejection region (for the standardized test statistic) for each hypothesis test. Identify the test as left-tailed, right-tailed, or two-tailed.

  1. H_{0}: \mu=141 vs. H_{a}: \mu < 141 @ \alpha=0.20.
  2. H_{0}: \mu=-54 vs. H_{a}: \mu < -54 @ \alpha=0.05.
  3. H_{0}: \mu=98.6 vs. H_{a}: \mu \neq 98.6 @ \alpha=0.05.
  4. H_{0}: \mu=3.8 vs. H_{a}: \mu > 3.8 @ \alpha=0.001

5. Compute the value of the test statistic for the indicated test, based on the information given.

  1. Testing H_{0}: \mu=72.2 vs. H_{a}: \mu > 72.2, \sigma unknown, n=55, \bar{x}=75.1, s=9.25
  2. Testing H_{0}: \mu=58 vs. H_{a}: \mu > 58, \sigma=1.22, n=40, \bar{x}=58.5, s=1.29
  3. Testing H_{0}: \mu=-19.5 vs. H_{a}: \mu < -19.5, \sigma unknown, n=30, \bar{x}=-23.2, s=9.55
  4. Testing H_{0}: \mu=805 vs. H_{a}: \mu \neq 805, \sigma=37.5, n=75, \bar{x}=818, s=36.2

7. Perform the indicated test of hypotheses, based on the information given.

  1. Test H_{0}: \mu=212 vs. H_{a}: \mu < 212 @ \alpha=0.10, \sigma unknown, n=36, \bar{x}=211.2, s=2.2
  2. Test H_{0}: \mu=-18 vs. H_{a}: \mu > -18 @ \alpha=0.05, \sigma=3.3, n=44, \bar{x}=-17.2, s=3.1
  3. Test H_{0}: \mu=24 vs. H_{a}: \mu \neq 24 @ \alpha=0.02, \sigma unknown, n=50, \bar{x}=22.8, s=1.9


APPLICATIONS

9. In the past the average length of an outgoing telephone call from a business office has been 143 seconds. A manager wishes to check whether that average has decreased after the introduction of policy changes. A sample of 100 telephone calls produced a mean of 133 seconds, with a standard deviation of 35 seconds. Perform the relevant test at the 1% level of significance.

11. The average household size in a certain region several years ago was 3.14 persons. A sociologist wishes to test, at the 5% level of significance, whether it is different now. Perform the test using the information collected by the sociologist: in a random sample of 75 households, the average size was 2.98 persons, with sample standard deviation 0.82 person.

13. An automobile manufacturer recommends oil change intervals of 3,000 miles. To compare actual intervals to the recommendation, the company randomly samples records of 50 oil changes at service facilities and obtains sample mean 3,752 miles with sample standard deviation 638 miles. Determine whether the data provide sufficient evidence, at the 5% level of significance, that the population mean interval between oil changes exceeds 3,000 miles.

15. A grocery store chain has as one standard of service that the mean time customers wait in line to begin checking out not exceed 2 minutes. To verify the performance of a store the company measures the waiting time in 30 instances, obtaining mean time 2.17 minutes with standard deviation 0.46 minute. Use these data to test the null hypothesis that the mean waiting time is 2 minutes versus the alternative that it exceeds 2 minutes, at the 10% level of significance.

17. Authors of a computer algebra system wish to compare the speed of a new computational algorithm to the currently implemented algorithm. They apply the new algorithm to 50 standard problems; it averages 8.16 seconds with standard deviation 0.17 second. The current algorithm averages 8.21 seconds on such problems. Test, at the 1% level of significance, the alternative hypothesis that the new algorithm has a lower average time than the current algorithm.


ADDITIONAL EXERCISES

19. The mean household income in a region served by a chain of clothing stores is $48,750. In a sample of 40 customers taken at various stores the mean income of the customers was $51,505 with standard deviation $6,852.

Test at the 10% level of significance the null hypothesis that the mean household income of customers of the chain is $48,750 against that alternative that it is different from $48,750.

The sample mean is greater than $48,750, suggesting that the actual mean of people who patronize this store is greater than $48,750. Perform this test, also at the 10% level of significance. (The computation of the test statistic done in part (a) still applies here).


LARGE DATA SET EXERCISES

Note: All of the data sets associated with these questions are missing, but the questions themselves are included here for reference.

21. Large Data Set 1 records the SAT scores of 1,000 students. Regarding it as a random sample of all high school students, use it to test the hypothesis that the population mean exceeds 1,510, at the 1% level of significance. (The null hypothesis is that \mu=1510).

