Introduction to the Practice of Statistics
Introduction to the Practice of Statistics
9th Edition
ISBN: 9781319055967
Author: Moore
Publisher: MAC HIGHER
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Chapter 8, Problem 94E
To determine

To find: The 95% margin of error for the difference of two proportions for sample size n=60.

Expert Solution
Check Mark

Answer to Problem 94E

Solution: The margin of error is 0.1717.

Explanation of Solution

Calculation: The assumption made that the sample size n is the common value of n1and n2 and hence n1=60andn2=60. The margin of error is defined as:

m=z*×SED

Where, z* is the critical value for the standard normal variable and SED is the standard error of the difference. Also, the standard error of D is defined as:

SED=p^1(1p^1)n1+p^2(1p^2)n2

Where,

p^1=Sample proportion from population 1p^2=Sample proportion from population 2n1=Sample size of group 1n2=Sample size of group 2

Substitute the provided values p^1=0.70,p^2=0.50 and n1=60,n2=60 in the above defined formula to calculate the standard error of D.

SED=0.70×(10.70)60+0.50×(10.50)60=0.0035+0.0042=0.0876

The value of z* for 95% confidence level is z*=1.96 from the standard normal table. So, the margin of error is obtained as:

m=z*×SED^=1.96×0.0876=0.1717

Therefore, the margin of error is obtained as 0.1717.

To determine

To find: The 95% margin of error for the difference of two proportions for sample size n=70.

Expert Solution
Check Mark

Answer to Problem 94E

Solution: The margin of error is 0.1589.

Explanation of Solution

Calculation: The assumption made that the sample size n is the common value of n1and n2 and hence n1=70,n2=70. The margin of error is defined as:

m=z*×SED

Where z* is the critical value for the standard normal variable and SED is the standard error of the difference. Also, the standard error of D is defined as:

SED=p^1(1p^1)n1+p^2(1p^2)n2

Where

p^1=Sample proportion from population 1p^2=Sample proportion from population 2n1=Sample size of group 1n2=Sample size of group 2

Substitute the provided values p^1=0.70,p^2=0.50 and n1=70,n2=70 in the above defined formula to calculate the standard error of D.

SED=0.70×(10.70)70+0.50×(10.50)70=0.003+0.0036=0.0811

The value of z* for 95% confidence level is z*=1.96 from the standard normal table. So, the margin of error is obtained as:

m=z*×SED^=1.96×0.0811=0.1589

Therefore, the margin of error is obtained as 0.1589.

To determine

To find: The 95% margin of error for the difference of two proportions for sample size n=80.

Expert Solution
Check Mark

Answer to Problem 94E

Solution: The margin of error is 0.1486.

Explanation of Solution

Calculation: The assumption made that the sample size n is the common value of n1and n2 and hence n1=80andn2=80. The margin of error is defined as:

m=z*×SED

Where z* is the critical value for the standard normal variable and SED is the standard error of the difference. Also, the standard error of D is defined as:

SED=p^1(1p^1)n1+p^2(1p^2)n2

Where

p^1=Sample proportion from population 1p^2=Sample proportion from population 2n1=Sample size of group 1n2=Sample size of group 2

Substitute the provided values p^1=0.70,p^2=0.50 and n1=80,n2=80 in the above defined formula to calculate the standard error of D.

SED=0.70×(10.70)80+0.50×(10.50)80=0.0026+0.0031=0.0758

The value of z* for 95% confidence level is z*=1.96 from the standard normal table. So, the margin of error is obtained as:

m=z*×SED^=1.96×0.0758=0.1486

Therefore, the margin of error is obtained as 0.1486.

To determine

To find: The 95% margin of error for the difference of two proportions for sample size n=100.

Expert Solution
Check Mark

Answer to Problem 94E

Solution: The margin of error is 0.1329.

Explanation of Solution

Calculation: The assumption made that the sample size n is the common value of n1and n2 and hence n1=100andn2=100. The margin of error for confidence level C is defined as:

m=z*×SED

Where z* is the value for the standard normal density curve and SED is the standard error of the difference. Also, the standard error of D is defined as:

SED=p^1(1p^1)n1+p^2(1p^2)n2

Where

p^1=Sample proportion from population 1p^2=Sample proportion from population 2n1=Sample size of group 1n2=Sample size of group 2

Substitute the provided values p^1=0.70,p^2=0.50 and n1=100,n2=100 in the above defined formula to calculate the standard error of D.

SED=0.70×(10.70)100+0.50×(10.50)100=0.0021+0.0025=0.0678

The value of z* for 95% confidence level is z*=1.96 from the standard normal table. So, the margin of error is obtained as:

m=z*×SED^=1.96×0.0678=0.1329

Therefore, the margin of error is obtained as 0.1329.

To determine

To find: The 95% margin of error for the difference of two proportions for sample size n=400.

Expert Solution
Check Mark

Answer to Problem 94E

Solution: The margin of error is 0.0665.

Explanation of Solution

Calculation: The assumption made that the sample size n is the common value of n1and n2 and hence n1=400andn2=400. The margin of error is calculated as,

m=z*×SED

Where z* is the critical value for the standard normal variable and SED is the standard error of the difference. The standard error of D is defined as:

SED=p^1(1p^1)n1+p^2(1p^2)n2

Where

p^1=Sample proportion from population 1p^2=Sample proportion from population 2n1=Sample size of group 1n2=Sample size of group 2

Substitute the provided values p^1=0.70,p^2=0.50 and n1=400,n2=400 in the above defined formula to calculate the standard error of D.

