BUSINESS ANALYTICS FINAL EXAM

BUSINESS ANALYTICS FINAL EXAM 2024/2025

PREP 200 REAL EXAM QUESTIONS AND

ANSWERS (100% CORRECT ANSWERS) BA

LATEST EXAM PREP (NEW!!)

If the expected production volume when there are 120 workers is approximately 131,958 units, which of the following equations would provide a reasonable estimate of the 68% prediction interval for the output of those 120 workers? - ANSWER-A reasonable estimate of the prediction interval is the point forecast (131,958) plus or minus the z-value times the standard error of the regression (14,994.93). As usual, the z-value is based on the desired level of confidence.Since we want a 68% prediction interval, the z-value is equal to one. Therefore 131,958±14,994.93 is the best option.The owner of a local health food store recently started a new ad campaign to attract more business and wants to test whether average daily sales have increased.Historically average daily sales were approximately $2,700. After the ad campaign, the owner took a random sample of forty-five days and found that daily average sales had increased to $2,984.What is store owner's null hypothesis? - ANSWER-The null hypothesis is the opposite of the hypothesis you are trying to substantiate. Since the owner wants to test for an increase, the null hypothesis is μ≤$2,700. Remember that the null hypothesis is always based on historical information.

The owner of a local health food store recently started a new ad campaign to attract more business and wants to test whether average daily sales have increased.Historically average daily sales were approximately $2,700. After the ad campaign, the owner took another random sample of forty-five days and found that average daily sales were $2,984 with a standard deviation of approximately $585. Calculate the upper bound of the 95% range of likely sample means for this one-sided hypothesis test using the CONFIDENCE.NORM function. - ANSWER-To construct a 95% range of likely sample means, calculate the margin of error using the function CONFIDENCE.NORM(alpha, standard_dev, size). However, 1 / 4

CONFIDENCE.NORM finds the margin of error for a two-sided hypothesis test and this question asks for the upper bound of a one-sided test. To find the upper bound for the one-sided test you must first determine what two-sided test would have a 5% rejection region on the right side. Since the distribution of sample means is symmetric, a two-sided test with a 10% significance level would have a 5% rejection region on the left side of the normal distribution and a 5% rejection region on the right side. Thus, the upper bound for a two-sided test with alpha=0.1 will be the same as the upper bound on a one-sided test with alpha=0.05. The margin of error is CONFIDENCE.NORM(alpha, standard_dev, size)=

CONFIDENCE.NORM(0.1,C3,C4)=CONFIDENCE.NORM(0.1,585,45)=$143.44

. The upper bound of the 95% range of likely sample means for this one-sided hypothesis test is the population mean plus the margin of error, which is approximately $2,700+$143.44=$2,843.44.

A college football coach has decided to recruit only the heaviest 15% of high school football players. He knows that high school players' weights are normally distributed and that this year, the mean weight is 225 pounds with a standard deviation of 43 pounds. Calculate the weight at which the coach should start recruiting players. - ANSWER-=NORM.INV(.85,B1,B2)

If an independent variable has a p-value of 0.0050, which of the following could represent the Lower 95% and the Upper 95% for that variable? - ANSWER-The p- value, 0.0050, is less than 0.05 so the independent variable is significant at the 5% significance level. Therefore, the 95% confidence interval for the coefficient of the independent variable does not include zero. The interval between -235.62 and - 5.64 does not contain zero.

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 2 / 4

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.) - ANSWER-The formula for calculating the minimum required sample size is n≥(z⁢sM)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 1.96⁢449,30050,0002=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 find the smallest integer that satisfies the equation. Therefore, the minimum required sample size is 311.

A store owner is interested in opening a second shop. She wants to estimate the true average daily revenue of her current shop to decide whether expanding her business is a good idea. The store owner takes a random sample of 60 days over a six-month period and finds that the mean revenue of those days is 3,472.00 dollars with variance 315,900.20 square dollars. Calculate a 95% confidence interval to estimate the true average daily revenue. - ANSWER-First calculate the sample standard deviation, which is equal to the square root of the variance. The sample standard deviation is $562.05. Then find the margin of error using the Excel function CONFIDENCE.NORM(alpha, standard_dev, size). Here,

CONFIDENCE.NORM(0.05,SQRT(B2),60)=CONFIDENCE.NORM(0.05,562.05,

60)=$142.22. The lower bound of the 95% confidence interval is the mean minus the margin of error, $3,472.00-$142.22=$3,329.78. The upper bound of the 95% confidence interval is the mean plus the margin of error, $3,472.00+$142.22=$3,614.22. You must link directly to values in order to obtain the correct answer.

Type II; the engineer deems the bridge safe and moves onto construction even though it is not actually safe - ANSWER-The type II error is that the engineer deems the bridge safe and moves onto construction even though it is not actually safe. This would be worse than presuming that a safe bridge is unsafe.

• / 4

IQ scores are known to be normally distributed. The mean IQ score is 100 and the standard deviation is 15. What percent of the population has an IQ over 115? - ANSWER-1-Norm.Dist(value,mean,sd,true)

=1-Norm.Dist(115,b1,b2,true)

IQ scores are known to be normally distributed. The mean IQ score is 100 and the standard deviation is 15. What percent of the population has an IQ between 85 and 105? - ANSWER-NORM.DIST(105,mean,sd,TRUE)- NORM.DIST(85,mean,sd,TRUE)

NORM.DIST(105,b1,b2,TRUE)-NORM.DIST(85,b1,b2,TRUE)

IQ scores are known to be normally distributed. The mean IQ score is 100 and the standard deviation is 15. The top 25% of the population (ranked by IQ score) have IQ's above what value? - ANSWER-Use the properties of the normal distribution to solve this problem. Since you are only interested in the top 25%, calculate the IQ at which 75% of people are below. The Excel function NORM.INV(probability, mean, standard_dev) returns the inverse of a normal cumulative distribution function.

Here, NORM.INV(0.75,B1,B2)=NORM.INV(0.75,100,15)=110 indicates that 75% of people have IQ's lower than 110. Hence, 25% of people have IQ's greater than 110.

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

• / 4