Question

A store owner surveyed 25 randomly selected customers and found the ages shown​ (in years). The mean is 33.24 and the standard deviation is 10.87. The owner wants to know if the mean age of all customers isis 29 years old. Use the given information to complete parts a through f.

Answer 1

The provided sample mean is Xbar=33.24 and the sample standard deviation is s=10.87, and the sample size is n=25.

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ = 29 mean age of customers is 29

Ha: μ ≠ 29 mean age of customers is significantly different from

This corresponds to a two-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=0.05, and the critical value for a two-tailed test is t_c = 2.064tc​=2.064.(using t table distribution)

The rejection region for this two-tailed test is R={t:∣t∣>2.064}

(3) Test Statistics

The t-statistic is computed as follows:

(4) Decision about the null hypothesis

Since it is observed that ∣t∣=1.95≤tc​=2.064, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p=0.0629, and since p=0.0629≥0.05, it is concluded that the null hypothesis is not rejected.

p vlaue is calculated using

p[|t|>1.95]= 2*p[t>1.95]

The number of degrees of freedom is df=24. We need to graph T≥1.95. The following is obtained:

The following is obtained graphically:

The following is obtained graphically:

p[|t|>1.95]= 2*0.0315 =2.064

(5) Conclusion

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean \muμ is different than 29, at the 0.05 significance level.

i.e mean age of customers in 29 years

Confidence Interval

The 95% confidence interval is 28.753<μ<37.727.