The Internal Revenue Service is studying contributions to charity. The mean calculated from a random sample of 36 returns is $150. Assume that the population standard deviation is known and equal to $20. Construct a 98 percent confidence interval for the population mean.

Formula:

CI98 = mean + or - z (sd/√n)

mean = 150; sd = 20; n = 36
Find z using a z-table to represent 98% confidence.

I'll let you take it from here.

what is answer

The Internal Revenue Service is studying contributions to charity. The mean calculated from a random sample of 36 returns is $150. Assume that the population standard deviation is known and equal to $20. Construct a 98 percent confidence interval for the population mean.

To construct a confidence interval for the population mean, we can use the formula:

Confidence Interval = Sample Mean ± (Z * (Population Standard Deviation / √(Sample Size)))

In this case, the sample mean is $150, the population standard deviation is $20, and the sample size is 36.

First, we need to determine the value of Z for a 98 percent confidence level. This value represents the number of standard deviations corresponding to the desired confidence level. We can use a standard normal distribution table or a statistical calculator to find this value.

For a 98 percent confidence level, the area in the tails of the standard normal distribution is (1 - 0.98) / 2 = 0.01. Looking up this area in the table, we find that the corresponding Z-value is approximately 2.33.

Now, we can substitute the values into the formula:

Confidence Interval = $150 ± (2.33 * ($20 / √36))

Simplifying this equation, we get:

Confidence Interval = $150 ± (2.33 * ($20 / 6))

Confidence Interval = $150 ± (2.33 * $3.33)

Confidence Interval = $150 ± $7.74

Finally, we can complete the calculation:

Lower Limit = $150 - $7.74 = $142.26
Upper Limit = $150 + $7.74 = $157.74

Therefore, the 98 percent confidence interval for the population mean is ($142.26, $157.74).