Question

Annual starting salaries for college graduates with degrees in business administration are generally expected to be between $20,000 and $40,000. Assume that a 95% confidence interval estimate of the population mean annual starting salary is desired. How large a sample should be taken if the desired margin of error is:

a.$300

b.$230

c.$80 Would you recommend trying to obtain the $80 margin of error? Explain.

A simple random sample of 80 items from a population with σ = 7 resulted in a sample mean of 33. If required, round your answers to two decimal places.

a. Provide a 90% confidence interval for the population mean. to

b. Provide a 95% confidence interval for the population mean. to

c. Provide a 99% confidence interval for the population mean. to

Question 2 A study showed that 67% of supermarket shoppers believe supermarket brands to be as good as national name brands. To investigate whether this result applies to its own product, the manufacturer of a national name-brand ketchup asked a sample of shoppers whether they believed that supermarket ketchup was as good as the national brand ketchup.

a. Formulate the hypotheses that could be used to determine whether the percentage of supermarket shoppers who believe that the supermarket ketchup was as good as the national brand ketchup differed from 67%. H0: p Select greater than or equal to 0.67 greater than 0.67 less than or equal to 0.67 less than 0.67 equal to 0.67 not equal to 0.67 Item 1 Ha: p Select greater than or equal to 0.67 greater than 0.67 less than or equal to 0.67 less than 0.67 equal to 0.67 not equal to 0.67 Item 2

b. If a sample of 100 shoppers showed 52 stating that the supermarket brand was as good as the national brand, what is the p-value (to 4 decimals)?

c. At α = .05, what is your conclusion? p-value Select greater than or equal to 0.05, reject greater than 0.05, do not reject less than or equal to 0.05, reject less than 0.05, reject equal to 0.05, do not reject not equal to 0.05, do not reject Item 4 H 0 d. Should the national brand ketchup manufacturer be pleased with this conclusion? Select Yes No Item 5

Question 3 Data on advertising expenditures and revenue (in thousands of dollars) for the Four Seasons Restaurant follow. Advertising Expenditures 1 2 4 6 10 14 20 Revenue 20 33 44 41 52 53 54

a. Let x equal advertising expenditures and y equal revenue. Complete the estimated regression equation below (to 2 decimals). y= _ + _x

b. compute the following to 1 decimal

SSE

SST

SSR

MSR

MSE

c. st whether revenue and advertising expenditures are related at a .05 level of significance. Compute the F test statistic (to 2 decimals). What is the p-value? Select less than .01 between .01 and .025 between .025 and .05 between .05 and .10 greater than .10 Item 9 What is the conclusion? Select Conclude revenue is related to advertising expenditure Cannot conclude revenue is related to advertising expenditure Item 10 What does the residual plot above suggest about the assumption of linearity between x and y? Select Despite the small sample size, there appears to be a linear relationship Question the assumption that a linear relationship exists

Question 4 Consider the following data for two variables,

x and y. xi 140 110 135 145 175 165 120

yi 150 100 120 120 130 130 110

a. Compute the standardized residuals for these data

Observation 1

Observation 2

Observation 3

Observation 4

Observation 5

Observation 6

Observation 7 Do the data include any outliers? Select Yes, there appear to be 3 outliers. Yes, there appear to be 2 outliers. Yes, there appears to be an outlier. No, there do not appear to be any outliers.

Question 5 The travel-to-work time for residents of the 15 largest cities in the United States is reported in the 2003 Information Please Almanac. Suppose that a preliminary simple random sample of residents of San Francisco is used to develop a planning value of 6.66 minutes for the population standard deviation. Round your answers to nearest whole number

A.If we want to estimate the population mean travel-to-work time for San Francisco residents with a margin of error of 5 minutes, what sample size should be used? Assume 95% confidence.

b. If we want to estimate the population mean travel-to-work time for San Francisco residents with a margin of error of 3 minutes, what sample size should be used? Assume 95% confidence.

Answer 1

1)

estimate of population Standard Deviation ,   σ = range/4 = (40000-20000)/4 = 5000

a)

sampling error ,    E =   300                  
Confidence Level ,   CL=   95%                  
                          
alpha =   1-CL =   5%                  
Z value =    Zα/2 =    1.960   [excel formula =normsinv(α/2)]              
                          
Sample Size,n = (Z * σ / E )² = (   1.960   *   5000   /   300   ) ² =   1067.072
                          
                          
So,Sample Size needed=       1068                  

b)

sampling error ,    E =   230                  
Confidence Level ,   CL=   95%                  
                          
alpha =   1-CL =   5%                  
Z value =    Zα/2 =    1.960   [excel formula =normsinv(α/2)]              
                          
Sample Size,n = (Z * σ / E )² = (   1.960   *   5000   /   230   ) ² =   1815.434
                          
                          
So,Sample Size needed=       1816                  

c)

sampling error ,    E =   80                  
Confidence Level ,   CL=   95%                  
                          
alpha =   1-CL =   5%                  
Z value =    Zα/2 =    1.960   [excel formula =normsinv(α/2)]              
                          
Sample Size,n = (Z * σ / E )² = (   1.960   *   5000   /   80   ) ² =   15005.699
                          
                          
So,Sample Size needed=       15006                  
NO, we do not  recommend trying to obtain the $80 margin of error because sample size required will be very large

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2)

a)

Level of Significance ,    α =    0.1          
'   '   '          
z value=   z α/2=   1.645   [Excel formula =NORMSINV(α/2) ]      
                  
Standard Error , SE = σ/√n =   7.0000   / √   80   =   0.7826
margin of error, E=Z*SE =   1.6449   *   0.7826   =   1.2873
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    33.00   -   1.287302   =   31.713
Interval Upper Limit = x̅ + E =    33.00   -   1.287302   =   34.287
90%   confidence interval is (   31.71   < µ <   34.29   )

b)

Level of Significance ,    α =    0.05          
'   '   '          
z value=   z α/2=   1.960   [Excel formula =NORMSINV(α/2) ]      
                  
Standard Error , SE = σ/√n =   7.0000   / √   80   =   0.7826
margin of error, E=Z*SE =   1.9600   *   0.7826   =   1.5339
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    33.00   -   1.533914   =   31.466
Interval Upper Limit = x̅ + E =    33.00   -   1.533914   =   34.534
95%   confidence interval is (   31.47   < µ <   34.53   )

c)

Level of Significance ,    α =    0.01          
'   '   '          
z value=   z α/2=   2.576   [Excel formula =NORMSINV(α/2) ]      
                  
Standard Error , SE = σ/√n =   7.0000   / √   80   =   0.7826
margin of error, E=Z*SE =   2.5758   *   0.7826   =   2.0159
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    33.00   -   2.015905   =   30.984
Interval Upper Limit = x̅ + E =    33.00   -   2.015905   =   35.016
99%   confidence interval is (   30.98   < µ <   35.02   )