## The Observed Significance of a Test

This section explains what the observed significance of a test is, including how to compute and use it in the p-value approach.

### EXERCISES

#### BASIC

1. Compute the observed significance of each test.

1. Testing $H_{0}: \mu=54.7$ vs. $H_{a}: \mu < 54.7$, test statistic $z=-1.72$.
2. Testing $H_{0}: \mu=195$ vs. $H_{a}: \mu \neq 195$, test statistic $z=-2.07$.
3. Testing $H_{0}: \mu=-45$ vs. $H_{a}: \mu > -45$, test statistic $z=2.54$.

3. Compute the observed significance of each test. (Some of the information given might not be needed).

1. Testing $H_{0}: \mu=27.5$ vs. $H_{a}: \mu > 27.5 ; n=49, \bar{x}=28.9, s=3.14$, test statistic $z=3.12$.
2. Testing $H_{0}: \mu=581$ vs. $H_{a}: \mu < 581 ; n=32, \bar{x}=560, s=47.8$, test statistic $z=-2.49$.
3. Testing $H_{0}: \mu=138.5$ vs. $H_{a}: \mu \neq 138.5 ; n=44, \bar{x}=137.6, s=2.45$, test statistic $z=-2.44$.

5. Make the decision in each test, based on the information provided.

1. Testing $H_{0}: \mu=82.9$ vs. $H_{a}: \mu < 82.9 @ \alpha=0.05$, observed significance $p=0.038$.
2. Testing $H_{0}: \mu=213.5$ vs. $H_{a}: \mu \neq 213.5 @ \alpha=0.01$, observed significance $p=0.038$.

#### APPLICATIONS

7. A lawyer believes that a certain judge imposes prison sentences for property crimes that are longer than the state average 11.7 months. He randomly selects 36 of the judge's sentences and obtains mean 13.8 and standard deviation 3.9 months.

1. Perform the test at the 1% level of significance using the critical value approach.
2. Compute the observed significance of the test.
3. Perform the test at the 1% level of significance using the p-value approach. You need not repeat the first three steps, already done in part (a).

9. The mean score on a 25-point placement exam in mathematics used for the past two years at a large state university is 14.3. The placement coordinator wishes to test whether the mean score on a revised version of the exam differs from 14.3. She gives the revised exam to 30 entering freshmen early in the summer; the mean score is 14.6 with standard deviation 2.4.

1. Perform the test at the 10% level of significance using the critical value approach.
2. Compute the observed significance of the test.
3. Perform the test at the 10% level of significance using the p-value approach. You need not repeat the first three steps, already done in part (a).

11. The mean yield for hard red winter wheat in a certain state is 44.8 bu/acre. In a pilot program a modified growing scheme was introduced on 35 independent plots. The result was a sample mean yield of 45.4 bu/acre with sample standard deviation 1.6 bu/acre, an apparent increase in yield.

1. Test at the 5% level of significance whether the mean yield under the new scheme is greater than 44.8 bu/acre, using the critical value approach.
2. Compute the observed significance of the test.
3. Perform the test at the 5% level of significance using the p-value approach. You need not repeat the first three steps, already done in part (a).