Question

In: Statistics and Probability

Exhibit: Costco Customers. Customers at Costco spend an average of $130 per trip (The Wall Street...

Exhibit: Costco Customers.

Customers at Costco spend an average of $130 per trip (The Wall Street Journal, October 6, 2010). One of Costco’s rivals would like to determine whether Costco's customers spend more per trip. A survey of the receipts of 25 customers found that the sample mean was $135.25. Assume that the population standard deviation of spending is $10.50 and the spending follows a normal distribution (use the significance level 0.07).

Round your solutions for this Exhibit to 4 decimal places.

1. Refer to the Exhibit Costco Customers.

Provide the null and the alternative hypotheses.

Group of answer choices

H0:μ≤130;H1:μ>130H0:μ≤130;H1:μ>130

H0:μ≥130;H1:μ<130H0:μ≥130;H1:μ<130

H0:μ≤135.25;H1:μ>135.25H0:μ≤135.25;H1:μ>135.25

H0:μ=130;H1:μ≠130H0:μ=130;H1:μ≠130

2. Refer to the Exhibit Costco Customers.

Compute the test statistic.

3. Refer to the Exhibit Costco Customers.

Calculate the p-value for the test.

4. Refer to the Exhibit Costco Customers.

State your conclusion for the test using the p-value.

Group of answer choices

p-value < 0.07, so we reject Ho. Therefore, there is enough evidence to conclude that Costco’s customers spend more than $130 per trip.

p-value < 0.07, so we reject Ho. Therefore, there is not enough evidence to conclude that Costco’s customers spend more than $130 per trip.

p-value < 0.07, so we cannot reject Ho. Therefore, there is enough evidence to conclude that Costco’s customers spend more than $130 per trip.

p-value < 0.07, so we cannot reject Ho. Therefore, there is not enough evidence to conclude that Costco’s customers spend more than $130 per trip.

Solutions

Expert Solution

Solution :

Given that ,

= 130

= 135.25

= 10.50

n = 25

The null and alternative hypothesis is ,

H0 :   = 130

H1 : > 130

This is the right tailed test .

Test statistic = z

= ( - ) / / n

= ( 135.25 - 130) / 10.50 / 25

= 2.5

The test statistic = 2.5

P - value = P(Z > 2.5 ) = 1 - P (Z < 2.5)

= 1 - 0.9938

= 0.0062

P-value = 0.0062

= 0.07

0.0062 < 0.07

P-value <

Reject the null hypothesis .

Conclusion : - p-value < 0.07, so we reject Ho. Therefore, there is enough evidence to conclude that Costco’s customers spend more than $130 per trip.


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