# Business Forecasting Study Set 1

## Quiz 1 :A Review of Basic Statistical Concepts

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Showing 1 - 20 of 22
Sandy James thinks that housing prices have stabilized in the past few months. To convince her boss, she intends to compare current prices with last year's prices. She collects 12 housing prices from the want ads: She then calculates the mean and standard deviation of the prices she has found. What are these two summary values?
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Essay

a.
Obtain the mean and standard deviation using MINITAB procedure.
MINITAB procedure:
Step 1: Choose Stat Basic Statistics Display Descriptive Statistics.
Step 2: In Variables enter the column of Prices.
Step 3: In Statistics , select Mean , StDev and N Number.
Step 4: Click OK.
MINITAB output:
From the MINITAB output, the mean price
is
and the standard deviation
of the prices is
.

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A large construction company is trying to establish a useful way to view typical profits from jobs obtained from competitive bidding. Because the jobs vary substantially in size and the final amount of the successful bid, the company has decided to express profits as percent earnings: When money is lost on a project, the earnings are negative and so is the resulting net profit. A sample of 30 jobs yields these percent earnings: a. Calculate an estimate of the mean percent earnings for the population of jobs or for all potential jobs. b. Construct a 95% confidence interval for the mean percent earnings for the population of jobs using a large-sample argument. c. Construct a 95% confidence interval for the mean percent earnings for the population of jobs assuming 30 is a small sample size. What additional assumption do you need to make in this case? d. Compare the two intervals in parts b and c. Explain why a sample size of 30 is often taken as the cutoff between large and small samples.
Free
Essay

a.
Obtain the mean percent earnings for the population of jobs by using MINITAB.
MINITAB procedure:
Step 1: Choose Stat Basic Statistics Display Descriptive Statistics.
Step 2: In Variables , enter Earnings.
Step 3: In Statistics , select Mean and StDev.
Step 4: Click OK.
MINITAB output:
From the above MINITAB output, the point estimate mean percent earning
for the population of jobs is
.
b.
Obtain a 95% confidence interval for the mean percent earnings for the population of jobs using a large-sample argument.
The level of significance is 0.95.
From the table B-2, the table value of
is given below:
The general formula for the confidence interval estimate for population mean using large- sample argument is
Obtain standard deviation by using MINITAB.
From the above MINITAB output in part (a), the standard deviation
is
.
The 95% confidence interval is obtained below:
Substitute,
,
,
and
.
Therefore, the 95% confidence interval for mean percent earning for population of jobs lies between
.
c.
Obtain a 95% confidence interval for the mean percent earnings for the population of jobs using a small-sample argument.
From the table B-3, the table value of
with
degrees of freedom is given below:
Thus, the value of
is
.
The 95% confidence interval is obtained below:
Thus, the 95% confidence interval for the mean percent earning for the population of jobs is
. The additional assumption that is made in this case is that the mean percent earnings are approximately normally distributed.
d.
Explanation:
From part (b) and (c), it is clear that the confidence intervals are approximately same. This is because the multipliers 1.96 and 2.045 are almost has the same magnitude.

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From data on a large sample of sales transactions, a small business owner reports that a 95% confidence interval for the mean profit per transaction, ? , is (23.41, 102.59). Use these data to determine a. A point estimate (best guess) of the mean, ? , and its 95% error margin. b. A 90% confidence interval for the mean, ?.
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Essay

a.
Obtain the point estimate of the mean
.
The sample mean
is the point estimate of the mean
.
Therefore, the point estimate of the mean
is
.
Obtain the 95% error margin.
It is given that 95% confidence interval for the mean profit per transaction is
Therefore, the 95% error margin is
.
b.
Obtain the 90% confidence interval for the mean
.
For, the two-tailed test,
From Table B-2, the required
value for 90% confidence level is,
Thus, the 90% confidence interval is obtained below:
Therefore, the 90% confidence interval for the mean profit per transaction is
.

