s to at loast 4 Customers at Costoo spend an average of $130 per trip (The Wall

Question

s to at loast 4 Customers at Costoo spend an average of $130 per trip (The Wall Stroet Jounal, October 8,2010). One of Costoo's rivals would like to determine whether its spend more per trip. A survey of the receipts customers found that the sample mean was $135.25. Assume that the population standard deviation s $10.50 and that spending follows a normal distribution. Use Table 1 of 25 a. Speaify the null and alternative hypotheses to test whether average spending at the rival's store is more than $130 Ho: z 130; HA:

Explanation / Answer

Answer to the question)

M = 130

Population standard deviation SD = 10.5

Sample size n = 25

Sample mean ( x bar)= 135.25

.

Claim : M > 130

.

Answer to part a)

Null hypothesis: Mean is less than or equal to 130

Ho; M < = 130

Alternate hypothesis: Mean is greater than 130

Ha: M > 130

Thus the third answer choice is Correct

.

Answer to part b-1)

Test statistic formula

Z = (x bar - M) / (SD / sqrt( n))

Z = (135.25 - 130) / ( 10.5 / sqrt(25))

Z = 2.5

.

Answer to part b-2)

P value is the probability of getting z > 2.5

We refer to the Z table and get : P(z > 2.5) = 1 - 0.99379 = 0.00621

Thus The P value < 0.01

Hence FIRST answer choice is correct

.

Answer to c)

At 5% significance level , we got P 0.00621 < 0.05 , we reject the null , and conclude that Mean > 130

Thus we reject the null , since the P value is smaller than alpha

Hence Second answer choice is correct

.

Answer to part d-1)

Z critical value for 5% significance level , right tailed test is : 1.645

.

Answer to part d-2)

We got Z statistic 2.5 > Z critical 1.645 , this implies that the Z statistic lies in the rejection regions

and thus we reject the null. We conclude that the mean is greater than 130

This implies that the customers spend more at the rival store

And hence First answer choice is correct

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