Description
Complete parts A - D of question 45 from Chapter 7, page 444, using file P08_07.xlsx
The file P08_07.xlsx contains a random sample of
200 service times during the busiest hour of the day
at a particular fast-food restaurant. Find a 95%
confidence interval for each of the following
population parameters. Then explain how each result
might be useful to the manager of the restaurant in
terms of improving service.
a. The mean service time
b. The standard deviation of service times
c. The proportion of service times longer than
90 seconds
d. The proportion of service times shorter than
60 seconds
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Explanation & Answer
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A. The Mean Service Time
Mean service time is 97.2 Sec
Sample size is 200
Standard Deviation is 67.3 Sec
Therefore, standard error of mean time value,
𝑠
𝑆𝐸 = √𝑛
67.3
= √200
=4.8
But 95% confidence interval is,
𝑥 ∓ 𝑡𝑎/2(𝑛−1) ×
𝑆
√𝑛
We require value of t for a/2=0.025 and (n-1) =199d.f. which,
T0.025.199 = TINV (0.025, 199)
=2.2585
On substitution,
97.2-2.2585×4.8≤ 𝜇 ≤ 97.2 − 2.2585×4.8
The answer: 86.4327≤𝜇 ≤ 107.9413
The average of service time lies between the interval 86.43 Sec and 108 Sec, which implies that
best average time is 86 Sec, whereas worst average time is 107 Sec. If the management needs
improvement, the process should be monitored closely and reduce the bottlenecks. This will
improve the efficiency.
B. Standard Deviation
Standard deviation = 67.3 Sec
95% confidence intervals for population standard deviation are;
(𝑛−1)𝑆 2
√𝑋 2 (𝑛−1)∝/2 ≤ 𝜎 ≤ √
(𝑛−1)𝑆 2
𝑋 2 (𝑛−1)(1−
∝
2
......... (1)
We can find Chi-square value using CHIINV function.
X2 (n-1).a/2 = X2199.0.025
= CHIINV
(0.025, 199)
= 235.9597
X2 (n-1)(1-.a/2 = X2199.0.975
= CHIINV (0.975, 199)
= 161.8262
After substituting we get;
...
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