Suppose the mass of college students can be modeled by the normal distribution. We believe the mean
of the normal distribution is 72 kg. We randomly sample 18 students and find an average of 69 kg and a
standard deviation of 4kg.
A. The standard error of óx is 0.9428
B. A t-statistic can be calculated and it is equal to −3.18
C. The degrees of freedom for the t-distribution will be 17
The correct statement(s) is(are)
(1) Only A
(2) All three statements are correct.
(3) B and C
(4) A and B
(5) A and C
To verify the correct statement(s), let's go through each option one by one:
A. The standard error of óx is 0.9428.
To calculate the standard error (SE) of a sample mean, you divide the standard deviation of the population by the square root of the sample size. In this case, the standard deviation (ó) given is 4 kg and the sample size is 18. Using the formula, we can calculate the standard error as follows:
SE = ó / √n
SE = 4 / √18
SE ≈ 4 / 4.2426
SE ≈ 0.9428
So, option A is correct.
B. A t-statistic can be calculated and it is equal to −3.18.
To calculate the t-statistic, you need to find the difference between the sample mean and the population mean, and then divide it by the standard error. The formula for calculating the t-statistic is:
t = (x̄ - μ) / SE
Given that the sample mean (x̄) is 69 kg and the population mean (μ) is 72 kg, and the standard error (SE) is 0.9428 (as calculated in option A), we can calculate the t-statistic as follows:
t = (69 - 72) / 0.9428
t ≈ -3.18
So, option B is correct.
C. The degrees of freedom for the t-distribution will be 17.
The degrees of freedom for a t-distribution in this case is equal to the sample size minus one (n - 1). Since the sample size is 18, the degrees of freedom would be 18 - 1 = 17.
So, option C is correct.
Based on the analysis, the correct statement is (2) All three statements are correct.
The correct statement is (5) A and C.
A. The standard error of σx (standard error of the mean) can be calculated using the formula:
Standard Error (σx) = Standard Deviation (σ) / √(Sample Size)
= 4 kg / √(18) ≈ 0.9428 kg
So statement A is correct.
C. The degrees of freedom for the t-distribution when calculating a t-statistic using a sample mean can be calculated by subtracting 1 from the sample size. In this case, the sample size is 18, so the degrees of freedom will be 18 - 1 = 17.
So statement C is also correct.