A national grocer’s magazine reports the typical shopper spends 7 minutes in lin
ID: 3393068 • Letter: A
Question
A national grocer’s magazine reports the typical shopper spends 7 minutes in line waiting to check out. A sample of 17 shoppers at the local Farmer Jack’s showed a mean of 6.4 minutes with a standard deviation of 4.3 minutes.
Is the waiting time at the local Farmer Jack’s less than that reported in the national magazine? Use the 0.050 significance level.
What is the decision rule? (Negative amount should be indicated by a minus sign. Round your answer to 3 decimal places.)
The value of the test statistic is . (Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.)
A national grocer’s magazine reports the typical shopper spends 7 minutes in line waiting to check out. A sample of 17 shoppers at the local Farmer Jack’s showed a mean of 6.4 minutes with a standard deviation of 4.3 minutes.
Is the waiting time at the local Farmer Jack’s less than that reported in the national magazine? Use the 0.050 significance level.
Explanation / Answer
A)
Formulating the null and alternative hypotheses,
Ho: u >= 7
Ha: u < 7
As we can see, this is a left tailed test.
Thus, getting the critical t,
df = n - 1 = 16
tcrit = - 1.745883676
Thus,
Reject H0: µ 7 and fail to reject H1: µ < 7 when the test statistic is LESS THAN -1.746. [ANSWER]
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B)
Getting the test statistic, as
X = sample mean = 6.4
uo = hypothesized mean = 7
n = sample size = 17
s = standard deviation = 4.3
Thus, t = (X - uo) * sqrt(n) / s = -0.575317064 = -0.58 [ANSWER]
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c)
As t > -1.746, we DO NOT REJECT HO. [ANSWER]
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