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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

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Answer:


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

Now we can calculate the degrees of freedom given by:


df=25+30-2=53

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


p_v =2*P(t_(53)>2.462) =0.0171

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

Explanation:

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


\sigma^2_A =\sigma^2_B =\sigma^2

And the statistic is given by this formula:


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

Where t follows a t distribution with
n_A+n_B -2 degrees of freedom

The system of hypothesis on this case are:

Null hypothesis:
\mu_A = \mu_B

Alternative hypothesis:
\mu_A \\eq \mu_B

The info given is:


n_A =25 represent the sample size for group A


n_B =30 represent the sample size for group B


\bar X_A =79 represent the sample mean for the group A


\bar X_B =71 represent the sample mean for the group B

And now we can calculate the statistic:


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

Now we can calculate the degrees of freedom given by:


df=25+30-2=53

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


p_v =2*P(t_(53)>2.462) =0.0171

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

User Nick Davies
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