Question

Customers of a large grocery chain who are members of the chain's incentives program spend an average of $45 per visit to the grocery. In an attempt to try and increase the amount spent by these regular customers at their stores, the chain institutes a new feature of the incentives program where customers receive an extra 5% discount on groceries when they spend $50 or more. A sample of 100 customers in the program is randomly selected and it is discovered that the next time they go to the store they spend an average of $53 with a standard deviation of $28.

a. Does this provide sufficient evidence that the true mean is now significantly more than $45? Conduct the appropriate hypothesis test. Be sure to specify your null and alternative hypothesis, test statistic, P-value, and conclusion.

b. Construct a 99% confidence interval for the true average amount customers spend when this new feature is added to the incentives program.

Answer 1

Since we are not given the data about the distribution, and population standard deviation is unknown, we use the students t test for degrees of freedom = n - 1 = 99

(a) Right Tailed t test, Single Mean

Given: = $45, = $53, s = $28, n = 100, = 0.01 (Since we are also finding the 99% CI)

The Hypothesis:

H0: = 45: The mean amount spent by customers after the introduction of the incentive program is equal to $45.

Ha: > 1.5: The mean amount spent by customers after the introduction of the incentive program is greater than $45.

This is a Right tailed test

The Test Statistic:

The test statistic is given by the equation:

t observed = 2.86

The p Value: The p value (Right tailed) for t = 2.86, for degrees of freedom (df) = 99, is; p value = 0.0026

The Critical Value: The critical value (Right Tail) at = 0.01, for df = 99, tcritical= +2.365

The Decision Rule:  

The Critical Value Method: If tobserved is > tcritical.

The p-value Method: If P value is < , Then Reject H0.

The Decision:

The Critical Value Method: Since tobserved (2.86) is > t critical (2.3.65), we Reject H0.

The p-value Method: Since P value (0.0026) is < (0.01) , We Reject H0.

The Conclusion: There is sufficient evidence at the 99% significance level to conclude that the mean amount spent by customers after the introduction of the incentive program is greater than $45.

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(b) 99% Confidence interval

From the data: = 53, s = 28, n = 100

The tcritical (2 tail) for = 0.01, for df = 99, is 2.626

The Confidence Interval is given by ME, where

The Lower Limit = 53 - 7.353 = 67.821

The Upper Limit = 53.167 + 7.353 = 45.647

The 99% Confidence Interval is (45.647 , 60.353)

Since the CI does not contain $45 and contains $54, we can reject the null hypothesis of = $45

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