23. Large Data Set 1 lists the SAT scores of 1,000 students.

  1. Regard the data as arising from a census of all students at a high school, in which the SAT score of every student was measured. Compute the population mean \mu.
  2. Regard the first 50 students in the data set as a random sample drawn from the population of part (a) and use it to test the hypothesis that the population mean exceeds 1,510, at the 10% level of significance. (The null hypothesis is that \mu=1510).
  3. Is your conclusion in part (b) in agreement with the true state of nature (which by part (a) you know), or is your decision in error? If your decision is in error, is it a Type I error or a Type II error?

ANSWERS

1. a. Z \leq-1.645

   b. Z \leq-1.96 or Z \geq 1.96

   c. Z \geq 1.28

   d. Z \leq-1.645 or Z \geq 1.645

3. a. Z \leq-0.84

   b. Z \leq-1.645

   c. Z \leq-1.96 or Z \geq 1.96

   d. Z \geq 3.1

5. a. Z=2.235

   b. Z=2.592

   c. Z=-2.122

   d. Z=3.002

7. a. Z=-2.18,-z_{0.10}=-1.28, reject H_{0}.

   b. Z=1.61, z_{0.05}=1.645, do not reject H_{0}.

   c. Z=-4.47,-z_{0.01}=-2.33, reject H_{0}.

9. Z=-2.86,-z_{0.01}=-2.33, reject H_{0}.

11. Z=-1.69,-z_{0.025}=-1.96, do not reject H_{0}.

13. Z=8.33, z_{0.05}=1.645, reject H_{0}.

15. Z=.2.02, z_{0.10}=1.28, reject H_{0}.

17. Z=-2.08,-z_{0.01}=-2.33, do not reject H_{0}.

19. a. Z=2.54, z_{0.05}=1.645, reject H_{0};

   b. Z=2.54, z_{0.10}=1.28, reject H_{0}.

21. H_{0}: \mu=1510 vs. H_{a}: \mu > 1510. Test Statistic: Z=2.7882. Rejection Region: [2.33, \infty). Decision: Reject H_{0}

23. a. \mu_{0}=1528.74

   b. H_{0}: \mu=1510 vs. H_{a}: \mu > 1510. Test Statistic: Z=-1.41. Rejection Region: [1.28, \infty). Decision: Fail to reject H_{0}.

   c. No, it is a Type II error.

Small Sample Tests for a Population Mean

LEARNING OBJECTIVE

  1. To learn how to apply the five-step test procedure for test of hypotheses concerning a population mean when the sample size is small.

In the previous section hypotheses testing for population means was described in the case of large samples. The statistical validity of the tests was insured by the Central Limit Theorem, with essentially no assumptions on the distribution of the population. When sample sizes are small, as is often the case in practice, the Central Limit Theorem does not apply. One must then impose stricter assumptions on the population to give statistical validity to the test procedure. One common assumption is that the population from which the sample is taken has a normal probability distribution to begin with. Under such circumstances, if the population standard deviation is known, then the test statistic \left(\bar{x}-\mu_{0}\right) /(\sigma / \sqrt{n}) still has the standard normal distribution, as in the previous two sections. If \sigma is unknown and is approximated by the sample standard deviation s, then the resulting test statistic \left(\bar{x}-\mu_{0}\right) /(s / \sqrt{n}) follows Student's t-distribution with n-1 degrees of freedom.


Standardized Test Statistics for Small Sample Hypothesis Tests Concerning a Single Population Mean

If \sigma is known: \quad Z=\frac{\bar{x}-\mu_{0}}{\sigma / \sqrt{n}}

If \sigma is unknown: \quad T=\frac{\bar{x}-\mu_{0}}{s / \sqrt{n}}

The first test statistic (\sigma known) has the standard normal distribution.

The second test statistic (\sigma unknown) has Student's t-distribution with n-1 degrees of freedom.

The population must be normally distributed.

The distribution of the second standardized test statistic (the one containing s) and the corresponding rejection region for each form of the alternative hypothesis (left-tailed, right-tailed, or two-tailed), is shown in Figure 8.11 "Distribution of the Standardized Test Statistic and the Rejection Region". This is just like Figure 8.4 "Distribution of the Standardized Test Statistic and the Rejection Region", except that now the critical values are from the t-distribution. Figure 8.4 "Distribution of the Standardized Test Statistic and the Rejection Region" still applies to the first standardized test statistic (the one containing \sigma) since it follows the standard normal distribution.

Figure 8.11 Distribution of the Standardized Test Statistic and the Rejection Region


The p-value of a test of hypotheses for which the test statistic has Student's t-distribution can be computed using statistical software, but it is impractical to do so using tables, since that would require 30 tables analogous to Figure 12.2 "Cumulative Normal Probability", one for each degree of freedom from 1 to 30. Figure 12.3 "Critical Values of" can be used to approximate the p-value of such a test, and this is typically adequate for making a decision using the p-value approach to hypothesis testing, although not always. For this reason the tests in the two examples in this section will be made following the critical value approach to hypothesis testing summarized at the end of Section 8.1 "The Elements of Hypothesis Testing", but after each one we will show how the p-value approach could have been used.


