3. Students Working Hours. Researchers are concerned about the impact of student
ID: 3134860 • Letter: 3
Question
3. Students Working Hours. Researchers are concerned about the impact of students working while they are enrolled in classes, and they’d like to know if students work too much and therefore are spending less time on their classes than they should be. First, the researchers need to find out, on average, how many hours a week students are working. A random sample of 300 students resulted in a sample mean of 6.5 hours worked per week. They know from previous studies that the population standard deviation of this variable is 5 hours.
(a) Calculate a 99% confidence interval, for the unknown population mean, . Round the upper and lower bound values to two decimal places. 2
(b) Free response submission. In the context of this question, provide an interpretation of the 99% confidence interval that you calculated.
(c) The researchers are not satisfied with the precision of the confidence interval. Assuming that the researchers have additional resources, what is the best option do they have to decrease the width of the current confidence interval? [Select the best choice.]
i. Increase the confidence level.
ii. Decrease the confidence level.
iii. Increase the sample size.
iv. Decrease the sample size.
v. None of the above.
(d) One of the researchers is interested in an 86% confidence interval for the unknown popula- tion mean, .
i. Find the positive z-score (z) for the corresponding confidence level, round it to two decimal places.
ii. Find the lower and the upper bound of the confidence interval, round each to two decimal places.
iii. Free response submission. Suppose that the number of hours that students work do not follow a Normal distribution. Is the confidence interval trustworthy? That is, do you think you can really be 86% confident? Explain. (State Yes or No first followed by your explanation).
Explanation / Answer
3.
A)
Note that
Margin of Error E = z(alpha/2) * s / sqrt(n)
Lower Bound = X - z(alpha/2) * s / sqrt(n)
Upper Bound = X + z(alpha/2) * s / sqrt(n)
where
alpha/2 = (1 - confidence level)/2 = 0.005
X = sample mean = 6.5
z(alpha/2) = critical z for the confidence interval = 2.575829304
s = sample standard deviation = 5
n = sample size = 300
Thus,
Margin of Error E = 0.743577871
Lower bound = 5.756422129
Upper bound = 7.243577871
Thus, the confidence interval is
( 5.756422129 , 7.243577871 ) [ANSWER]
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b)
We are 99% confident that the true mean number of hours worked per week is between 5.756422129 and 7.243577871 hours.
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c)
As increasing sample size decreases the margin of error, they can
OPTION III: INCREASE THE SAMPLE SIZE [ANSWER]
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d)
i.
Here,
alpha/2 = (1 - confidence level)/2 = 0.07
Hence,
z(alpha/2) = critical z for the confidence interval = 1.475791028 [ANSWER]
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ii.
Note that
Margin of Error E = z(alpha/2) * s / sqrt(n)
Lower Bound = X - z(alpha/2) * s / sqrt(n)
Upper Bound = X + z(alpha/2) * s / sqrt(n)
where
alpha/2 = (1 - confidence level)/2 = 0.07
X = sample mean = 6.5
z(alpha/2) = critical z for the confidence interval = 1.475791028
s = sample standard deviation = 5
n = sample size = 300
Thus,
Margin of Error E = 0.426024174
Lower bound = 6.073975826
Upper bound = 6.926024174
Thus, the confidence interval is
( 6.073975826 , 6.926024174 ) [ANSWER]
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iii.
Yes, because we have a large enough sample size, n = 300.
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