Sample satisfaction scores about the two retailers by the customers are shown below. In the report, 100 means the respondent is completely satisfied. Assume that a population standard deviation of 12 is a reasonable assumption for both retailers. Conduct the hypothesis test and report the p-value. At a .05 level of significance what is your conclusion? Include the solution process.
Retailer A Sample size = 25 Sample mean = 79
Retailer B sample size = 30 sample mean = 71

For this case the p value is lower than the significance so then we have enough evidence to reject the null hypothesis and we can conclude that the true means are different

Step-by-step explanation:

We assume that the population deviation is the same for both cases

[tex]\sigma^2_A =\sigma^2_B =\sigma^2[/tex]

And the statistic is given by this formula:

[tex]t=\frac{(\bar X_A -\bar X_B)-(\mu_{A}-\mu_B)}{\sigma \sqrt{\frac{1}{n_A}+\frac{1}{n_B}}}[/tex]

Where t follows a t distribution with [tex]n_A+n_B -2[/tex] degrees of freedom

The system of hypothesis on this case are:

Null hypothesis: [tex]\mu_A = \mu_B[/tex]

Alternative hypothesis: [tex]\mu_A \neq \mu_B[/tex]

The info given is:

[tex]n_A =25[/tex] represent the sample size for group A

[tex]n_B =30[/tex] represent the sample size for group B

[tex]\bar X_A =79[/tex] represent the sample mean for the group A

[tex]\bar X_B =71[/tex] represent the sample mean for the group B

And now we can calculate the statistic:

[tex]t=\frac{(79 -71)-(0)}{12\sqrt{\frac{1}{25}+\frac{1}{30}}}=2.462[/tex]

Now we can calculate the degrees of freedom given by:

[tex]df=25+30-2=53[/tex]

And now we can calculate the p value using the altenative hypothesis:

[tex]p_v =2*P(t_{53}>2.462) =0.0171[/tex]

For this case the p value is lower than the significance so then we have enough evidence to reject the null hypothesis and we can conclude that the true means are different