124,247 unique listeners per day. The manager finds that the p-value for the hypothesis
test is approximately 0.0743. How would you interpret the p-value? - ️️If the
average number of unique daily listeners per day is still 131,520, the likelihood of
obtaining a sample with a mean at least a...
HBX-Business Analytics
124,247 unique listeners per day. The manager finds that the p-value for the hypothesis
test is approximately 0.0743. How would you interpret the p-value? - ✔️✔️If the
average number of unique daily listeners per day is still 131,520, the likelihood of
obtaining a sample with a mean at least as extreme as 124,247 is 7.43%.
is correct
The null hypothesis is that the average number of unique daily listeners per day has not
changed, that is, it is still 131,520. Therefore, the p-value of 0.0743 indicates that if the
average number of unique daily listeners is still 131,520, the likelihood of obtaining a
sample with a mean at least as extreme as 124,247 is 7.43%%.
A professor presents a summary of grades to her students. A student asks why the
mean of the data set is so much smaller than the median. Which of the following is most
likely? - ✔️✔️The distribution of the data is skewed to the left
is correct
When the distribution of data is skewed to the left, the mean is most likely less than the
median. The extreme values in the left tail pull the mean towards them.
Consider the four outliers in the 2012 revenue data: companies with revenue of $237
billion, $246 billion, $447 billion, and $453 billion. If we removed these companies from
the data set, what would happen to the standard deviation - ✔️✔️The standard
deviation would decrease.
is correct
The standard deviation gives more weight to observations that are further from the
mean. Therefore, removing the outliers would decrease the standard deviation.
A journalist wants to determine the average annual salary of CEOs in the S&P 1,500.
He does not have time to survey all 1,500 CEOs but wants to be 95% confident that his
estimate is within $50,000 of the true mean. The journalist takes a preliminary sample
and estimates that the standard deviation is approximately $449,300. What is the
minimum number of CEOs that the journalist must survey to be within $50,000 of the
true average annual salary? Remember that the z-value associated with a 95%
confidence interval is 1.96.
Please enter your answer as an integer; that is, as a whole number with no decimal
point. - ✔️✔️The formula for calculating the minimum required sample size is
n≥(z*s/M)^2, where M=50.000 is the desired margin of error for the confidence interval,
s=$449,300 is the sample standard deviation, and z=1.96. Using these data we find that
1.96449,30050,0002=310.201.96449,30050,0002=310.20 Since n must be an integer
(let's not even think of what 0.20 CEOs would look like!) and n must be greater than or
equal to 310.20, we must round up to 311. Since 310.20 is closer to 310 than to 311, we
would normally round 310.20 down to 310. However, in this case we must round up to
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