Question

The idle time for taxi drivers in a day are normally distributed with an unknown population mean and standard deviation. If a random sample of 23 taxi drivers is taken and results in a sample mean of 172 minutes and sample standard deviation of 16 minutes, find a 98% confidence interval estimate for the population mean using the Student's t-distribution.

Answer 1

`172-2.51(16)/(√(23))=163.626`

`172+2.51(16)/(√(23))=180.374`

So on this case the 98% confidence interval would be given by (163.626;180.374)

Explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

`\bar X=172` represent the sample mean

`\mu` population mean (variable of interest)

s=16 represent the sample standard deviation

n=23 represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

`\bar X \pm t_(\alpha/2)(s)/(√(n))` (1)

In order to calculate the critical value
`t_(\alpha/2)` we need to find first the degrees of freedom, given by:

`df=n-1=23-1=22`

Since the Confidence is 0.98 or 98%, the value of
`\alpha=0.02` and
`\alpha/2 =0.01`, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.01,22)".And we see that
`t_(\alpha/2)=2.51`

Now we have everything in order to replace into formula (1):

`172-2.51(16)/(√(23))=163.626`

`172+2.51(16)/(√(23))=180.374`

So on this case the 98% confidence interval would be given by (163.626;180.374)