Help me understand this problem, I have several more like this & want to get an understanding on to solve it:

The mean score,x, on an aptitude test for a random sample of 9 students was 64. Assuming that standard deviation=16, construct a 95.44% confidence interval for the mean score of all students taking the test.

A. 53.3 to 74.7 B. 60.4 to 67.6 C. 32 to 96 D. 56.0 to 72.0

The standard error of the mean (SEM) is the standard deviation divided by the square root of the sample size - which in this instance is 9 - so the SEM here is 16/sqrt(9)=5.33.

The question actually contains a little clue that you're probably on the right track in that very specific figure of 95.44% for a confidence interval - because if you look up that figure in a set of Normal probability tables, you should find that it corresponds to a very convenient Z value.

Your confidence interval will then range from (M - 5.33*Z) to (M + 5.33*Z).

To construct a confidence interval, we need to use the formula:

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

In this case, we are given the mean score (x) of the sample, which is 64. The standard deviation (σ) is given as 16, and the sample size (n) is 9.

The critical value (Z) for a 95.44% confidence interval can be obtained by finding the Z-score corresponding to the desired confidence level. In this case, the confidence level is 95.44%, which corresponds to a 2-tailed Z-score of 1.96.

Plugging the values into the formula, we get:

Confidence Interval = 64 ± (1.96 * (16 / √9))

Simplifying the expression inside the brackets:

Confidence Interval = 64 ± (1.96 * 5.333)

Confidence Interval = 64 ± 10.465

Now, to determine the lower and upper bounds of the confidence interval, we subtract and add this value, respectively, to the mean:

Lower Bound = 64 - 10.465 = 53.535
Upper Bound = 64 + 10.465 = 74.465

So, the 95.44% confidence interval for the mean score of all students taking the test is approximately 53.535 to 74.465.

Therefore, the correct answer is A. 53.3 to 74.7.