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 87.6 seconds. A manager devises a new drive-through system that he believes will decrease wait time.
As a test, he initiates the new system at his restaurant and measures the wait time for 10 randomly selected orders. The wait times are provided in the table to the right. Complete parts (a) and (b) below. 104.2 81.4 66.9 95.0 56.2 85.6 75.2 69.3 67.7 78.5 LOADING... Click the icon to view the table of correlation coefficient critical values.
(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.
The normal probability plot is shown below and the sample correlation coefficient is known to be r equals 0.986. Are the conditions for testing the hypothesis satisfied? ▼ the conditions ▼ satisfied.
The normal probability plot ▼ linear enough, since the correlation coefficient is ▼ less greater than the critical value. 60 75 90 105 -2 -1 0 1 2 Time (sec) Expected z-score A normal probability plot has a horizontal axis labeled Time (seconds) from 50 to 115 in increments of 5 and a vertical axis labeled Expected z-score from negative 2 to 2 in increments of 0.5. Ten plotted points closely follow the pattern of a line that rises from left to right through (56, negative 1.55) and (95, 1). All coordinates are approximate.
(b) Is the new system effective? Conduct a hypothesis test using the P-value approach and a level of significance of alpha equals 0.05. First determine the appropriate hypotheses.
Upper H 0: ▼ p mu sigma ▼ less than equals not equals greater than 87.6
Upper H 1: ▼ p mu sigma ▼ equals greater than less than not equals 87.6
Find the test statistic. t 0 equals nothing (Round to two decimal places as needed.) Find the P-value. The P-value is nothing. (Round to three decimal places as needed.)
Use the alpha equals 0.05 level of significance. What can be concluded from the hypothesis test?
A. The P-value is greater than the level of significance so there is not 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 less than the level of significance so there is sufficient evidence to conclude the new system is effective.
Click to select your answer(s).
(a)
Yes, the conditions are satisfied. The normal probability plot is linear enough since the correlation coefficient is greater than the critical value.
(b)
H0: = 87.6
H1: 87.6
From the given data, the following statistics are calculated:
n = 10
= 780/10 =78
s= 14.2715
SE = s/
= 14.2715/
= 4.5130
Test statistic is given by:
t0 = (78 - 87.6)/4.5130
= - 2.12
ndf = n - 1 = 10 - 1 = 9
One Tail- Left Side Test
By Technology, P -Value = 0.032
Correct option:
D. The P - Value is less than the level of significance so there is sufficient evidence to conclude the new system is effective.
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