# Standard Deviation for Nielsen Media Research

PROBLEM SET II ALTERNATE 1

9-11. The owner of Britten's Egg Farm wants to estimate the mean number of eggs laid per chicken. A sample of 20 chickens shows they laid an average of 20 eggs per month with a standard deviation of 2 eggs per month.

a. What is the value of the population mean? What is the best estimate of this value?

b. Explain why we need to use the t distribution. What assumption do you need to make?

c. For a 95 percent confidence interval, what is the value of t?

d. Develop the 95 percent confidence interval for the population mean.

e. Would it be reasonable to conclude that the population mean is 21 eggs? What about 25 eggs?

9-30. Past surveys reveal that 30 percent of tourists going to Las Vegas to gamble during a

weekend spend more than $1,000. Management wants to update this percentage.

a. The new study is to use the 90 percent confidence level. The estimate is to be within 1

percent of the population proportion. What is the necessary sample size?

b. Management said that the sample size determined above is too large. What can be done

to reduce the sample? Based on your suggestion recalculate the sample size.

CHAPTER 10

10-7. A recent national survey found that high school students watched an average (mean) of 6.8

DVDs per month. A random sample of 36 college students revealed that the mean number

of DVDs watched last month was 6.2, with a standard deviation of 0.5. At the .05 significance

level, can we conclude that college students watch fewer DVDs a month than high

school students?

10-14. Research at the University of Toledo indicates that 50 percent of the students change their major area of study after their first year in a program. A random sample of 100 students in the College of Business revealed that 48 had changed their major area of study after their first year of the program. Has there been a significant decrease in the proportion of students who change their major after the first year in this program? Test at the .05 level of significance.

CHAPTER 11

11-25. Fry Brothers Heating and Air Conditioning, Inc. employs Larry Clark and George Murnen to make service calls to repair furnaces and air conditioning units in homes. Tom Fry, the owner, would like to know whether there is a difference in the mean number of service calls they make per day. A random sample of 40 days last year showed that Larry Clark made an average of 4.77 calls per day, with a standard deviation of 1.05 calls per day. For a sample of 50 days George Murnen made an average of 5.02 calls per day, with a standard deviation of 1.23 calls per day. At the .05 significance level, is there a difference in the mean number of calls per day between the two employees? What is the p-value?

Here average call of Larry Clark = 4.77,

Average call of George Murnen = 5.02, and

Then z value = < 1.96 thus there is a difference in the mean number of calls per day between the two employees.

And p value is 0.3531 from normal table

11-39. The manufacturer of an MP3 player wanted to know whether a 10 percent reduction in price is enough to increase the sales of their product. To investigate, the owner randomly selected eight outlets and sold the MP3 player at the reduced price. At seven randomly selected outlets, the MP3 player was sold at the regular price. Reported below is the number of units sold last month at the sampled outlets. At the .01 significance level, can the manufacturer conclude that the price reduction resulted in an increase in sales?

Regular price 138 121 88 115 141 125 96

Reduced price 128 134 152 135 114 106 112 120

Using MS EXCEL DATA ANALYSIS

Variance of regular prize = 396.5714

Variance of reduced prize = 277.8393

z-Test: Two Sample for Means

Variable 1 Variable 2

Mean 117.7143 125.125

Known Variance 396.5714 227.8393

Observations 7 8

Hypothesized Mean Difference 0

z -0.80318

P(Z<=z) one-tail 0.210936

z Critical one-tail 1.644854

P(Z<=z) two-tail 0.421873

z Critical two-tail 1.959964

Thus z value is less than 0.05 significant levels. Hence the manufacturer can conclude that the price reduction resulted in an increase in sales.

CHAPTER 12

12-30. There are four auto body shops in a community and all claim to promptly serve customers. To check if there is any difference in service, customers are randomly selected from each repair shop and their waiting times in days are recorded. The output from a statistical software package is:

Summary

Groups Count Sum Average Variance

Body Shop A 3 15.4 5.133333 0.323333

Body Shop B 4 32 8 1.433333

Body Shop C 5 25.2 5.04 0.748

Body Shop D 4 25.9 6.475 0.595833

ANOVA

Source of Variation SS df MS F p-value

Between Groups 23.37321 3 7.791069 9.612506 0.001632

Within Groups 9.726167 12 0.810514

Total 33.09938 15

Is there evidence to suggest a difference in the mean waiting times at the four body shops? Use the .05 significance level.

Since the p-value is less than F-value. Thus there is no evidence to suggest a difference in the mean waiting times at the four body shops

CHAPTER 13

13-56. A consumer buying cooperative tested the effective heating area of 20 different electric

space heaters with different wattages. Here are the results.

Heater Wattage Area

1 1,500 205

2 750 70

3 1,500 199

4 1,250 151

5 1,250 181

6 1,250 217

7 1,000 94

8 2,000 298

9 1,000 135

10 1,500 211

11 1,250 116

12 500 72

13 500 82

14 1,500 206

15 2,000 245

16 1,500 219

17 750 63

18 1,500 200

19 1,250 151

20 500 44

a. Compute the correlation between the wattage and heating area. Is there a direct or an

Indirect relationship?

According to correlation formula

Now using MS-excel data analysis, r = 0.939287

Positive values of r shows direct relationship.

b. Conduct a test of hypothesis to determine if it is reasonable that the coefficient is greater

than zero. Use the .05 significance level.

= 0.94/(1 - (.88)^2 / 19 ) = 0.96265

c. Develop the regression equation for effective heating based on wattage.