EXAMPLE 10

The price of a popular tennis racket at a national chain store is $179. Portia bought five of the same racket at an online auction site for the following prices:

\begin{array}{lllll}155 & 179 & 175 & 175 & 161\end{array}

Assuming that the auction prices of rackets are normally distributed, determine whether there is sufficient evidence in the sample, at the 5% level of significance, to conclude that the average price of the racket is less than $179 if purchased at an online auction.


Solution:

  • Step 1. The assertion for which evidence must be provided is that the average online price \mu is less than the average price in retail stores, so the hypothesis test is

\begin{aligned} H_{0}: \mu &=179 \\ \text { vs. } H_{a}: \mu & < 179 @ \alpha=0.05 \end{aligned}

  • Step 2. The sample is small and the population standard deviation is unknown. Thus the test statistic is

T=\frac{\bar{x}-\mu_{0}}{s / \sqrt{n}}

and has the Student t-distribution with n-1=5-1=4 degrees of freedom.

  • Step 3. From the data we compute \bar{x}=169 and s=10.39. Inserting these values into the formula for the test statistic gives

T=\frac{\bar{x}-\mu_{0}}{s / \sqrt{n}}=\frac{169-179}{10.39 / \sqrt{5}}=-2.152

  • Step 4. Since the symbol in H_{a} is "<" this is a left-tailed test, so there is a single critical value, -t_{\alpha}=-t_{0.05}[d f=4]. Reading from the row labeled d f=4 in Figure 12.3 "Critical Values of " its value is -2.132. The rejection region is (-\infty,-2.132].
  • Step 5. As shown in Figure 8.12 "Rejection Region and Test Statistic for" the test statistic falls in the rejection region. The decision is to reject H_{0}. In the context of the problem our conclusion is:

The data provide sufficient evidence, at the 5% level of significance, to conclude that the average price of such rackets purchased at online auctions is less than $179.

Figure 8.12

Rejection Region and Test Statistic for Note 8.42 "Example 10"


To perform the test in Note 8.42 "Example 10" using the p-value approach, look in the row in Figure 12.3 "Critical Values of" with the heading d f=4 and search for the two t-values that bracket the unsigned value 2.152 of the test statistic. They are 2.132 and 2.776, in the columns with headings t_{0.050} and t_{0.025}. They cut off right tails of area 0.050 and 0.025, so because 2.152 is between them it must cut off a tail of area between 0.050 and 0.025. By symmetry -2.152 cuts off a left tail of area between 0.050 and 0.025, hence the p-value corresponding to t=-2.152 is between 0.025 and 0.05. Although its precise value is unknown, it must be less than \alpha=0.05, so the decision is to reject H_{\mathrm{0}}.


EXAMPLE 11

A small component in an electronic device has two small holes where another tiny part is fitted. In the manufacturing process the average distance between the two holes must be tightly controlled at 0.02 mm, else many units would be defective and wasted. Many times throughout the day quality control engineers take a small sample of the components from the production line, measure the distance between the two holes, and make adjustments if needed. Suppose at one time four units are taken and the distances are measured as

\begin{array}{llll}0.021 & 0.019 & 0.023 & 0.020\end{array}

Determine, at the 1% level of significance, if there is sufficient evidence in the sample to conclude that an adjustment is needed. Assume the distances of interest are normally distributed.


Solution:

  • Step 1. The assumption is that the process is under control unless there is strong evidence to the contrary. Since a deviation of the average distance to either side is undesirable, the relevant test is

 \begin{aligned} H_{0} &: \mu=0.02 \\ \text { vs. } H_{a}: \mu & \neq 0.02 @ \alpha=0.01\end{aligned}

where \mu denotes the mean distance between the holes.

  • Step 2. The sample is small and the population standard deviation is unknown. Thus the test statistic is

T=\frac{\bar{x}-\mu_{0}}{s / \sqrt{n}}

and has the Student t-distribution with n-1=4-1=3 degrees of freedom.

  • Step 3. From the data we compute \bar{x}=0.02075 and s=0.00171. Inserting these values into the formula for the test statistic gives

T=\frac{\bar{x}-\mu_{0}}{s / \sqrt{n}}=\frac{0.02075-0.02}{0.00171 / \sqrt{4}}=0.877

  • Step 4. Since the symbol in H_{a} is "\neq" this is a two-tailed test, so there are two critical values, \pm t_{\alpha / 2}=-t_{0.005}[d f=3]. Reading from the row in \underline{\text { Figure }} 12.3 "Critical Values of " labeled d f=3 their values are \pm 5.841. The rejection region is (-\infty,-5.841] \cup[5.841, \infty).
  • Step 5. As shown in Figure 8.13 "Rejection Region and Test Statistic for " the test statistic does not fall in the rejection region. The decision is not to reject H_{0}. In the context of the problem our conclusion is:
    The data do not provide sufficient evidence, at the 1% level of significance, to conclude that the mean distance between the holes in the component differs from 0.02 mm.

