Question

a regional manager is comparing customer satisfaction ratings at two of its stores. 100 customer rattings from each store are randomly selected "the manager wants to see if there is a significant difference in customer satisfaction ratings at the two stores. what is the P-value and the conclusion at a signficicance level of 0.05"

Answer 1

Answer:

Yes, there is a significant difference in customer satisfaction ratings at the two stores

The P-value is P=0.000.

Step-by-step explanation:

The question is incomplete:

The store A has a sample mean of 4.05 and a sample standard deviation of 0.039. The store B has a sample mean of 3.99 and a sample standard deviation of 0.025.

We have to perform an hypothesis test for the difference between means.

The claim will state that satisfaction ratings differ. So, the null and alternative hypothesis are:

$H_0: \mu_1-\mu2=0\\\\H_a: \mu_1-\mu2\neq0$

being μ1: the actual staisfaction rating of store A, and μ2: the actual satisfaction rating of store B.

The significance level is 0.05.

We know that, for both samples, the sample size is n=100.

The difference between means is:

$M_d=\mu_1-\mu_2=4.05-3.99=0.06$

The standard error of the diffence between means is calculated as:

$s_M=\sqrt{\dfrac{s_1^2+s_2^2}{n}}=\sqrt{\dfrac{0.039^2+0.025^2}{100}}=\sqrt{\dfrac{0.0021}{100}}=\sqrt{0.000021}=0.0046$

The z-statistic can be calculated as:

$z=\dfrac{M_d-(\mu_1-\mu_2)}{s_M}=\dfrac{0.06-0}{0.0046}=13.04$

The P-value for this test statistic in a two tailed test is

$P-value=2P(z>13.04)=0.000$

The P-value is smaller than the significance level, so the effect is significant. The null hypothesis is rejected.

There is enough evidence that the customer satisfaction rating differs from store A to store B.