2. The manager of the William Jessup Shack has developed a new system to reduce

Question

2. The manager of the William Jessup Shack has developed a new system to reduce the time students spend waiting to be served during peak business hours. The mean waiting time during peak business hours under the current system is roughly 9 to 10 minutes. The manager hopes that the new system will have a mean waiting time that is less than five minutes. The mean of the sample of 100 students, waiting times is 4.46. If we let denote the mean ofall possible student waiting times using the new system and assume that the population standard deviation equals 2.47: a) Calculate 95 percent and 99 percent confidence intervals for Using the 95 percent confidence interval, can the manager be 95 percent confident that is less than five minutes? Explain b) Using the 99 percent confidence interval, can the Shack manager be 99 percent confident that is less than five minutes? Explain c) Based on your answers to parts b and c, how convinced are you that the new mean waiting time is less than five minutes? d)

Explanation / Answer

Quetion 2

Here sample mean x = 4.46 mins

Population standard deviation = 2.47

standard error of sample mean se0= /sqrt(n) = 2.47/sqrt(100) = 2.47/10 = 0.247

95% confidence interval = x +- Z0.05 se0= 4.46 +- 1.96 * 0.247 = (3.976, 4.944)

99% confidence interval = x +- Z0.01 se0= 4.46 +2.576 * 0.247 = (3.824, 5.096)

(b) Here using the 95% confidence interval, the manager can be 95% confident that is less than five minutes as the 95% confidence interval doesn't consist the value of 5 minutes and it has its range below the value of 5 minutes.

(c) Here using the 99% confidence interval, the manager can't be 99% confident that is less than five minutes as the 99% confidence interval does consist the value of 5 minutes and it has its range that conisist the value of 5 minutes in its range.

(d) Here we have now to calculate p - value about how confident we are about it.

Z = (5 - 4.46)/0.247 = 2.1862

so here we are NORMSINV(2.1862) = 0.971or say 97.1% confident that waiting time is less than 5 minutes.

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