Question

Construct a 98% confidence interval for the population mean. Assume the population has a normal distribution. A random sample of 40 college students has mean annual earnings of $3200 with a standard deviation of $665.

Answer 1

Here as we are given the sample standard deviation and not the population standard deviation, therefore t distribution would be used for confidence interval here.

The confidence interval is thus computed as:

For n-1 = 39 degrees of freedom, we get from the t distribution tables that:

P( -2.426 < t₃₉ < 2.426 ) = 0.98

Therefore the t value used here should be 2.426

The confidence interval is thus computed here as:

This is the required 98% confidence interval for the population mean here.