Answer The length of time to do a cable installation by Multi-Cable Inc. is normally distributed with a mean of 42.8 minutes and a standard deviation of 6.2 minutes. What percentage of installations take less than 30 minutes?

x = The length of time to do a cable installation = 42.8 = 6.20

P(x < 30) = P(z < -2.06) = 0.5 - 0.4803 = 0.0197 or 1.97%

Answer Accident claims are checked for completeness by branch offices of Fortune Insurance before they are sent to a regional office for payment. Historically 80% of the claims are complete when they reach the regional office. You select a random sample of 20 claims that have arrived at the regional office. Find the probability that:

X = Number of completed forms in samplen = 20p = 0.80

a. At least 16 claims are complete,

P(X 16) = P(X = 16) + … + P(X = 20)= 0.6296

b. All 20 are complete,

P(X = 20) = 0.0115

c. Fewer than 12 are complete,

P(X < 12) = P(X 11) = P(X = 0) + … + P(X = 11)

= 0.0100

d. Exactly 12 are complete.

P(X = 12) = 0.0222

Answer Until this year the mean braking distance of a Nikton automobile moving at 60 miles per hour was 175 feet. Nikton engineers have developed what they consider a better braking system. They test the new brake system on a random sample of 81 cars and determine the sample mean braking distance. The results are:

a. Compute the 90% confidence interval for the mean braking

distance. Interpret this interval.

We are 90% confident that the population mean breaking distance is between 162.1 ft and 171.9 ft.

b. How many cars should be tested if Nikton wants to be 90% confident of being within 2 feet of the population mean braking distance? Assume the sample size will be larger than 30.

Answer The manager at Trion Electric Co-op is concerned about the rise in the number of late payments. If more than 10% of the accounts are behind in their payments, the manager will initiate a costly monitoring program. A random sample of 75 accounts shows that 9 were behind in their payments.

a. Does the sample data provide evidence to conclude that more than 10% of all accounts are delinquent (using =0.01)? Use the 5 step hypothesis testing procedure.

p = population proportion of accounts behind in payments.

1. Formulate the null and alternative hypotheses:

H0: p = 0.10

Ha: p > 0.10

2. Determine the criterion for rejection or non-rejection of the null hypothesis:

= .01 n = 75 Ha: p > .10

Rejection region: z-distribution, one-tail, upper z > z.01 z> 2.33

3. Calculate the test statistic.

1

2

4. Compare the test statistic to the rejection region and make a judgment about the null and alternative hypotheses:

Because the calculated z is less than 2.33, we cannot reject the null hypothesis (H0) and, therefore, we cannot accept the alternative hypothesis (Ha) with at least 99% confidence.

5. Interpret the statistical decision in terms of the problem.

We are unable to conclude with 99% confidence that more than 10% of accounts are behind in payments.

b. Compute the observed p value in the hypothesis test, and interpret this value. What does this mean?

observed p value = 0.5 - 0.2190 = 0.2810

Type I Error Probability

We have a 28.10% chance of being wrong in rejecting H0 and accepting Ha.

We are 71.90% confident in rejecting H0 and accepting Ha.

We are 71.90% confident that more than 10% of accounts are behind in payments.

Answer A manufacturer of athletic footwear claims that the mean life of his product will exceed 50 hours. A random sample of 36 pairs of shoes leads to the following results in terms of useful life:

a. Does the sample data provide evidence to conclude that the manufacturer's claim is correct (using =0.05)? Use the 5 step hypothesis testing procedure.

1. Formulate the null and alternative hypotheses.

H0: = 50

Ha: > 50

2. Determine the criterion for rejection or non-rejection of the null hypothesis.

=0.05

n = 36 Ha: > 50

z-distribution, one tail, upper z > z.05 z > 1.645

3. Calculate the test statistic.

4. Compare the test statistic to the rejection region and make a judgment about the null and alternative hypotheses.

Because the calculated z is less than 1.645, we cannot reject H0 and we cannot accept Ha with at least 95% confidence.

5. Interpret the statistical decision in terms of the problem.

We are unable to conclude that the manufacturer's claim is correct. We were unable to be 95% sure that the mean life exceeds 50 hours based on this sample.

b. Compute the observed p value in the hypothesis test, and interpret this value. What does this mean?

observed p value = 0.5 - 0.4251 = 0.0749

15105

7500

6500

5500

4500

3500

2500

1500

Bidders

Sa

l ep

ri ce

R-Squared = 0.948

Y = 1007.56 + 324.11X

Regres sion Plot

Type I Error Probability

We have a 7.49% chance of being wrong in rejecting H0 and accepting Ha.

We are 92.51% confident in rejecting H0 and accepting Ha.

We are 92.51% confident that the mean lifetime exceeds 50 hours.

Answer AJ AUCTION sells art drawings of little known artist at auction. The dealer is interested in the relationship between the final sale price ($) (Y) and the number of persons bidding on the drawing (X). A random sample of 20 drawing is selected.

