Question

A news report stated that the average car thief can break into a car in 30 seconds. You think you
can do it faster, so you break into 16 cars. Your average time was 25 seconds with a sample
standard deviation of 6 seconds. Based on your results, test the claim that the average time a car
thief can break into a car is less than 30 seconds. Use a 0.10 level.

Answer 1

The provided sample mean is 25 and the sample standard deviation is s = 6 and the sample size is n = 16

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

average car thief can break into a car in 30 seconds

average car thief can break into a car in less than 30 seconds

This corresponds to a left-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.1, and the critical value for a left-tailed test is t_c = -1.341

(3) Test Statistics

The t-statistic is computed as follows:

(4) Decision about the null hypothesis

Since it is observed that t = -3.333 <tc​=−1.341, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p = 0.0023 , and since p = 0.0023 <0.1, it is concluded that the null hypothesis is rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean μ is less than 30, at the 0.1 significance level.