Figure 8.13

Rejection Region and Test Statistic for Note 8.43 "Example 11"


To perform the test in Note 8.43 "Example 11" using the p-value approach, look in the row in Figure 12.3 "Critical Values of" with the heading d f=3 and search for the two t-values that bracket the value 0.877 of the test statistic. Actually 0.877 is smaller than the smallest number in the row, which is 0.978, in the column with heading t_{0.200}. The value 0.978 cuts off a right tail of area 0.200, so because 0.877 is to its left it must cut off a tail of area greater than 0.200. Thus the p-value, which is the double of the area cut off (since the test is two-tailed), is greater than 0.400. Although its precise value is unknown, it must be greater than \alpha=0.01, so the decision is not to reject H_{0}.

KEY TAKEAWAYS

  • There are two formulas for the test statistic in testing hypotheses about a population mean with small samples. One test statistic follows the standard normal distribution, the other Student's t-distribution.
  • The population standard deviation is used if it is known, otherwise the sample standard deviation is used.
  • Either five-step procedure, critical value or p-value approach, is used with either test statistic.

EXERCISES

BASIC

1. Find the rejection region (for the standardized test statistic) for each hypothesis test based on the information given. The population is normally distributed.

a. H_{0}: \mu=27 vs. H_{a}: \mu < 27 @ \alpha=0.05, n=12, \sigma=2.2.

b. H_{0}: \mu=52 vs. H_{a}: \mu \neq 52 @ \alpha=0.05, n=6, \sigma unknown.

c. H_{0}: \mu=-105 vs. H_{a}: \mu > -105 @ \alpha=0.10, n=24, \sigma unknown.

d. H_{0}: \mu=78.8 vs. H_{a}: \mu \neq 78.8 @ \alpha=0.10, n=8, \sigma=1.7

3. Find the rejection region (for the standardized test statistic) for each hypothesis test based on the information given. The population is normally distributed. Identify the test as left-tailed, right-tailed, or two-tailed.

a. H_{0}: \mu=141 vs. H_{a}: \mu < 141 @ \alpha=0.20, n=29, \sigma unknown.

b. H_{0}: \mu=-54 vs. H_{a}: \mu < -54 @ \alpha=0.05, n=15, \sigma=1.9

c. H_{0}: \mu=98.6 vs. H_{a}: \mu \neq 98.6 @ \alpha=0.05, n=12, \sigma unknown.

d. H_{0}: \mu=3.8 vs. H_{a}: \mu > 3.8 @ \alpha=0.001, n=27, \sigma unknown.

5. 5. A random sample of size 20 drawn from a normal population yielded the following results: \bar{x}=49.2, s= 1.33.

a. Test H_{0}: \mu=50 vs. H_{a}: \mu \neq 50 @ \alpha=0.01.

b. Estimate the observed significance of the test in part (a) and state a decision based on the p-value approach to hypothesis testing.

7. A random sample of size 8 drawn from a normal population yielded the following results: \bar{x}=289, s= 46.

a. Test H_{0}: \mu=250 vs. H_{a}: \mu > 250 @ \alpha=0.05.

b. Estimate the observed significance of the test in part (a) and state a decision based on the p-value approach to hypothesis testing.


APPLICATIONS

9. Researchers wish to test the efficacy of a program intended to reduce the length of labor in childbirth. The accepted mean labor time in the birth of a first child is 15.3 hours. The mean length of the labors of 13 first-time mothers in a pilot program was 8.8 hours with standard deviation 3.1 hours. Assuming a normal distribution of times of labor, test at the 10% level of significance test whether the mean labor time for all women following this program is less than 15.3 hours.

11. Six coins of the same type are discovered at an archaeological site. If their weights on average are significantly different from 5.25 grams then it can be assumed that their provenance is not the site itself. The coins are weighed and have mean 4.73 g with sample standard deviation 0.18 g. Perform the relevant test at the 0.1% (1/10th of 1%) level of significance, assuming a normal distribution of weights of all such coins.

13. The recommended daily allowance of iron for females aged 19–50 is 18 mg/day. A careful measurement of the daily iron intake of 15 women yielded a mean daily intake of 16.2 mg with sample standard deviation 4.7 mg.

