Question

Guest ages at a ski mountain resort typically have a right-skewed distribution. Assume the standard deviation (σ) of age is 14.5 years. From a random sample of 40 guests the sample mean is 36.4 years. Calculate a 99 percent confidence interval for µ, the true mean age of guests.

Answer 1

Answer: (30.49 years, 42.31 years)

Step-by-step explanation:

The formula to find the confidence interval is given by :-

$\overline{x}\pm z^*\dfrac{\sigma}{\sqrt{n}}.$

, where $\overline{x}$ = Sample mean

z* = Critical value.

$\sigma$ = Population standard deviation.

n= Sample size.

As per given , we have

$\overline{x}=36.4$

$\sigma=14.5$

n= 40

We know that the critical value for 99% confidence interval : z* = 2.576 (By z-table)

A 99 percent confidence interval for µ, the true mean age of guests will be :

$36.4\pm (2.576)\dfrac{14.5}{\sqrt{40}}\\\\ 36.4\pm (2.576)2.29265130362\\\\=36.4\pm5.90586975813\\\\\approx36.4\pm5.91\\\\=(36.4-5.91,\ 36.4+5.91)\\\\=(30.49,\ 42.31)$

∴ a 99 percent confidence interval for µ, the true mean age of guests  = (30.49 years, 42.31 years)