Question

A marketing researcher wants to estimate the mean amount spent per year​ ($) on a web site by membership member shoppers. Suppose a random sample of 100 membership member shoppers who recently made a purchase on the web site yielded a mean amount spent of $57 and a standard deviation of ​$ 54 Complete parts​ (a) and​ (b) below.

a. Is there evidence that the population mean amount spent per year on the web site by membership member shoppers is different from %51 (Using a .01 level of significance)

Identify the critical​ value(s).

Determine the test statistic.

State the conclusion.

Determine the​ p-value and interpret its meaning

Answer 1

Solution:

The provided sample mean is x̄ = 57 and the Sample standard deviation is s = 54, and the sample size is n = 100.
(1) Null and Alternative Hypotheses:
The following null and alternative hypotheses need to be tested:

Ho: μ = 51
Ha: μ ≠ 51
This corresponds to a Two tail test, for which a t-test for one mean, with unknown population standard deviation will be used.
(2) Rejection Region:
Given significance level = α = 0.01
So Critical Value for the test is, tc = 2.6264   .....Using excel formula
=TINV(0.01,99)

3) Test statistic:

4. Decision: Since it is observed that |t| = 1.1111 < tc = 2.6264
It is then concluded that the Null Hypothesis is not rejected.

Using the P-value approach: The p-value is p = 0.2692…Using excel formula =TINV(0.01,99)
and since p = 0.2692 >0.01, it is concluded that the Null Hypothesis is not rejected

5. Conclusion: It is concluded that the Null Hypothesis is not rejected. Therefore, there is not enough evidence to claim that the population mean μ is different than 51, at the 0.01 significance level.

Done