Question

In a random sample of 8 people, the mean commute time to work was 35.5 minutes and the standard deviation was 7.3 minutes. A 98% confidence interval using the t-distribution was calculated to be (27.8,43.2). After researching commute times to work, it was found that the population standard deviation is 8.9 minutes. Find the margin of error and construct 98% confidence interval using the standard normal distribution with the appropriate calculations for a standard deviation that is known. Compare the results.

The margin of error of ? is:

A 98% confidence interval using the standard normal distribution is:

Answer #1

We are given the sample standard deviation as s = 7.3 minutes.

The population standard deviation here is
= 8.9 minutes.

For n - 1 = 8 - 1 = 7 degrees of freedom, we get from the t
distribution tables:

P( -2.998 < t_{7} < 2.998 ) = 0.98

From standard normal tables we get:

P( -2.326 < Z < 2.326 ) = 0.98

**T distribution tables:
**

Therefore sample mean is computed here as:

**Standard normal Tables:
**

**So the 98% confidence interval using standard normal
distribution would be given as:**

**This is the required confidence interval
here.**

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