Question #337534

A supermarket boasts that checkout times for customers are never more than 15 minutes. A random sample of 36 costumers reveals a mean checkout time of 17 minutes with a standard deviation of 3 minutes. What can you conclude about the supermarket’s boast at the 0.05 level? 


1
Expert's answer
2022-05-05T17:01:39-0400

The following null and alternative hypotheses need to be tested:

H0:μ15H_0:\mu\le15

Ha:μ>15H_a:\mu>15

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

Based on the information provided, the significance level is α=0.05,\alpha = 0.05, df=n1=35df=n-1=35 degrees of freedom, and the critical value for a right-tailed test is tc=1.689572.t_c = 1.689572.

The rejection region for this right-tailed test is R={t:t>1.689572}R = \{t: t > 1.689572\}

The t-statistic is computed as follows:


t=xˉμs/n=17153/36=4t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{17-15}{3/\sqrt{36}}=4

Since it is observed that t=4>1.689572=tc,t = 4 >1.689572= t_c, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value for right-tailed, df=35df=35 degrees of freedom, t=4t=4 is p=0.000156,p = 0.000156, and since p=0.000156<0.05=α,p= 0.000156<0.05=\alpha, it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean μ\mu

is greater than 15, at the α=0.05\alpha = 0.05 significance level.



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