Question

A sample of 100 information systems managers had an average hourly income of $40.00 with a standard deviation of $8.00. If we want to determine a 95% confidence interval for the average hourly income of the population, the value of t is:________.

Answer 1

Answer:

The t-score is 1.96

Step-by-step explanation:

The margin of error is given as

(Margin of Error) = (critical value) × (standard deviation of the sample mean)

The critical value is usually obtained from the t-score or the z-score at the given confidence level.

With small sample sizes (sample sizes less than 30) and/or information about the population standard deviation is not known, the t-distribution is used to find critical values.

We convert the confidence level to significance level and use the sample size to trace out the t-score.

Significance level = (1 - confidence level)/2

For example, a confidence level of 95% mean that 5% (spread equally at the top and bottom of the distribution as 2.5% and 2.5%) is still room for error, hence, the significance level = (5%/2) = 2.5%

This information, with the sample size, directs one exactly to where to obtain the t-score for the distribution on the t-score table.

But, with large sample sizes, with known information about its standard deviation, we typically use critical values on the Z-distribution to obtain the margin of error.

And as the sample size increases, the t-score approximates the z-score.

So, for a sample size of 100, the t-score can be simply obtained from the z-score table for a confidence level of 95%; and that is 1.96.

Hope this Helps!!!