Question

From a population that is not normally distributed and whose standard deviation is not known, a sample of 50 items is selected to develop an interval estimate for mean. Which of the following statements are true?
a) the standard normal distribution can be used
b) the t distribution with 50 degrees of freedom must be used
c) the t distribution with 49 degrees of freedom must be used
d) the sample size must be increased in order to develop an interval estimate

Answer 1

Answer:

Correct option: c) the t distribution with 49 degrees of freedom must be used

Explanation:

The (1 - α)% confidence interval for population mean (μ) is:

$CI=\bar x\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$

OR

$CI=\bar x\pm t_{\alpha/2, (n-1)}\frac{s}{\sqrt{n}}$

A t-interval is used when the there is no information provided about the population standard deviation and when the population is normally distributed.

In this case there is no information about the population standard deviation and the population is also not normally distributed.

But as the t-distribution is derived from the normal distribution, to construct a t-interval the sample drawn must be large, i.e. n > 30.

Because for large sample sizes the sampling distribution of sample means will follow a Normal distribution with mean μ of the population and standard deviation $\frac{s}{\sqrt{n}}$.

In this case the sample size is, n = 50 > 30.

So a t distribution with n - 1 = 50 - 1 = 49 degrees of freedom will be used to construct the confidence interval for mean.