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 90% confidence interval for this group is ____________________.

a. 100; (100.26; 103.74)
b. 102; (100.26; 103.74)
c. 100; (98.26; 101.74)

90% confidence interval leaves .05 on each side. Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability (.05) and its Z score.

90% = mean ± 1.645 SEm

SEm = SD/√n

To find the point estimate of the mean for this group, we can simply take the sample mean, which is given as 102. So the point estimate of the mean for this group is 102.

To find the 90% confidence interval for this group, we can use the formula:
Confidence interval = (sample mean) ± (z-score) * (standard deviation/square root of sample size)

The z-score for a 90% confidence level is 1.645. We are given the population standard deviation, which is 15, and the sample size, which is 200.

Plugging these values into the formula, we have:
Confidence interval = 102 ± (1.645) * (15/√200)

Calculating this expression, we get:
Confidence interval = 102 ± 1.645 * (15/√200) ≈ (100.26, 103.74)

So the 90% confidence interval for this group is approximately (100.26, 103.74).

Based on the calculations, the correct answer is a. 100; (100.26; 103.74).