HBX Business Analytics Exam
Preparation
Use the descriptive statistics tool to calculate the summary statistics for the data set
provided below.
Use C1 as the output range so that your response is graded correctly and include a
label for the data. - Answer The Input Range is A1:A13. You must check the Labels in
first row box since we included a label in cell A1 to ensure that the output table is
appropriately labeled, and you must select Summary Statistics to produce the output
table.
You need to use C1 as your output range so that your calculations are in the blue cells
for grading.
Below are data showing students' grades on a statistics quiz and the number of hours
they spent studying. Create a scatterplot to show the relationship. - Answer The Input Y
Range is Grade B1:B25 and the Input X Range is Hours Studied A1:A25. You must
check the Labels in first row box since we included labels in cells A1 and B1 to ensure
that the scatter plot's axes are appropriately labeled.
For the following scenario, determine whether it would be better to analyze cross-
sectional or time series data.
We want to see if the Red Sox performance changes over the course of the baseball
season. - Answer Time Series
Since we are interested in comparing the Red Sox performance at different points in
time during the baseball season, we should analyze time series data.
For the following scenario, determine whether it would be better to analyze cross-
sectional or time series data.
We want to compare the daily sales of stores in a mall during a day-long mall-wide
event. - Answer Cross-Sectional
Since we are interested in the sales of different stores on a single day (a single point in
time), we should analyze a cross-section of the stores in the mall.
If the average IQ is 100 and the standard deviation is 15, approximately what
percentage of people have IQs above 130? - Answer 130 is two standard deviations
above the mean (130-100=30=2*15=2*stdev). We know that approximately 95% of the
distribution is within 2 standard deviations of the mean. Therefore 5% must fall beyond
2 standard deviations, 2.5% at the top and 2.5% at the bottom.
If a particular standardized test has a mean score of 500 and standard deviation of 100,
what percentage of test-takers score between 500 and 600? - Answer 100 is one
standard deviation above the mean (600-500 =100= 1*100 = 1*stdev). We know that
approximately 68% of the distribution is within 1 standard deviation of the mean.
Therefore 34% must fall beyond 1 standard deviation above the mean.
, Recall that the z-value associated with a value measures the number of standard
deviations the value is from the mean.
If a particular standardized test has an average score of 500 and a standard deviation of
100, what z-value corresponds to a score of 350? - Answer z=(x-µ)/σ. Here x= 350,
µ=500, the population mean, and σ=100, the population standard deviation. Thus z =
(350-500)/100 = (-150)/100 = -1.5
For a normal distribution with mean 100 and standard deviation 10, find the probability
of obtaining a value less than or equal to 118. - Answer The cumulative probability
associated with the value 118 is NORM.DIST(118,B1,B2,TRUE)=0.96, or 96%.
Approximately 96% of the population has values less than or equal to 118.
Note that because the normal distribution is continuous, the probability of an outcome
being equal to single, discrete value (such as 118) is 0. Thus the probability of obtaining
a value less than 118 is equivalent to obtaining a value less than or equal to 118. You
must link directly to cells to obtain the correct answer.
For a normal distribution with mean 100 and standard deviation 10, find the probability
of obtaining a value less than or equal to 118. - Answer The cumulative probability
associated with the value 118 is NORM.DIST(118,Mean,Standard
Deviation,TRUE)=0.96, or 96%. Approximately 96% of the population has values less
than or equal to 118.
Note that because the normal distribution is continuous, the probability of an outcome
being equal to single, discrete value (such as 118) is 0. Thus the probability of obtaining
a value less than 118 is equivalent to obtaining a value less than or equal to 118. You
must link directly to cells to obtain the correct answer.
Calculate the 95% confidence interval for the true population mean based on a sample
with mean x¯=15, standard deviation s=2, and sample size n=100. - Answer The margin
of error is based on the significance level (1-confidence level, or 1-0.95=0.05), the
standard deviation (in B2) and the sample size (in B3). We can compute the margin of
error using the Excel function CONFIDENCE.NORM(0.05,B2,B3).
The lower bound of the 95% confidence interval is the sample mean minus the margin
of error, that is B1-CONFIDENCE.NORM(0.05,B2,B3)=15-0.39=14.61.
The upper bound of the 95% confidence interval is the sample mean plus the margin of
error, that is B1+CONFIDENCE.NORM(0.05,B2,B3)=15+0.39=15.39.
Calculate the 99% confidence interval for the true population mean for the BMI data.
Recall that the new sample contains 15 people and has a mean of 25.97 kg/m2 and a
standard deviation of 7.10 kg/m2. - Answer Because our sample has fewer than 30
cases, we cannot assume that the distribution of sample means will be normal, and
must use the t-distribution. The margin of error is based on the significance level (1-
confidence level, or 1-0.99=0.01), the standard deviation (in B2) and the sample size (in
Preparation
Use the descriptive statistics tool to calculate the summary statistics for the data set
provided below.
