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 its 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 is $10.50 and that spending follows a normal distribution. Use standard normal curve area table. A, Specify the appropriate null and alternative hypotheses to test whether average spending at the rival�s store is more than $130.B. Calculate the value of the test statistic. C. Calculate the p-value. D. Calculate the critical value

Step-by-step explanation:

Given the following :

Average amount spent (U) = $130 per trip

Sample size (n) = 25

Sample mean (m) = $135.25

Population standard deviation (sd) = $10.50

Assume a standard distribution.

A.) Null hypothesis ; Alternative hypothesis

Hnull : U ≤ 130 ; Ha : U > 130

B.) value of test statistic :

Z = (m - U) / standard error

Standard Error = sd / sqrt(n)

Standard Error( S.E)= 10.50 / sqrt(25)

S.E = 10.50 / 5 = 2.1

Hence,

Z = (135.25 - 130) / 2.1

Z = 5.25 / 2.1

Z = 2.5

C. Calculate the p-value.

Using the p value calculator by entering the z-score above, the p value obtained is 0.0124

P value = 0.01

D. Calculate the critical value

answer: c i think

step-by-step explanation:

60.9774

step-by-step explanation:

because when you add 47.6 plus 7.32 equals 54.92

and when you add .0574 + 6= 6.0574

and when you add 54.92 and 6.0574 you get the answer of 60.9774.

plzz mark brainlest last one got deleted. cause i didn't explain it but i did but i understand what the mod meant