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We want to forecast whether the mean number of absent days per year has increased for our large workforce. A year ago the mean was known to be 12.1. A recent sample of 100 employees reveals a sample mean of 13.5 with a sample standard deviation of 1.7 days. Test at the.05 significance level to determine if the population mean has increased or if the difference between 13.5 and 12.1 simply represents sampling error.
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New Horizons Airlines wants to forecast the mean number of unoccupied seats per flight to Germany next year. To develop this forecast, the records of 49 flights are randomly selected from the files for the past year, and the number of unoccupied seats is noted for each flight. The sample mean and standard deviation are 8.1 seats and 5.7 seats, respectively. Develop a point and 95% interval estimate of the mean number of unoccupied seats per flight during the past year. Forecast the mean number of unoccupied seats per flight to Germany next year. Discuss the accuracy of this forecast.
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Over a period of years, a toothpaste has received a mean rating of 5.9, on a 7-point scale, for overall customer satisfaction with the product. Because of a minor unadvertised change in the product, there is concern that the customer satisfaction may have changed. Suppose the satisfaction ratings from a sample of 60 customers have a mean of 5.60 and a standard deviation of.87. Do these data indicate that the mean satisfaction rating is different from 5.9? Test with ? =.05. What is the p -value for the test?
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The manager of a frozen yogurt store claims that a medium-size serving contains an average of more than 4 ounces of yogurt. From a random sample of 14 servings, she obtains a mean of 4.31 ounces and a standard deviation of.52 ounce. Test, with ? =.05, the manager's claim. Find the p -value for the test. Assume that the distribution of weight per serving is normal.
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Based on past experience, the California Power Company forecasts that the mean residential electricity usage per household will be 700 kwh next January. In January, the company selects a simple random sample of 50 households and computes a mean and standard deviation of 715 and 50, respectively. Test at the.05 significance level to determine whether California Power's forecast is reasonable. Calculate and interpret the p -value for the test.
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Population experts indicate that family size has decreased in the last few years. Ten years ago the average family size was 2.9. Consider the population of 200 family sizes given in Table P-10. Randomly select a sample of 30 family sizes and test the hypothesis that the average family size has not changed in the last 10 years. TABLE P-10
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James Dobbins, maintenance supervisor for the Atlanta Transit Authority, would like to determine whether there is a positive relationship between the annual maintenance cost of a bus and its age. If a relationship exists, James feels that he can do a better job of predicting the annual bus maintenance budget. He collects the data shown in Table P-11. a. Plot a scatter diagram. b. What kind of relationship exists between these two variables? c. Compute the correlation coefficient. TABLE P-11
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Anna Sheehan is the manager of the Spendwise supermarket chain. She would like to be able to predict paperback book sales (books per week) based on the amount of shelf display space (feet) provided. Anna gathers data for a sample of 11 weeks, as shown in Table P-12. a. Plot a scatter diagram. b. What kind of relationship exists between these two variables? c. Compute the correlation coefficient. Determine the equation of the least squares line by calculating the slope and Y -intercept. Use this equation to forecast the number of books sold if 5.2 feet of shelf space is used (i.e., X = 5.2). TABLE P-12
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Consider the population of 200 weekly observations presented in Table P-13. The independent variable X is the average weekly temperature of Spokane, Washington. The dependent variable Y is the number of shares of Sunshine Mining Stock traded on the Spokane exchange in a week. Randomly select data for 16 weeks and compute the coefficient of correlation. ( Hint: Make sure your sample is randomly drawn from the population.) Then determine the least squares line and forecast Y for an average weekly temperature of 63. TABLE P-13
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A real estate investor collects the following data on a random sample of apartments on the west side of College Station, Texas. a. Plot the data as a scatter diagram with Y = rent and X = size. b. Determine the equation of the fitted line relating rent to size. c. What is the estimated increase in rent for an additional square foot of space? d. Forecast the monthly rent for an apartment with 750 square feet.
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Abbott Sons needs to forecast the mean age, ? , of its hourly workforce. A random sample of personnel files is pulled with the results below. Prepare both a point estimate and a 98% confidence interval for the mean age of the entire workforce. Test the hypothesis H 0 : ? = 44 versus H 1 : ? ? 44 at the 2% level. Are the results of the hypothesis test consistent with the confidence interval for ? ? Would you expect them to be?
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In each of the following situations, state an appropriate null hypothesis, H 0 , and alternative hypothesis, H 1. Identify the parameter that you use to state the hypotheses. a. Census Bureau data show that the mean household income in the area served by a shopping mall is \$63,700 per year. A market research firm surveys shoppers at the mall to find out whether the mean household income of mall shoppers is higher than that of the general population. b. Last year the local fire department took an average of 4.3 minutes to respond to calls. Do this year's data show a different average response time? c. The mean area of several thousand apartments in a new development is advertised to be 1,300 square feet. A tenant group thinks that the apartments are smaller than advertised. They hire an engineer to measure a sample of apartments to test their suspicion.
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An investor with a substantial stock portfolio sued her broker and brokerage firm because lack of diversification in her portfolio led to poor performance. The rates of return for the 39 months that the account was managed by the broker produced these summary statistics: , S = 5.99%. Consider the 39 monthly returns as a random sample from the population of returns the brokerage would generate if it managed the account forever. Using the sample results, construct a 95% confidence interval for the mean monthly market return. Let the S P 500 represent the market, and suppose the mean S P 500 return for the same period is 0.94%. Is this a realistic value for the population mean of the client's account? Explain.
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Table P-18 gives weekly wages (WAGES) in dollars and length of service (LOS) in months, at a specific point in time, for 16 women who hold customer service jobs in Texas banks. TABLE P-18 a. Plot the data in Table P-18 as a scatter diagram with WAGES along the vertical ( Y ) axis and LOS along the horizontal ( X ) axis. b. Calculate the sample correlation coefficient, r. Using the sign and magnitude of r , describe the nature of the linear association between WAGES and LOS. Can you think of other variables that might affect weekly wages in addition to length of service? c. Calculate the fitted line that can be used to predict WAGES from LOS. If LOS is 80 months, what is the predicted WAGES?
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