It is known that the population mean for the Full Scale IQ of the WAIS is 100 with a standard deviation of 15. A researcher assesses a sample of 200 adults and find that they have a mean Full Scale IQ of 102. The point estimate of the mean for this group is _ and the 99% confidence interval for this group is __________.

Question

It is known that the population mean for the Full Scale IQ of the WAIS is 100 with a standard deviation of 15. A researcher assesses a sample of 200 adults and find that they have a mean Full Scale IQ of 102. The point estimate of the mean for this group is _ and the 99% confidence interval for this group is __________.

a. 100; (97.27; 102.73)
b. 102; (97.27, 102.73)
c. 100; (99.27; 104.73)
d. 102; (99.27, 104.73)

Answer 1

99% = mean ± 2.575 SEm

SEm = SD/√n

"mean for this group" = sample mean

Answer 2

To calculate the point estimate of the mean for this group, we can use the formula:

Point Estimate = Sample Mean

In this case, the sample mean is given as 102. Therefore, the point estimate of the mean for this group is 102.

To calculate the 99% confidence interval for this group, we can use the formula:

Confidence Interval = Sample Mean ± (Critical Value × Standard Error)

The critical value corresponds to the level of confidence and the sample size. As the level of confidence is 99%, we need to find the critical value corresponding to a 99% confidence level.

The standard error can be calculated using the formula:

Standard Error = Standard Deviation / √(Sample Size)

In this case, the population standard deviation is given as 15 and the sample size is 200. Therefore:

Standard Error = 15 / √(200)

Once we have the critical value and the standard error, we can calculate the confidence interval as:

Confidence Interval = 102 ± (Critical Value × Standard Error)

To find the critical value, we can consult a Z-table or use a statistical calculator. At a 99% confidence level, the critical value is approximately 2.576.

Now, substituting the values into the confidence interval formula:

Confidence Interval = 102 ± (2.576 × (15 / √(200)))

Simplifying this expression:

Confidence Interval = 102 ± (2.576 × 1.06)

Confidence Interval = 102 ± 2.73256

Finally, rounding the interval to two decimal places:

Confidence Interval ≈ (99.27, 104.73)

Therefore, the point estimate of the mean for this group is 102, and the 99% confidence interval for this group is (99.27, 104.73).

The correct answer is d. 102; (99.27, 104.73).