Question

Question 92.52 pts

This question has 9 parts. Each of the 9 parts (Part A - Part I) has a dropdown list of possible answers. Choose the best answer from the dropdown list for EACH part of the question below.

Investigators want to assess if there is any difference in total cholesterol level between those on a new exercise regimen and those taking a new oral cholesterol medication. They take a sample of 30 participants on a new exercise regimen, and find they have a mean total cholesterol level of 202 mg/dL and sample standard deviation of 6 mg/dL. They compare these results to a sample of 32 participants on the new oral cholesterol medication (with a mean total cholesterol level of 196 mg/dL and sample standard deviation of 10 mg/dL).

Your research question of interest is:

Does the population of people on the new exercise regimen have a different mean total cholesterol level than the population of people on the new medication?

You will use the information above to complete the question parts below for a two independent samples t-test.

Part A: Which of the following represents the appropriate null hypothesis (H0), given this research question of interest?

PART A ANSWER: H0:                            [ Select ]                       ["µ1 - µ2 = 196", "µ1 - µ2 = 202", "µ1 - µ2 = 0", "µ = 0", "x̄ = 0"]      

Part B: Which of the following represents the appropriate alternative hypothesis (H1), given this research question of interest?

PART B ANSWER: H1:                            [ Select ]                       ["µ ≠ 0", "µ1 - µ2 = 0", "µ1 - µ2 ≠ 0", "µ ≠ 202", "x̄ ≠ 196"]      

Part C: Which of the following represents the appropriate critical value(s), testing at an alpha level (α) of 0.01?

PART C ANSWER:                            [ Select ]                       ["-2.431 and 2.431", "-2.660 and 2.660", "-2.000 and 2.000", "3.460", "-1.259"]      

Part D: What is the pooled variance (s2p) associated with your test? (rounded to the nearest hundredth)

PART D ANSWER:                             [ Select ]                       ["69.07", "42.70", "90.82", "9.84", "2.62"]      

Part E: What is the estimated standard error of x̄1 - x̄2 associated with your test? (rounded to the nearest hundredth)

PART E ANSWER:                             [ Select ]                       ["0.53", "2.42", "2.11", "6.19", "14.90"]      

Part F: What is the t-statistic associated with your test? (rounded to the nearest hundredth)

PART F ANSWER:                             [ Select ]                       ["1.24", "2.84", "4.23", "1.39", "2.11"]      

Part G: Given your test results, what is your decision about the null hypothesis?

PART G ANSWER:                             [ Select ]                       ["Reject the null", "Fail to reject (i.e., retain) the null"]      

Part H: The best interpretation of the appropriate decision regarding the null hypothesis would be: "Based on our study, we                           [ Select ]                       ["have", "do NOT have"]         enough evidence to conclude that the population of people on the new exercise regimen appears to have a different mean total cholesterol level than the population of people on the new medication."

Part I: Which of the below represents a 99% confidence intervalfor the difference between the unknown population mean cholesterol levels of people on the new exercise regimen and people on the new medication?

PART I ANSWER:                             [ Select ]                       ["[-1.502 – 4.261]", "[0.387 – 11.613]", "[-4.893 – 12.019]", "[2.819 – 4.512]", "[8.201 – 15.624]"]      

Flag this Question

Question 100.28 pts

Use the data above examining the research question of interest: "Does the population of people on the new exercise regimen have a different mean total cholesterol level than the population of people on the new medication?"

Which of the below represents the best interpretation of the 99% confidence interval you calculated on the previous question for the difference between the unknown population mean cholesterol levels of people on the new exercise regimen and people on the new medication?

*note: the actual values for the confidence interval limits are left blank intentionally

About 99% of people on the new exercise regimen have a cholesterol level between ___ and ___.

We are 99% confident that the true difference between the unknown population mean cholesterol levels of people on the new exercise regimen and people on the new medication falls between ___ and ___.

It is guaranteed that the difference between the unknown population mean cholesterol levels of people on the new exercise regimen and people on the new medication falls between ___ and ___.

We are 99% confident that difference between the sample mean cholesterol levels of people on the new exercise regimen and people on the new medication falls between ___ and ___.

About 99% of people on the new medication have a cholesterol level between ___ and ___.

Answer 1

Solution:

Here, we have to use two sample t test for population mean assuming pooled variance.

We are given

X1bar = 202

X2bar = 196

S1 = 6

S2 = 10

n1 = 30

n2 = 32

α = 0.01

Part A

H₀: µ₁ - µ₂ = 0

Part B

H₁: µ₁ - µ₂ ≠ 0

Part C

We are given α = 0.01

df = n1 + n2 – 2 = 30 + 32 – 2 = 60

Critical value = ±2.6603 (by using t-table)

Answer: "-2.660 and 2.660"

Part D

Pooled variance is given as below:

S_p² = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)

S_p² = [(30 – 1)*6^2 + (32 – 1)*10^2]/(30 + 32 – 2)

S_p² = 69.0667

S_p² = 69.07

Part E

Estimated Standard error is given as below:

SE = sqrt[S_p²*((1/n1)+(1/n2))]

SE = sqrt[69.07*((1/30)+(1/32))]

SE = 2.1120

SE = 2.11

Part F

Test statistic formula is given as below:

t = (X1bar – X2bar) / sqrt[S_p²*((1/n1)+(1/n2))]

t = (202 – 196) / 2.11

t = 6/2.11

t = 2.84

t-statistic = 2.84

Part G

Reject the null (because absolute value of t-statistic is greater than absolute critical value.)

Part H

"Based on our study, we "have" enough evidence to conclude that the population of people on the new exercise regimen appears to have a different mean total cholesterol level than the population of people on the new medication."

Part I

Answer: "[0.387 – 11.613]"

Explanation:

Confidence interval = (X1bar – X2bar) ± t*SE

(X1bar – X2bar) = 6

Confidence level = 99%

df = 60

Critical t value = 2.6603 (by using t-table)

SE = 2.1120

Confidence interval = 6 ± 2.6603 *2.1120

Confidence interval = 6 ± 5.6185

Lower limit = 6 - 5.6185 = 0.387

Upper limit = 6 + 5.6185 = 11.613

We are 99% confident that the true difference between the unknown population mean cholesterol levels of people on the new exercise regimen and people on the new medication falls between 0.387 and 11.613.