  1. Assuming that daily iron intake in women is normally distributed, perform the test that the actual mean daily intake for all women is different from 18 mg/day, at the 10% level of significance.
  2. The sample mean is less than 18, suggesting that the actual population mean is less than 18 mg/day. Perform this test, also at the 10% level of significance. (The computation of the test statistic done in part (a) still applies here).

15. The average number of days to complete recovery from a particular type of knee operation is 123.7 days. From his experience a physician suspects that use of a topical pain medication might be lengthening the recovery time. He randomly selects the records of seven knee surgery patients who used the topical medication. The times to total recovery were:

\begin{array}{lllllll}128 & 135 & 121 & 142 & 126 & 151 & 123\end{array}

  1. Assuming a normal distribution of recovery times, perform the relevant test of hypotheses at the 10% level of significance.
  2. Would the decision be the same at the 5% level of significance? Answer either by constructing a new rejection region (critical value approach) or by estimating the p-value of the test in part (a) and comparing it to \alpha.

17. Pasteurized milk may not have a standardized plate count (SPC) above 20,000 colony-forming bacteria per milliliter (cfu/ml). The mean SPC for five samples was 21,500 cfu/ml with sample standard deviation 750 cfu/ml. Test the null hypothesis that the mean SPC for this milk is 20,000 versus the alternative that it is greater than 20,000, at the 10% level of significance. Assume that the SPC follows a normal distribution.


ADDITIONAL EXERCISES

19. A calculator has a built-in algorithm for generating a random number according to the standard normal distribution. Twenty-five numbers thus generated have mean 0.15 and sample standard deviation 0.94. Test the null hypothesis that the mean of all numbers so generated is 0 versus the alternative that it is different from 0, at the 20% level of significance. Assume that the numbers do follow a normal distribution.

21. A manufacturing company receives a shipment of 1,000 bolts of nominal shear strength 4,350 lb. A quality control inspector selects five bolts at random and measures the shear strength of each. The data are:

\begin{array}{lllll}4,320 & 4,290 & 4,360 & 4,350 & 4,320\end{array}

  1. Assuming a normal distribution of shear strengths, test the null hypothesis that the mean shear strength of all bolts in the shipment is 4,350 lb versus the alternative that it is less than 4,350 lb, at the 10% level of significance.
  2. Estimate the p-value (observed significance) of the test of part (a).
  3. Compare the p-value found in part (b) to \alpha=0.10 and make a decision based on the p-value approach. Explain fully.

Answers

1. a. Z \leq-1.645
    b. T \leq-2.571 or T \geq 2.571
    c. T \geq 1.319
    d. Z \leq-1645 or Z \geq 1.645

3. a. T \leq-0.855
    b. Z \leq-1.645
    c. T \leq-2.201 or T \geq 2.201
    d. T \geq 3.435

5. a. T=-2.690, d f=19,-t_{0.005}=-2.861, do not reject H_{0}.
    b. 0.01 < p-value  < 0.02, \alpha=0.01, do not reject H_{0}.

7. a. T=2.398, d f=7, t_{0.05}=1.895, reject H_{0}.
    b. 0.01 < p-value  < 0.025, \alpha=0.05, reject H_{0}

9. T=-7.560, d f=12,-t_{0.10}=-1.356, reject H_{0}.

11. T=-7.076, d f=5,-t_{0.0005}=-6.869, reject H_{0}

13. a. T=-1.483, d f=14,-t_{0.05}=-1.761, do not reject H_{0};
      b. T=-1.483, d f=14,-t_{0.10}=-1.345, reject H_{0};

15. a. T=2.069, d f=6, t_{0.10}=1.44, reject H_{0};
      b. T=2.069, d f=6, t_{0.05}=1.943, reject H_{0}.

17. T=4.472, d f=4, t_{0.10}=1.533, reject H_{0}.

19. T=0.798, d f=24, t_{0.10}=1.318, do not reject H_{0}

21. a. T=-1.773, d f=4,-t_{0.05}=-2.132, do not reject H_{0}.
      b. 0.05 < p-value  < 0.10
      c. \alpha=0.05, do not reject \mathrm{H}_{0}

Large Sample Tests for a Population Proportion

LEARNING OBJECTIVES

  1. To learn how to apply the five-step critical value test procedure for test of hypotheses concerning a population proportion.
  2. To learn how to apply the five-step p-value test procedure for test of hypotheses concerning a population proportion.

Both the critical value approach and the p-value approach can be applied to test hypotheses about a population proportion p. The null hypothesis will have the form H_{0}: p=p_{0} for some specific number p_{\mathrm{o}} between \mathrm{o} and 1. The alternative hypothesis will be one of the three inequalities p < p_{0}, p > p_{0}, or p \neq p_{0} for the same number p_{0} that appears in the null hypothesis.