By MS-Excel data analysis

Regression equation is

Y = 5.92x + 276.55

d. Which heater looks like the "best buy" based on the size of the residual?

RESIDUAL OUTPUT

Observation Predicted Y Residuals

1 1491.295307 8.704692718

2 691.35128 58.64871999

3 1455.742239 44.2577606

4 1171.317696 78.68230363

5 1349.083036 -99.08303577

6 1562.401443 -312.401443

7 833.5635515 166.4364485

8 2042.367859 -42.3678594

9 1076.509515 -76.50951536

10 1526.848375 -26.84837516

11 963.9248004 286.0751996

12 703.2023026 -203.2023026

13 762.4574158 -262.4574158

14 1497.220819 2.779181405

15 1728.31576 271.6842402

16 1574.252466 -74.25246567

17 649.8727008 100.1272992

18 1461.667751 38.33224928

19 1171.317696 78.68230363

20 537.2879859 -37.28798587

Heater no.14 should buy on the basis of RESIDUAL.

CHAPTER 15

15-12. For many years TV executives used the guideline that 30 percent of the audience were watching each of the prime-time networks and 10 percent were watching cable stations on a weekday night. A random sample of 500 viewers in the Tampa-St. Petersburg, Florida, area last Monday night showed that 165 homes were tuned in to the ABC affiliate, 140 to the CBS affiliate, 125 to the NBC affiliate, and the remainder were viewing a cable station. At the .05 significance level, can we conclude that the guideline is still reasonable?

There are sample of 500 viewers. 30 percent of the audience was watching each of the prime-time networks and 10 percent were watching cable stations on a weekday night. Then probability of watching prime time networks 0.3 and probability of watching cable stations 0.1. A random sample of 500 viewers in the Tampa-St. Petersburg, Florida, and area last Monday night showed that 165 homes were tuned in to the ABC affiliate, 140 to the CBS affiliate, 125 to the NBC affiliate, and the remainder was viewing a cable station. Then probability of watching prime time channel 430/500 = 0.86

And for cable = 70/500 = 0.14

At 0.05 significance level using Z- test,

For prime time > 1.96

For cable

Thus we can conclude that the guideline is still reasonable

15-25. Nielsen Media Research wants to pretest a questionnaire to be mailed to several thousand

viewers. One question involves the ranking of male and female college students with

respect to the popularity of programs. The composite rankings of a small group of college

students are:

Ranking by Ranking by

Program Males Females

"Monday Night Football" 1 5

"Tonight Show with Jay Leno" 4 1

"60 Minutes" 3 2

"Late Show with David Letterman" 2 4

"Sports Center on ESPN" 5 3

a. Draw a scatter diagram. Let the rankings by males be X.

b. Compute Spearman's rank-order correlation coefficient. Interpret.

= 1 - 34*6/5(25-1) = 1 - 34/20 = -0.7

This shows negative correlation

© BrainMass Inc. brainmass.com March 21, 2019, 5:51 pm ad1c9bdddfhttps://brainmass.com/statistics/hypothesis-testing/standard-deviation-235431

#### Solution Preview

CHAPTER 9

9-11. The owner of Britten's Egg Farm wants to estimate the mean number of eggs laid per chicken. A sample of 20 chickens shows they laid an average of 20 eggs per month with a standard deviation of 2 eggs per month.

a. What is the value of the population mean? What is the best estimate of this value?

Population mean is 20. The best estimate of this value is also 20.

b. Explain why we need to use the t distribution. What assumption do you need to make?

Since here sample size is small (< 30 ) therefore we cannot use here Z- DISTRIBUTION. Thus for small sample size the t distribution is only matter.

We assume that the data is normally distributed.

c. For a 95 percent confidence interval, what is the value of t?

Using t- distribution table , n = 20 therefore degree of freedom is 19

d. Develop the 95 percent confidence interval for the population mean.

For confidence interval =

=

e. Would it be reasonable to conclude that the population mean is 21 eggs? What about 25 eggs?

Since here confidence interval lies 19 - 21 (approximately) therefore for 21 eggs it can be better estimate. For 25 eggs it cannot be better estimate.

9-30. Past surveys reveal that 30 percent of tourists going to Las Vegas to gamble during a

weekend spend more than $1,000. Management wants to update this percentage.

a. The new study is to use the 90 percent confidence level. The estimate is to be within 1

percent of the population proportion. What is the necessary sample size?

The probability of tourists going to Las Vegas to gamble during a

Weekend spend more than $1,000 (p) = 0.3

For 90% sample size (z value) = 1.645

b. Management said that the sample size determined above is too large. What can be done

to reduce the sample? Based on your suggestion recalculate the sample size.

Since sample size is too large therefore to reduce sample size, with in 90 percent confidence interval, management should increase percent of proportion. For example if we take, estimate with in 3% of proportion. Then

CHAPTER 10

10-7. A recent national survey found that high school students watched an average (mean) of 6.8

DVDs per month. A random sample of 36 college students revealed that the mean number

of DVDs watched last month was 6.2, with a standard deviation of 0.5. At the .05 significance

level, can we conclude that college students watch fewer DVDs a month than high

school students?

Since there are total N = 36 student

Null Hypothesis: None of college students watch fewer DVDs a month than high school students

Alternative Hypothesis: college students watch fewer DVDs a month than high

school students

Z test =

= (6.8 ...

#### Solution Summary

The standard deviation for Nielsen Media Research is analyzed.