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

Answer:

$172-2.51\frac{16}{\sqrt{23}}=163.626$    

$172+2.51\frac{16}{\sqrt{23}}=180.374$

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

Step-by-step 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:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$   (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):

$172-2.51\frac{16}{\sqrt{23}}=163.626$    

$172+2.51\frac{16}{\sqrt{23}}=180.374$

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