Correlations (Pearson)Correlation of salepric and bidders = 0.974

Regression Analysis

The regression equation issalepric = 1008 + 324 bidders

Predictor Coef Stdev t-ratio pConstant 1007.6 214.5 4.70 0.000bidders 324.11 17.90 18.11 0.000

s = 320.2 R-sq = 94.8% R-sq(adj) = 94.5%

Fit Stdev.Fit 95.0% C.I. 95.0% P.I. 4248.7 75.3 (4090.4, 4406.9) (3557.4, 4940.0)

a. State the equation of the best fit line. State the values of b0 and b1.

salepric = 1008 + 324 bidders

b0 = 1008

b1 = 324

b. Graph the best fit line on the scatter diagram.

See regression line on above scatter diagram.

c. Determine s, the estimated standard deviation of , and interpret this value.

s = 320.2

s measures the spread of the distribution of y values about the least-squares line. Most of the observations lie within 2s or 2(320.2) = 640.4.

d. Determine the index of determination and interpret this value.

R2 = 94.8%

94.8% of the variation in final sale price can be explained by the number of bidders.

e. Construct the 95% confidence interval for . Interpret this interval.

t0.025 (18 df)= 2.101

324.11 ± 2.101(17.90)324.11 ± 37.6079324 ± 38

We are 95% confident that each additional bidder increases the final sale price by between $286 and $362.

f. Does this data provide significant evidence ( = 0.05) that final sale price is associated with the number of bidders? Test the utility of this model using a two tailed test. Find the observed p-value and interpret.

We are almost 100% sure that ß1 is not equal to 0.

We are almost 100% confident that final sale price is associated with the number of bidders.

g. Find the 95% prediction interval for the final sale price when the number of bidders is 10. Interpret this interval.

(3,557.40, 4,940.00)

If the number of bidders in a particular instance is 10, then we are 95% confident that the final price is between $3,557 and $4,940.

h. Find the 95% confidence interval for the final sale price when the number of bidders is 10. Interpret this interval.

(4090.4, 4406.9)

We are 95% sure that in all situations where the number of bidders is 10, the mean final sale price is between $4,090.40 and $4,406.90.

Answer The personnel department of a large industrial corporation would like to develop a model to predict the weekly salary based upon the length of employment and age of its managerial employees. Salary is modeled as a linear function of length of employment and age.

The regression equation isWklySlry = 662 + 0.671 LengEmpl - 1.38 Age

Predictor Coef Stdev t-ratio pConstant 661.85 57.17 11.58 0.000LengEmpl 0.6712 0.1475 4.55 0.000Age -1.383 1.772 -0.78 0.449

s = 39.39 R-sq = 81.4% R-sq(adj) = 78.5%

Analysis of Variance

SOURCE DF SS MS F pRegression 2 88297 44149 28.45 0.000Error 13 20175 1552Total 15 108472

SOURCE DF SEQ SSLengEmpl 1 87352Age 1 946

Fit Stdev.Fit 95.0% C.I. 95.0% P.I.857.01 16.70 (820.91, 893.10) (764.54, 949.47)

a. Using the regression output, specify the leastsquares prediction equation.

WklySlry = 662 + 0.671 LengEmpl - 1.38 Age

b. Perform a global usefulness test of the model (use the =.05 level) and interpret the result.

SOURCE DF SS MS F pRegression 2 88297 44149 28.45 0.000Error 13 20175 1552Total 15 108472

As can be seen from the ANOVA portion of the regression output, the F value for the model is 28.45 which is highly significant (.000). This means that the model can be useful in predicting salary as a function of length of employment and age.

c. Find and interpret the multiple coefficient of determination.

R-sq(adj) = 78.5%

The multiple coefficient of correlation indicates that 78.5% of the variation in salary can be explained in terms of length of employment and age.

d. Perform tests on the independent variables (=.05, twotailed) and interpret the results.

Predictor Coef Stdev t-ratio pConstant 661.85 57.17 11.58 0.000LengEmpl 0.6712 0.1475 4.55 0.000Age -1.383 1.772 -0.78 0.449

As can be seen in the above portion of the regression output, one variablelength of employmentis significant and therefore contributes to the model. Length of employment has a t value of 4.55 and is significant at the .000 level. Age, however, has a t value of -.78 and is not significant at the .05 level (p-value equals 0.449). Therefore, the variable age should be dropped from the model.

e. Find the 95% prediction interval for weekly income when the length of employment is 400 months and the age is 53 and interpret it in the context of the problem.

Fit Stdev.Fit 95.0 % C.I. 95.0% P.I.857.01 16.70 (820.91, 893.10) (764.54, 949.47)

The 95% prediction interval can be seen to be 764.54 to 949.47. This means that the weekly income of a particular individual who has been employed for 400 months and is 53 years old would be between about $765 and $950.

f. Find the 95% confidence interval for weekly income when the length of employment is 400 months and the age is 53 and interpret it in the context of the problem.

Fit Stdev.Fit 95.0% C.I. 95.0% P.I.857.01 16.70 (820.91, 893.10) (764.54, 949.47)

The 95% confidence interval can be seen to be 820.91 to 893.10. This means that across all situations for individuals who have been employed for 400 months and are 53 years old, the average weekly income would be between about $820 and $893.