Exam (elaborations)
HBS Business CORE - HBX CORe Practice_Quiz_3.
- Course
- HBS Business CORE - HBX CORe
- Institution
- Harvard University
HBS Business CORE - HBX CORe
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Practice Quiz 31) A business school professor is interested to know if watching a video about the
Central Limit Theorem helps students understand it. To assess this, the
professor tests students’ knowledge both immediately before they watch the
video and immediately after. The professor takes a sample of students, and for
each one compares their test score after the video to their score before the
video. Using the data below, calculate the p-value for the following hypothesis
test:
H0:μafter≤μbeforeH0:μafter≤μbefore
Ha:μafter>μbefore
The p-value of the one-sided hypothesis test is T.TEST(array1, array2, tails,
type)=T.TEST(B2:B31,C2:C31,1,1), which is approximately 0.0128. You must designate
this test as a one-sided test (that is, assign the value 1 to the tails argument) and as a
type 1 (a paired test) because you are testing the same students on the same
knowledge at two points in time. You must link directly to values in order to obtain the
correct answer.
, 2) A curious student in a large economics course is interested in calculating the
percentage of his classmates who scored lower than he did on the GMAT; he
scored 490. He knows that GMAT scores are normally distributed and that the
average score is approximately 540. He also knows that 95% of his classmates
scored between 400 and 680. Based on this information, calculate the
percentage of his classmates who scored lower than he did.
Since GMAT scores are normally distributed, we know that P(μ–1.96σ ≤ x ≤ μ+1.96σ) =
95%. Thus, to find the standard deviation, subtract the lower bound from the mean and
divide by 1.96. The standard deviation of the distribution is (B1-B2)/1.96 = (540-
400)/1.96 = 71.4. (Note that because the normal curve is symmetrical, we could
calculate the same value using (B3-B1)/1.96 = (680-540)/1.96 = 71.4). To find the
cumulative probability, P(x ≤ 490), use the Excel function NORM.DIST(x, mean,
standard_dev, TRUE). Here, NORM.DIST(B4,B1,71.4,TRUE) =
NORM.DIST(490,540,71.4,TRUE) = 0.24, or 24%. Approximately 24% of his classmates
scored lower than he did. You must link directly to the values in order to obtain the
correct answer
, 3) An airport shuttle company forecasts the number of hours its drivers will work
based on the distance to be driven (in miles) and the number of jobs (each job
requires the pickup and drop-off of one set of passengers) using the following
regression equation:
Travel time=-0.60+0.05(distance)+0.75(number of jobs)
On a given day, Victor and Sofia drive approximately the same distance but Sofia has
two more jobs than Victor. If Victor worked for 4 hours, for how long can the company
expect Sofia to work?
Please enter your answer rounded to one digit to the right of the decimal point. For
example, if you think Sofia would work 236.7134 hours, enter 236.7.
The only difference between the workloads of the two drivers is the number of jobs each
has; Sofia has two additional jobs. Therefore the company can expect Sofia to work the
four hours Victor worked, plus an additional 0.75 hours for each of the two additional
jobs, that is, 4+0.75(2)=5.5 hours.
4) If the mean of a normally distributed population is -10 with a standard deviation of
2, what is the likelihood of obtaining a value less than or equal to -7?
To calculate the likelihood of obtaining a value less than or equal to -7, P(x≤-7), use the
Excel function NORM.DIST(x, mean, standard_dev, TRUE). Here, NORM.DIST(-
7,B1,B2,TRUE)=NORM.DIST(-7,-10,2,TRUE)=0.93, or 93%. Approximately 93% of the
population falls in the area under the curve less than or equal to -7.
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