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)

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Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability (.005 for two-tailed) and its Z score.

Mean for this group = 102

99% = mean ± Z(SEm)

SEm = SD/√n

You need to do the calculations.

To find the point estimate of the mean for this group, you simply take the mean of the sample, which is 102. So the point estimate is 102.

To find the confidence interval, we need to use the formula:

Confidence Interval = Mean ± (Z * (Standard Deviation / √n))

where:
Mean = Sample mean (102 in this case)
Z = Z-score corresponding to the desired confidence level (99% in this case)
Standard Deviation = Population standard deviation (15 in this case)
n = Sample size (200 in this case)

To find the Z-score corresponding to a 99% confidence level, you can use a standard normal distribution table or a statistical calculator. The Z-score for a 99% confidence level is approximately 2.576.

Plugging these values into the formula, we have:

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

Simplifying the calculation, we get:

Confidence Interval = 102 ± (2.576 * 1.06)

Calculating the upper and lower bounds of the confidence interval, we find:

Upper Bound = 102 + (2.576 * 1.06) ≈ 104.73
Lower Bound = 102 - (2.576 * 1.06) ≈ 99.27

So the 99% confidence interval for this group is approximately (99.27, 104.73).

Therefore, the correct answer is option d. 102; (99.27, 104.73).