Question

The mean waiting time at the drive-through of a fast-food restaurant from the time an order is placed to the time the order is received is 84.3 seconds. A manager devises a new drive-through system that he believes will decrease wait time. He initiates the new system at his restaurant and measures the wait time for ten randomly selected orders. The wait times are provided in the table below. Based on the given data, is the new system effective? Use the α = 0.10 level of significance.

106.5	67.4	58.0	75.9	65.1
80.4	95.5	87.3	70.9	71.0

On a separate sheet of paper, write down the hypotheses (H₀ and H_a) to be tested.

Conditions:
Use Minitab Express to perform a normality test on the given data.
The P-value for the Anderson-Darling test of normality is ______ (Report this value exactly as it appears in Minitab Express. Do not round.)
Based on both the normal probability plot and this P-value, the t-test for means  (is / is not) valid for the given data.

Rejection Region:
To test the given hypotheses, we will use a (left / right / two) -tailed test. The appropriate critical value(s) for this test is/are _________ .  (Report your answer exactly as it appears in Table VI. For two-tailed tests, report both critical values in the answer blank separated by only a single space.)

Answer 1

hypothesis:-

conditions:-

i am using to test the normality of data.

steps :-

copy the data in a column of minitab named wait time graph probability plot single ok in graph variables select wait times in distribution select normal okok.

the probability plot so obtained be:-

The P-value for the Anderson-Darling test of normality is 0.553

Based on both the normal probability plot and this P-value, the t-test for means is  valid for the given data.

[ p value = 0.553 >0.10..so we fail to reject the null hypothesis and conclude that the data are normally distributed.]

Rejection Region:
To test the given hypotheses, we will use a left  -tailed test. The appropriate critical value(s) for this test is/are -1.383

[interpretation:-

THE t - TEST BE:-

necessary data for calculation:-

mean () = 77.80

sd (s) = 14.88

sample size (n) = 10

hypothesized mean(M) = 84.3

test statistic:-

df = 10-1 = 9

t critical value = -1.383

decision:-

t calc = -1.381 > t critical = -1.383.

so, we fail to reject the null hypothesis. hence, we conclude that the claim of the can not be supported.]

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