University of the People MATH 1281 - Graded Quiz Unit 3

A case study that deals with the effect of prior reputation on the evaluation of an instructor is
presented in Question 11.1 in the book (Section 11.5). Students were exposed to either one of
two conditions, C and P, and were asked to rate a lecture given by the instructor. It was found
that the average rating among the 25 students that were exposed to condition C was 2.613 and
the average rating among the 24 students that were exposed to condition P was 2.236. The
sample standard deviations were 0.533 and 0.543, respectively.

Question 1
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You want to construct a confidence interval for the expected rating of students that were




co
exposed to condition P. The confidence level of the interval should be 85%. The quantile of the




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t-distribution that should be used for the construction is: (Assume a Normal distribution for the
rating)




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Select one:
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a. qt(0.85,24)

b. qt(0.925,24)
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c. qt(0.85,23)
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d. qt(0.925,23)

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ed d




The central part should include 85% of the distribution. The tails should include 15%. Half of this
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probability, or 7.5%, goes to the upper tail and half to the lower tail. The probability above the quantile is
0.075. Therefore, the probability below the quantile should be 0.925.
The number of degrees of freedom is the number of observations minus 1. The number of observations is
24. Hence, the number of degrees of freedom is 23.
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Combining the two arguments we get that quantile should be qt(0.925,23) .
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The correct answer is: qt(0.925,23)

Question 2
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Question text
The confidence interval with confidence level of 85% for the expected rating of students that
were exposed to condition P is: (Assume a Normal distribution for the rating. Choose the
interval closest to the answer)
Select one:

a. [2.07, 2.40]


b. [ 2.12, 2.35]




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c. [2.50, 2.73]




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d. [2.59, 2.64]




o.
Feedback
rs e
The sample average is 2.236 and the sample standard deviation is 0.543. The estimated standard
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deviation of the sample average is 0.543/√24, where 24 is the sample size. We apply the formula for
computing a confidence interval for the expectation of a Normal measurement: 2.236 +
qt(c(0.075,0.925),23)*0.543/sqrt(24) . The output of the computation is: [2.070929, 2.401071],
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which after rounding up gives [2.07,2.40].
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The correct answer is: [2.07, 2.40]
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Question 3
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ar stu
is




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Th




Question text
You want to construct a confidence interval for the variance of the rating of students that were
exposed to condition P. The confidence level of the interval should be 80%. A quantile of the chi-
square distribution that should be used for the construction is: (Assume a Normal distribution for
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the rating)
Select one:

a. qchisq(0.85,23)




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