In: Math
A group of psychiatrists want to investigate if there is any difference in depression levels (measured on a scale from [0=not at all depressed] to [100=severely depressed]) in patients who have been treated for depression by being told to regularly exercise versus those who were given an oral medication, Contentica, to treat depression.
The researchers selected a sample of 15 patients who were told to regularly exercise and found that their mean depression level was a score of 48 with a standard deviation of 9.
The researchers also selected a sample of 14 patients who were given the medication Contentica and found that their mean depression level was a score of 54 with a standard deviation of 10.
Conduct an appropriate two-tailed hypothesis test (two independent samples t-test) on whether or not patients who regularly exercise have a differentdepression level than those who are prescribed the medication Contentica using an alpha (α) level of 0.01.
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 = 15", "μ1 - μ2 = 0", "μ1 - μ2 = 54", "μ1 - μ2 = 48", "μ1 - μ2 ≠ 48"]
Part B: Which of the following represents the appropriate alternative hypothesis (H1), given this research question of interest?
PART B ANSWER: H1:
[ Select ] ["μ1 - μ2 ≠ 48", "μ1 - μ2 ≠ 14", "μ1 - μ2 ≠ 54", "μ1 ≠ 15", "μ1 - μ2 ≠ 0"]
Part C: What is the degrees of freedom (df)associated with your test?
PART C ANSWER:
[ Select ] ["21", "45", "10", "27", "54"]
Part D: Which of the following represents the appropriate critical value(s), testing at an alpha level (α) of 0.01?
PART D ANSWER:
[ Select ] ["-2.892 and 2.892", "-2.771 and 2.771", "-2.312 and 2.312", "-1.251 and 1.251", "3.85"]
Part E: What is the pooled variance(s2p) associated with your test? (rounded to the nearest hundredth)
PART E ANSWER:
[ Select ] ["45.12", "315.81", "90.15", "76.43", "12.86"]
Part F: What is the estimated standard error of x̄1 - x̄2 associated with your test? (rounded to the nearest hundredth)
PART F ANSWER:
[ Select ] ["3.53", "4.16", "7.89", "2.98", "1.05"]
Part G: What is the t-statistic associated with your test? (rounded to the nearest hundredth)
PART G ANSWER:
[ Select ] ["1.21", "5.47", "3.96", "2.85", "-1.70"]
Part H: Given your test results, what is your decision about the null hypothesis?
PART H ANSWER:
[ Select ] ["Reject the null", "Fail to reject (i.e., retain) the null"]
Part I: 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 patients who regularly exercise do appear to have a different depression level than those who are prescribed the medication Contentica."
Part J: Which of the below represents a 99% confidence interval for the difference between the unknown population mean cholesterol levels of people on the new exercise regimen and people on the new medication?
PART J ANSWER:
[ Select ] ["[-18.28 to -3.16]", "[-15.78 to 3.78]", "[-7.39 to 1.05]", "[-12.81 to 10.45]", "[3.45 to 9.82]"]
PART A:
Correct option:
H0: μ1 - μ2 = 0
PART B:
Correct option:
H1: μ1 - μ2 0
PART C:
Correct option:
"27"
Explanation:
Degrees of Freedom = n1 + n2 - 2 = 15 + 14 - 2 = 27
PART D:
Correct option:
"-2.771 and 2.771"
Explanation:
Degrees of Freedom = 27
= 0.01
Two Tail Test
From Table, critical values of t = 2.771
PART E:
Correct option:
"90.15"
Explanation:
PART F:
Correct option:
"3.53"
Explanation:
PART G:
Correct option:
"-1.70"
Explanation:
t = (48 - 54)/3.53
= - 1.70
PART H:
Correct option:
"Fail to reject (i.e., retain) the null"
Explanation:
Since calculated value of t = - 1.70 is greater than critical value of t = - 2.771, the difference is not significant. Fail to reject null hypothesis.
PART I:
Correct option:
"do NOT have"
PART J:
Correct option:
"[-15.78 to 3.78]"
Explanation:
ndf = 27
= 0.01
From Table, critical values of t = 2.7707
Confidence Interval:
(48 - 54) (2.7707 X 3.53)
= - 6 9.7806
= ( - 15.78 ,3.78)