Use C1 as the output range so that your response is graded correctly and include a
label for the data. - Answer The Input Range is A1:A13. You must check the Labels in
first row box since we included a label in cell A1 to ensure that the output table is
appropriately labeled, and you must select Summary Statistics to produce the output
table.
You need to use C1 as your output range so that your calculations are in the blue cells
for grading.
Below are data showing students' grades on a statistics quiz and the number of hours
they spent studying. Create a scatterplot to show the relationship. - Answer The Input Y
Range is Grade B1:B25 and the Input X Range is Hours Studied A1:A25. You must
check the Labels in first row box since we included labels in cells A1 and B1 to ensure
that the scatter plot's axes are appropriately labeled.
For the following scenario, determine whether it would be better to analyze cross-
sectional or time series data.
We want to see if the Red Sox performance changes over the course of the baseball
season. - Answer Time Series
Since we are interested in comparing the Red Sox performance at different points in
time during the baseball season, we should analyze time series data.
For the following scenario, determine whether it would be better to analyze cross-
sectional or time series data.
We want to compare the daily sales of stores in a mall during a day-long mall-wide
event. - Answer Cross-Sectional
Since we are interested in the sales of different stores on a single day (a single point in
time), we should analyze a cross-section of the stores in the mall.
If the average IQ is 100 and the standard deviation is 15, approximately what
percentage of people have IQs above 130? - Answer 130 is two standard deviations
above the mean (130-100=30=2*15=2*stdev). We know that approximately 95% of the
distribution is within 2 standard deviations of the mean. Therefore 5% must fall beyond
2 standard deviations, 2.5% at the top and 2.5% at the bottom.
If a particular standardized test has a mean score of 500 and standard deviation of 100,
what percentage of test-takers score between 500 and 600? - Answer 100 is one
standard deviation above the mean (600-500 =100= 1*100 = 1*stdev). We know that
approximately 68% of the distribution is within 1 standard deviation of the mean.
Therefore 34% must fall beyond 1 standard deviation above the mean.
, Recall that the z-value associated with a value measures the number of standard
deviations the value is from the mean.
If a particular standardized test has an average score of 500 and a standard deviation of
100, what z-value corresponds to a score of 350? - Answer z=(x-µ)/σ. Here x= 350,
µ=500, the population mean, and σ=100, the population standard deviation. Thus z =
(350-500)/100 = (-150)/100 = -1.5
For a normal distribution with mean 100 and standard deviation 10, find the probability
of obtaining a value less than or equal to 118. - Answer The cumulative probability
associated with the value 118 is NORM.DIST(118,B1,B2,TRUE)=0.96, or 96%.
Approximately 96% of the population has values less than or equal to 118.
Note that because the normal distribution is continuous, the probability of an outcome
being equal to single, discrete value (such as 118) is 0. Thus the probability of obtaining
a value less than 118 is equivalent to obtaining a value less than or equal to 118. You
must link directly to cells to obtain the correct answer.
For a normal distribution with mean 100 and standard deviation 10, find the probability
of obtaining a value less than or equal to 118. - Answer The cumulative probability
associated with the value 118 is NORM.DIST(118,Mean,Standard
Deviation,TRUE)=0.96, or 96%. Approximately 96% of the population has values less
than or equal to 118.
Note that because the normal distribution is continuous, the probability of an outcome
being equal to single, discrete value (such as 118) is 0. Thus the probability of obtaining
a value less than 118 is equivalent to obtaining a value less than or equal to 118. You
must link directly to cells to obtain the correct answer.
Calculate the 95% confidence interval for the true population mean based on a sample
with mean x¯=15, standard deviation s=2, and sample size n=100. - Answer The margin
of error is based on the significance level (1-confidence level, or 1-0.95=0.05), the
standard deviation (in B2) and the sample size (in B3). We can compute the margin of
error using the Excel function CONFIDENCE.NORM(0.05,B2,B3).
The lower bound of the 95% confidence interval is the sample mean minus the margin
of error, that is B1-CONFIDENCE.NORM(0.05,B2,B3)=15-0.39=14.61.
The upper bound of the 95% confidence interval is the sample mean plus the margin of
error, that is B1+CONFIDENCE.NORM(0.05,B2,B3)=15+0.39=15.39.
Calculate the 99% confidence interval for the true population mean for the BMI data.
Recall that the new sample contains 15 people and has a mean of 25.97 kg/m2 and a
standard deviation of 7.10 kg/m2. - Answer Because our sample has fewer than 30
cases, we cannot assume that the distribution of sample means will be normal, and
must use the t-distribution. The margin of error is based on the significance level (1-
confidence level, or 1-0.99=0.01), the standard deviation (in B2) and the sample size (in