The information in Section 6.3 "The Sample Proportion" in Chapter 6 "Sampling Distributions" gives the following formula for the test statistic and its distribution. In the formula p_{0} is the numerical value of p that appears in the two hypotheses, q_{0}=1-p_{0}, \hat{p} is the sample proportion, and n is the sample size. Remember that the condition that the sample be large is not that n be at least 30 but that the interval

\left[\hat{p}-3 \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}, \hat{p}+3 \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\right]

lie wholly within the interval [0,1].

Standardized Test Statistic for Large Sample Hypothesis Tests Concerning a Single Population Proportion

Z=\dfrac{\hat{p}-p_{0}}{\sqrt{\dfrac{p_{0} q_{0}}{n}}}

The test statistic has the standard normal distribution.

The distribution of the standardized test statistic and the corresponding rejection region for each form of the alternative hypothesis (left-tailed, right-tailed, or two-tailed), is shown in Figure 8.14 "Distribution of the Standardized Test Statistic and the Rejection Region".

Figure 8.14 Distribution of the Standardized Test Statistic and the Rejection Region



EXAMPLE 12

A soft drink maker claims that a majority of adults prefer its leading beverage over that of its main competitor's. To test this claim 500 randomly selected people were given the two beverages in random order to taste. Among them, 270 preferred the soft drink maker's brand, 211 preferred the competitor's brand, and 19 could not make up their minds. Determine whether there is sufficient evidence, at the 5% level of significance, to support the soft drink maker's claim against the default that the population is evenly split in its preference.


Solution:

We will use the critical value approach to perform the test. The same test will be performed using the p-value approach in Note 8.49 "Example 14".

We must check that the sample is sufficiently large to validly perform the test. Since \hat{p}=270 / 500=0.54

\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}=\sqrt{\frac{(0.54)(0.46)}{500}} \approx 0.02

and

\left[\hat{p}-3 \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}, \hat{p}+3 \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\right]

=[0.54-(3)(0.02), 0.54+(3)(0.02)]

=[0.48,0.60] \subset[0,1]

so the sample is sufficiently large.

  • Step 1. The relevant test is

\begin{aligned} H_{0}: p &=0.50 \\ \text { vs. } H_{a}: p & > 0.50 @ \alpha=0.05\end{aligned}

where p denotes the proportion of all adults who prefer the company's beverage over that of its competitor's beverage.

  • Step 2. The test statistic is

Z=\frac{\hat{p}-p_{0}}{\sqrt{\frac{p_{0} q_{0}}{n}}}

and has the standard normal distribution.

  • Step 3. The value of the test statistic is

Z=\frac{\hat{p}-p_{0}}{\sqrt{\frac{p_{0} q_{0}}{n}}}=\frac{0.54-0.50}{\sqrt{\frac{(0.50)(0.50)}{500}}}=1.789

  • Step 4. Since the symbol in H_{a} is ">" this is a right-tailed test, so there is a single critical value, z_{\alpha}=z_{0.05}. Reading from the last line in Figure 12.3 "Critical Values of" its value is 1.645. The rejection region is [1.645, \infty).
  • Step 5. As shown in Figure 8.15 "Rejection Region and Test Statistic for " the test statistic falls in the rejection region. The decision is to reject H_{0}. In the context of the problem our conclusion is:

The data provide sufficient evidence, at the 5% level of significance, to conclude that a majority of adults prefer the company's beverage to that of their competitor's.

Figure 8.15

Rejection Region and Test Statistic for Note 8.47 "Example 12"



EXAMPLE 13

Globally the long-term proportion of newborns who are male is 51.46%. A researcher believes that the proportion of boys at birth changes under severe economic conditions. To test this belief randomly selected birth records of 5,000 babies born during a period of economic recession were examined. It was found in the sample that 52.55% of the newborns were boys. Determine whether there is sufficient evidence, at the 10% level of significance, to support the researcher's belief.


Solution:

We will use the critical value approach to perform the test. The same test will be performed using the p-value approach in Note 8.50 "Example 15".

The sample is sufficiently large to validly perform the test since

\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}=\sqrt{\frac{(0.5255)(0.4745)}{5000}} \approx 0.01

hence

\begin{aligned} &{\left[\hat{p}-3 \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}, \hat{p}+3 \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\right]} \\ &=[0.5255-0.03,0.5255+0.03] \\ &=[0.4955,0.5555] \subset[0,1] \end{aligned}

  • Step 1. Let p be the true proportion of boys among all newborns during the recession period. The burden of proof is to show that severe economic conditions change it from the historic long-term value of 0.5146 rather than to show that it stays the same, so the hypothesis test is

\begin{aligned} H_{0}: p &=0.5146 \\ \text { vs. } H_{a}: p & \neq 0.5146 @ \alpha=0.10 \end{aligned}

  • Step 2. The test statistic is

Z=\frac{\hat{p}-p_{0}}{\sqrt{\frac{p_{0} q_{0}}{n}}}

and has the standard normal distribution.

