The Survey of Study Habits and Attitudes (SSHA) is a psychological test that measures students' study habits and attitudes toward school. Scores range from 0 to 200. The mean score for college students is about 115.8, and the standard deviation is about 32.9. A teacher suspects that the mean μ for older students is higher than 115.8. She gives the SSHA to an SRS of 25 students who are at least 30 years old. Suppose we know that scores in the population of older students are Normally distributed with standard deviation σ = 32.9.

We seek evidence against the claim that μ = 115.8. What is the sampling distribution of the mean score of a sample of 25 students if the claim is true?

The distribution is

Binomial


Normal

with mean =________ and standard deviation_________ (±0.0001) =

The sampling distribution of the mean score of a sample of 25 students, if the claim is true, is a normal distribution.

The mean of the sampling distribution is equal to the population mean, which is μ = 115.8.

The standard deviation of the sampling distribution, also known as the standard error, is determined by dividing the population standard deviation by the square root of the sample size. In this case, the population standard deviation is σ = 32.9 and the sample size is 25.

Therefore, the standard deviation of the sampling distribution is:

Standard deviation = σ / √(n)
Standard deviation = 32.9 / √(25)

Calculating this value:

Standard deviation = 32.9 / 5
Standard deviation ≈ 6.58

Hence, the sampling distribution has a mean of 115.8 and a standard deviation of approximately 6.58.

The sampling distribution of the mean score of a sample of 25 students follows a normal distribution.

The mean of this sampling distribution is equal to the population mean, which is μ = 115.8.

The standard deviation of the sampling distribution, denoted as the standard error (SE), can be calculated using the following formula:

SE = σ / √n

where σ is the population standard deviation (32.9 in this case), and n is the sample size (25 in this case).

Plugging in the values, we have:

SE = 32.9 / √25

SE = 32.9 / 5

SE ≈ 6.58 (±0.0001)

Therefore, the mean of the sampling distribution is 115.8 and the standard deviation is approximately 6.58.