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 < t_{39} < 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.**

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