  • Step 3. The value of the test statistic is

Z=\frac{\hat{p}-p_{0}}{\sqrt{\frac{p_{0} q_{0}}{n}}}=\frac{0.5255-0.5146}{\sqrt{\frac{(0.5146)(0.4854)}{5000}}}=1.542

  • Step 4. Since the symbol in H_{a} is "\neq" this is a two-tailed test, so there are a pair of critical values, \pm z_{\alpha / 2}=\pm z_{0.05}=\pm 1.645. The rejection region is (-\infty,-1.645] \cup[1.645, \infty).
  • Step 5. As shown in Figure 8.16 "Rejection Region and Test Statistic for " the test statistic does not fall in the rejection region. The decision is not to reject H_{0}. In the context of the problem our conclusion is:

The data do not provide sufficient evidence, at the 10% level of significance, to conclude that the proportion of newborns who are male differs from the historic proportion in times of economic recession.

Figure 8.16

Rejection Region and Test Statistic for Note 8.48 "Example 13"



EXAMPLE 14

Perform the test of Note 8.47 "Example 12" using the p-value approach.


Solution:

We already know that the sample size is sufficiently large to validly perform the test.

  • Steps 1–3 of the five-step procedure described in Section 8.3.2 "The " have already been done in Note 8.47 "Example 12" so we will not repeat them here, but only say that we know that the test is right-tailed and that value of the test statistic is Z = 1.789.
  • Step 4. Since the test is right-tailed the p-value is the area under the standard normal curve cut off by the observed test statistic, z = 1.789, as illustrated in Figure 8.17. By Figure 12.2 "Cumulative Normal Probability" that area and therefore the p-value is 1−0.9633=0.0367.
  • Step 5. Since the p-value is less than \alpha = 0.05 the decision is to reject H_{0}.

Figure 8.17

P-Value for Note 8.49 "Example 14"



EXAMPLE 15

Perform the test of Note 8.48 "Example 13" using the p-value approach.


Solution:

We already know that the sample size is sufficiently large to validly perform the test.

  • Steps 1–3 of the five-step procedure described in Section 8.3.2 "The " have already been done in Note 8.48 "Example 13". They tell us that the test is two-tailed and that value of the test statistic is Z = 1.542.
  • Step 4. Since the test is two-tailed the p-value is the double of the area under the standard normal curve cut off by the observed test statistic, z = 1.542. By Figure 12.2 "Cumulative Normal Probability" that area is 1−0.9382=0.0618, as illustrated in Figure 8.18, hence the p-value is 2 \times 0.0618=0.1236.
  • Step 5. Since the p-value is greater than \alpha=0.10 the decision is not to reject H_{0}.

Figure 8.18

P-Value for Note 8.50 "Example 15"


KEY TAKEAWAYS

  • There is one formula for the test statistic in testing hypotheses about a population proportion. The test statistic follows the standard normal distribution.
  • Either five-step procedure, critical value or p-value approach, can be used.

EXERCISES

BASIC

On all exercises for this section you may assume that the sample is sufficiently large for the relevant test to be validly performed.

1. Compute the value of the test statistic for each test using the information given.

  1. Testing H_{0}: p=0.50 vs. H_{a}: p 0.50, n=360, \hat{p}=0.56.
  2. Testing H_{0}: p=0.50 vs. H_{a}: p \neq 0.50, n=360, \hat{p}=0.56.
  3. Testing H_{0}: p=0.37 vs. H_{a}: p < 0.37, n=1200, \hat{p}=0.35.

3. For each part of Exercise 1 construct the rejection region for the test for \alpha=0.05 and make the decision based on your answer to that part of the exercise.

5. For each part of Exercise 1 compute the observed significance (p-value) of the test and compare it to \alpha=0.05 in order to make the decision by the p-value approach to hypothesis testing.

7. Perform the indicated test of hypotheses using the critical value approach.

  1. Testing H_{0}: p=0.55 vs. H_{a}: p > 0.55 @ \alpha=0.05, n=300, \hat{p}=0.60.
  2. Testing H_{0}: p=0.47 vs. H_{a}: p \neq 0.47 @ \alpha=0.01, n=9750, \hat{p}=0.46.

9. Perform the indicated test of hypotheses using the p-value approach.

  1. Testing H_{0}: p=0.37 vs. H_{a}: p \neq 0.37 @ \alpha=0.005, n=1300, \hat{p}=0.40.
  2. Testing H_{0}: p=0.94 vs. H_{a}: p > 0.94 @ \alpha=0.05, n=1200, \hat{p}=0.96.


