The mean waiting time at the drivethrough of a fast food res

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 86.1 seconds.  A manager devises a new drive-through system that she believes will decrease wait time.  As a test, she initiates the new system at her restaurant and measures wait time per 10 randomly selected orders.  The wait times are shown in the table below:

109.8, 67.1, 56.0, 74.2, 66.7, 80.2, 93.4, 86.5, 72.9, 83.2

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A) Because the sample size is small, the manager must verify that the wait time is normally distributed and the sample does not contain any outliers.  Are the conditions satisfied?

a) Yes

b) No, it contains outliers

c) No, it is not normally distributed and contains outliers

d) No, the sample is not normally distributed

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B) Is the new system effective? Conduct a hypothesis test using the P-value approach and a level of significance of a = 0.1.

Choose the correct hypothesis:

     \\(Ho: (choose \\ option: \\ p, \\ \\mu, \\ or \\ \\sigma),(choose \\ option: \\ <, \\ >, \\ =, \\ or \\ does \\ not \\ equal) \\ 84.1\\)     

     \\(H1: (choose \\ option: \\ p, \\ \\mu, \\ or \\ \\sigma),(choose \\ option: \\ <, \\ >, \\ =, \\ or \\ does \\ not \\ equal) \\ 84.1\\)     

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C) Find the test statistic: t0 = ___.

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D) Fine the P-Value: P-value = ___.

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E) Use the a = 0.1 level of significance.  What can be concluded from the hypothesis test?

a) The P-value is less than the level of significance so there is sufficient evidence to conclude the new system is effective.

b) The P-value is greater than the level of significance so there is sufficient evidence to conclude the new system is effective.

c) The P-value is less than the level of significance so there is not sufficient evidence to conclude the new system is effective.

d) The P-value is greater than the level of significance so there is not sufficient evidence to conclude the new system is effective.

Solution

A)

a) YES. There are no outliers.

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b)

Formulating the null and alternative hypotheses,              
              
Ho:   u   >=   86.1  
Ha:    u   <   86.1   [ANSWER]

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c)
              
As we can see, this is a    left   tailed test.      
              
              
Getting the test statistic, as              
              
X = sample mean =    79          
uo = hypothesized mean =    86.1          
n = sample size =    10          
s = standard deviation =    15.32420018          
              
Thus, t = (X - uo) * sqrt(n) / s =    -1.465144747 [ANSWER]

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d)

As
  
df = n - 1 =    9          
      
              
Then the p value is, for left tailed test,
              
p =    0.088461315   [ANSWER]

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e)

As P <    0.1,

OPTION a) The P-value is less than the level of significance so there is sufficient evidence to conclude the new system is effective. [ANSWER, A]
              
Comparing t and tcrit (or, p and significance level), we   REJECT THE NULL HYPOTHESIS.