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] ± [tex]Z_{1-\alpha /2}[/tex] * [tex]\frac{Sigma}{\sqrt{n} }[/tex]]
The value of the statistic for the 1 - α: 0.98 interval is:
[tex]Z_{1-\alpha /2}= Z_{0.99}= 2.334[/tex]
Next is to calculate the sample mean:
X[bar]= ∑X/n= 304.5/13= 23.42 minutes
[23.42 ± 2.334 * ([tex]\frac{4.2}{\sqrt{13} }[/tex])]
[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!
