Question

In: Statistics and Probability

The mean waiting time at the drive-through of a fast-food restaurant from the time an order...

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 eleven 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.

102.5 67.4 58.0 75.9 65.1 68.3
80.4 95.5 87.3 70.9 71.0

Conditions:
Use Minitab Express to perform a normality test on the given data.
The P-value for the Anderson-Darling test of normality is [a] (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 [b] (is / is not) valid for the given data.

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

On a separate sheet of paper, sketch the rejection region(s) for this test. You will need this sketch in Question 7.

Refer to the data and Minitab Express output from Question 6. Note that a table of summary statistics for the given data appears beneath the normal probability plot in Minitab Express. Summary statistics can also be calculated in Minitab Express via Descriptive Statistics -> Describe.

The sample mean wait time is [a] seconds. (Report this value exactly as it appears in Minitab Express. Do not round.)
The sample standard deviation of wait times is [b] seconds.  (Report this value exactly as it appears in Minitab Express. Do not round.)
The test statistic for this test is t0=[c].  (Calculate this value in a single step in your calculator using the sample statistics reported above, and report your answer rounded to 3 decimal places.)

Label the test statistic in your sketch from Question 6. Use this sketch to conclude the hypothesis test.

We [d] (reject / fail to reject) H0.
The given data [e] (does / does not) provide significant evidence that the new drive-through system is effective in reducing the mean reducing wait time.


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

Solutions

Expert Solution

Hypotheses

Ho=u=84.3

Ha=u<84.3 ( claim)

Using Minitab<stats<descriptive statistics<1 sample t-test

Here is the output:

One-Sample T: data

Descriptive Statistics

N Mean StDev SE Mean 90% Upper Bound
for μ
11 76.57 13.61 4.10 82.20

μ: mean of data

Test

Null hypothesis H₀: μ = 84.3
Alternative hypothesis H₁: μ < 84.3
T-Value P-Value
-1.88 0.045

a) P value is 0.045

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

Rejection Region:
To test the given hypotheses, we will use a (left ) [c]-tailed test.

d) Critical value= -1.37218

Critical value is find using excel command =T.inv(0.10,10)

Degree of freedom =n-1=11-1=10

The sample mean wait time is 76.57 [a] seconds.

The sample standard deviation of wait times is 13.61 [b] seconds

The test statistic for this test is t0= -1.372

Since the test statistic (-1.88)< to (critical)

We reject the null hypothesis. there is sufficient evidence.

We [d] (reject  H0.
The given data [e] (does  provide significant evidence that the new drive-through system is effective in reducing the mean reducing wait time.

Please do the comment for any doubt or clarification. Thank You!


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