Question

Assume the distribution of commute times to a major city follows the normal probability distribution, and the standard deviation is 4.2 minutes. A random sample of 13 commute times is given below in minutes. Find the 98% confidence interval for the mean travel time in minutes. Round your answers to two decimal places and use ascending order.

Answer 1

Answer:

The 98% CI for the mean travel time to the major city is [20.70; 26.14]minutes

Step-by-step explanation:

Hello!

The variable of interest is

X: commute time to a major city.

This variable has a normal distribution

X~N(μ;σ²)

The standard deviation is known to be:

σ= 4.2 minutes

A random sample of n= 13 commute times was taken:

11.5, 13.2, 14.7, 17.1, 18.7, 21.8, 22.4, 25, 26.9, 27.6, 31.1, 35.9, 38.6

You need to estimate the population mean travel time by calculating a 98% CI.

Since the variable has a normal distribution and the population standard deviation is known ,the statistic to use for this interval is the standard normal, then the formula for the interval is:

[X[bar] ± $Z_{1-\alpha /2}$ * $\frac{Sigma}{\sqrt{n} }$]

The value of the statistic for the 1 - α: 0.98 interval is:

$Z_{1-\alpha /2}= Z_{0.99}= 2.334$

Next is to calculate the sample mean:

X[bar]= ∑X/n= 304.5/13= 23.42 minutes

[23.42 ± 2.334 * ($\frac{4.2}{\sqrt{13} }$)]

[20.70; 26.14]minutes

With a 98% confidence level you'd expect that the interval [20.70; 26.14]minutes will contain the true mean travel time to the major city.

I hope this helps!