Question

In: Statistics and Probability

A random sample of 19 airline passengers at Hartsfield Jackson International airport showed that the mean...

A random sample of 19 airline passengers at Hartsfield Jackson International airport showed that the mean time spent waiting in line to pass though the TSA checkpoint was 31 minutes with a standard deviation of 7 minutes. Construct a 95 % confidence interval for the mean waiting time by all passengers in the TSA checkpoint at this airport. Assume that such waiting time for all passengers is normally distributed.

A) 31+1.65(7/P19)

B) 31+1.746(7/P19)

C) 31+1.96(7/P19)

D ) 31+2.101(7/P19)

Solutions

Expert Solution

The answer is:

One-Sample t-test Confidence Interval
The provided sample mean is Xˉ=31 and the sample standard deviation is s=7. The size of the sample is n = 19 and the required confidence level is 95%.

Degree of freedom
The number of degrees of freedom are df = 19 - 1 = 18, and the significance level is α=0.05.

Critical Value
Based on the provided information, the critical t-value for α=0.05 and df=18 degrees of freedom is tc​=2.1009.

Margin of Error

Therefore, based on the information provided, the 95% confidence for the population mean μ is calculated as:


Therefore, the 95% confidence interval for the population mean μ is 27.6261<μ<34.3739, which indicates that we are 95% confident that the true population proportion μ is contained by the interval (27.6261,34.3739)

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orchestra answered 2 years ago
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