Question

1. A new program in Kentucky is designed to reduce the proportion of gas tank leaks at service stations.  Prior to the program, it was estimated that as many as 30% of gas tanks at Kentucky services stations had leaks.  They want to demonstrate that the program was effective.  After implementation, the state took a random sample of n = 40 service stations, and found that only 9 had leaks ( = 0.225). All conditions for this hypothesis test are met.

Calculate the p-value for the test.  Show how numbers are plugged into the formula in order to receive full credit.  Show additional work for partial credit.  

Answer 1

Note that all the conditions for the hypotheses test are met.

Null and alternative hypotheses:

The null and alternative hypotheses are as follows:

Decision rule:

If p-value is less than the significance level, reject the null hypothesis; otherwise, fail to reject the null hypothesis.

Test statistic:

P-value:

Using Excel formula “=NORM.S.DIST(–1.0351,TRUE)”, the p-value is obtained as 0.1503.

Conclusion:

Since p-value is greater than 0.05, the null hypothesis is not rejected at 5% level of significance (Fail to reject H0). Therefore, there is enough evidence to suggest the leaks have reduced from 30%.