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**
**(H****0****)**, 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 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 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

H_{0}: µ_{1} - µ_{2} = 0

Part B

H_{1}: µ_{1} - µ_{2} ≠ 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_{p}^{2} = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1
+ n2 – 2)

S_{p}^{2} = [(30 – 1)*6^2 + (32 – 1)*10^2]/(30 +
32 – 2)

S_{p}^{2} = 69.0667

S_{p}^{2} = 69.07

Part E

Estimated Standard error is given as below:

SE = sqrt[S_{p}^{2}*((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_{p}^{2}*((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.

This question has 10
parts. Each of the 10 parts (Part A - Part J) has
a dropdown list of possible answers. Choose the best answer from
the dropdown list for EACH part of the question
below.
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...

A new drug to lower total cholesterol is being tested. A study
is designed to evaluate the efficacy of the drug. Sixteen patients
agree to participate in the study and each is asked to take the new
drug for 4 weeks. Before starting the treatment, each patient's
total cholesterol level is measured. After taking the drug for 4
weeks, each patient's total cholesterol level is measured again.
The 99% confidence interval for the change in total cholesterol
(After less Before)...

A serum cholesterol level above 250 mg/dl of blood is a risk
factor for cardiovascular disease in humans. At a
medical center in St. Louis, a study to test the effectiveness of
new cholesterol-lowering drug was conducted. One hundred
people with cholesterol levels between 250 mg/dl and 300 mg/dl were
available for this study. Fifty people were assigned at
random to each of two treatment groups. One group
received the standard cholesterol-lowering medication and the other
group received the new drug. After taking the drug...

Can someone please explain the last three parts of this
question parts 7-9.
A cholesterol level over 200 is considered high. The
manufacturers of a certain brand of cereal are interested in
whether eating their cereal for breakfast daily will result in a
decrease in cholesterol levels for people with high cholesterol.
They randomly select ten people with high cholesterol to
participate in the study. Each subject eats the cereal every
morning for a month. The cholesterol levels of each...

This question has 8 parts. Each of the 8 parts (Part A - Part H)
has a dropdown list of possible answers. Choose the best answer
from the dropdown list for EACH part of the question below. Open
the "Lab Dataset" (HSCI390.sav) you have been using for lab
assignments in SPSS. Your analysis will focus on the variable
"Weight in pounds" (NQ50).
Researchers are interested in examining how the average weight
of students at CSUN compares to the average weight...

Question 12 (1 point)
The owner of a golf course wants to determine if his golf course
is more difficult than the one his friend owns. He has 9 golfers
play a round of 18 holes on his golf course and records their
scores. Later that week, he has the same 9 golfers play a round of
golf on his friend's course and records their scores again. The
average difference in the scores (treated as the scores on his
course...

A group of eight individuals with high cholesterol levels were
given a new drug that was designed to lower cholesterol levels.
Cholesterol levels, in mg/dL, were measured before and after
treatment for each individual, with the following results:
Subject
Before
After
1
283
215
2
299
206
3
274
179
4
284
212
5
240
178
6
275
212
7
293
192
8
277
196
Can you conclude that the reduction in mean cholesterol level
after treatment is greater than...

home / study / math / statistics and probability / statistics
and probability questions and answers / To Test The Efficacy Of A
New Cholesterol-lowering Medication, 10 People Are Selected At
Random. ...
Question: To test the efficacy of a new
cholesterol-lowering medication, 10 people are selected at random.
...
To test the efficacy
of a new cholesterol-lowering medication, 10 people are selected at
random. Each has their LDL levels measured (shown below as Before),
then take the medicine for...

A food manufacturer claims that eating its new cereal as part of
a daily diet lowers total blood cholesterol levels. The table shows
the total blood cholesterol levels (in milligrams per deciliter of
blood) of seven patients before eating the cereal and after one
year of eating the cereal as part of their diets. Use technology to
test the mean difference. Assume the samples are random and
dependent, and the population is normally distributed. At αs=0.05 ,
can you conclude...

A group of eight individuals with high cholesterol levels were
given a new drug that was designed to lower cholesterol levels.
Cholesterol levels, in mg/dL, were measured before and after
treatment for each individual, with the following results:
Subject
Before
After
1
283
215
2
299
206
3
274
179
4
284
212
5
240
178
6
275
212
7
293
192
8
277
196
Can you conclude that the mean cholesterol level after treatment
is less than the mean...

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 2 minutes ago

asked 48 minutes ago

asked 50 minutes ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 2 hours ago

asked 2 hours ago