APPLICATIONS

11. Five years ago 3.9% of children in a certain region lived with someone other than a parent. A sociologist wishes to test whether the current proportion is different. Perform the relevant test at the 5% level of significance using the following data: in a random sample of 2,759 children, 119 lived with someone other than a parent.

13. Two years ago 72% of household in a certain county regularly participated in recycling household waste. The county government wishes to investigate whether that proportion has increased after an intensive campaign promoting recycling. In a survey of 900 households, 674 regularly participate in recycling. Perform the relevant test at the 10% level of significance.

15. A report five years ago stated that 35.5% of all state-owned bridges in a particular state were "deficient". An advocacy group took a random sample of 100 state-owned bridges in the state and found 33 to be currently rated as being "deficient". Test whether the current proportion of bridges in such condition is 35.5% versus the alternative that it is different from 35.5%, at the 10% level of significance.

17. According to the Federal Poverty Measure 12% of the U.S. population lives in poverty. The governor of a certain state believes that the proportion there is lower. In a sample of size 1,550, 163 were impoverished according to the federal measure.

  1. Test whether the true proportion of the state's population that is impoverished is less than 12%, at the 5% level of significance.
  2. Compute the observed significance of the test.

19. A special interest group asserts that 90% of all smokers began smoking before age 18. In a sample of 850 smokers, 687 began smoking before age 18.

  1. Test whether the true proportion of all smokers who began smoking before age 18 is less than 90%, at the 1% level of significance.
  2. Compute the observed significance of the test.


ADDITIONAL EXERCISES

21. A rule of thumb is that for working individuals one-quarter of household income should be spent on housing. A financial advisor believes that the average proportion of income spent on housing is more than 0.25. In a sample of 30 households, the mean proportion of household income spent on housing was 0.285 with a standard deviation of 0.063. Perform the relevant test of hypotheses at the 1% level of significance. Hint: This exercise could have been presented in an earlier section.


LARGE DATA SET EXERCISES

23. Large Data Sets 4 and 4A list the results of 500 tosses of a die. Let p denote the proportion of all tosses of this die that would result in a five. Use the sample data to test the hypothesis that p is different from 1/6, at the 20% level of significance.

http://www.gone.2012books.lardbucket.org/sites/all/files/data4.xls

http://www.gone.2012books.lardbucket.org/sites/all/files/data4A.xls

25. Lines 2 through 536 in Large Data Set 11 is a sample of 535 real estate sales in a certain region in 2008. Those that were foreclosure sales are identified with a 1 in the second column. Use these data to test, at the 10% level of significance, the hypothesis that the proportion p of all real estate sales in this region in 2008 that were foreclosure sales was less than 25%. (The null hypothesis is H_{0}: p=0.25).

http://www.gone.2012books.lardbucket.org/sites/all/files/data11.xls

ANSWERS

1. a. Z=2.277
    b. Z=2.277
   c. Z=-1.435

3. a. Z \geq 1.645; reject H_{0}.
    b. Z \leq-1.96 or Z \geq 1.96; reject H_{0}.
    c. Z \leq-1.645 ; do not reject H_{0}.

5. a. p-value =0.0116, \alpha=0.05; reject H_{0}.
    b. p-value =0.0232, \alpha=0.05 ; reject H_{0}.
    c. p-value =0.0749, \alpha=0.05; do not reject H_{0}.

7. a. Z=1.74, z_{0.05}=1.645, reject H_{0}.
    b. Z=-1.98,-z_{0.005}=-2.576, do not reject H_{0}

9. a. Z=2.24, p-value =0.025, \alpha=0.005, do not reject H_{0}.
    b. Z=2.92, p-value =0.0018, \alpha=0.05, reject H_{0}.

11. z=1.11, z_{0.025}=1.96, do not reject H_{0}.

13. Z=1.93, z_{0.10}=1.28, reject H_{0}.

15. Z=-0.523, \pm z_{0.05}=\pm 1.645, do not reject H_{0}.

17. a. Z=-1.798,-z_{0.05}=-1.645, reject H_{0};
    b. p--value =0.0359.

19. a. Z=-8.92,-z_{0.01}=-2.33, reject H_{0};
    b. p-value \approx 0.

21. Z=3.04, z_{0.01}=2.33, reject H_{0}.

23. H_{0}: p=1 / 6 vs. H_{a}: p \neq 1 / 6. Test Statistic: Z=-0.76. Rejection Region: (-\infty,-1.28] \cup[1.28, \infty). Decision: Fail to reject H_{0}.

25. H_{0}: p=0.25 vs. H_{a}: p < 0.25. Test Statistic: Z=-1.17. Rejection Region: (-\infty,-1.28]. Decision: Fail to reject H_{0}.