SED=0.70×(10.70)400+0.50×(10.50)400=0.000525+0.000625=0.0339

The value of z* for 95% confidence level is z*=1.96 from the standard normal table. So, the margin of error is obtained as:

m=z*×SED^=1.96×0.0339=0.0665

Therefore, the margin of error is obtained as 0.0665.

To determine

To find: The 95% margin of error for the difference of two proportions for sample size n=500.

Expert Solution
Check Mark

Answer to Problem 94E

Solution: The margin of error is 0.0595.

Explanation of Solution

Calculation: The assumption made that the sample size n is the common value of n1and n2 and hence n1=500andn2=500. The margin of error for confidence level C is defined as:

m=z*×SED

Where z* is the critical value for the standard normal variable and SED is the standard error of the difference. The standard error of D is defined as:

SED=p^1(1p^1)n1+p^2(1p^2)n2

Where

p^1=Sample proportion from population 1p^2=Sample proportion from population 2n1=Sample size of group 1n2=Sample size of group 2

Substitute the provided values p^1=0.70,p^2=0.50 and n1=500,n2=500 in the above defined formula to calculate the standard error of D.

SED=0.70×(10.70)500+0.50×(10.50)500=0.00042+0.0005=0.0303

The value of z* for 95% confidence level is z*=1.96 from the standard normal table. So, the margin of error is obtained as:

m=z*×SED^=1.96×0.0303=0.0595

Therefore, the margin of error is obtained as 0.0595.

To determine

To find: The 95% margin of error for the difference of two proportions for sample size n=1000.

Expert Solution
Check Mark

Answer to Problem 94E

Solution: The margin of error is 0.0420.

Explanation of Solution

Calculation: The assumption made that the sample size n is the common value of n1and n2 and hence n1=1000andn2=1000. The margin of error is calculated as:

m=z*×SED

Where z* is the critical value for the standard normal variable and SED is the standard error of the difference. The standard error of D is defined as:

SED=p^1(1p^1)n1+p^2(1p^2)n2

Where

p^1=Sample proportion from population 1p^2=Sample proportion from population 2n1=Sample size of group 1n2=Sample size of group 2

Substitute the provided values p^1=0.70,p^2=0.50 and n1=1000,n2=1000 in the above defined formula to calculate the standard error of D.

SED=0.70×(10.70)1000+0.50×(10.50)1000=0.00021+0.00025=0.0214

The value of z* for 95% confidence level is z*=1.96 from the standard normal table. So, the margin of error is obtained as:

m=z*×SED^=1.96×0.0214=0.0420

Therefore, the margin of error is obtained as 0.0420.

To determine

The obtained results of margin of error for different sample sizes in a table.

Expert Solution
Check Mark

Answer to Problem 94E

Solution: The obtained results of 95% margins of error for different sample sizes are tabulated as:

N

m

60

0.1717

70

0.1589

80

0.1486

100

0.1329

400

0.0665

500

0.0595

1000

0.0420

Explanation of Solution

The 95% margins of error for the difference between two proportions for different sample sizes are calculated as in the previous parts and are tabulated as:

n

m

60

0.1717

70

0.1589

80

0.1486

100

0.1329

400

0.0665

500

0.0595

1000

0.0420

So, it is observed that with an increase in the sample size the margin of error reduces. Hence, a larger sample results in more accurate results.

To determine

To graph: The 95% margins of error obtained for different sample sizes.

Expert Solution
Check Mark

Explanation of Solution

Graph: To obtain the graph of 95% margins of error for different sample sizes, Minitab is used. The steps followed to construct the graph are:

Step 1: Enter the data of the sample sizes and the associated margins of error values in the worksheet.

Step 2: Select Graph-->Scatterplot.

Step 3: Select Simple from the opened dialog and click on OK.

Step 4: Select the Y variables as margin of error column and X variable as sample size column and click on OK.

The graph is obtained as:

Introduction to the Practice of Statistics, Chapter 8, Problem 94E

To determine

To explain: A short summary of the obtained results.

Expert Solution
Check Mark

Answer to Problem 94E

Solution: The obtained results of margins of error and the graph shows that the margin of error decreases as the sample size increases. But the rate of decrease of margin of error is very less for large n.

Explanation of Solution

The margins of error for different sample sizes are obtained and tabulated. The results show that the margin of error decreases as the sample size increases. This is also shown by the graph.

Hence, a larger sample is desirable for a lower error.

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Chapter 8 Solutions

Introduction to the Practice of Statistics

Ch. 8.1 - Prob. 11UYKCh. 8.1 - Prob. 12UYKCh. 8.1 - Prob. 13UYKCh. 8.1 - Prob. 14ECh. 8.1 - Prob. 15ECh. 8.1 - Prob. 16ECh. 8.1 - Prob. 17ECh. 8.1 - Prob. 18ECh. 8.1 - Prob. 19ECh. 8.1 - Prob. 20ECh. 8.1 - Prob. 21ECh. 8.1 - Prob. 22ECh. 8.1 - Prob. 23ECh. 8.1 - Prob. 24ECh. 8.1 - Prob. 25ECh. 8.1 - Prob. 26ECh. 8.1 - Prob. 27ECh. 8.1 - Prob. 28ECh. 8.1 - Prob. 29ECh. 8.1 - Prob. 30ECh. 8.1 - Prob. 31ECh. 8.1 - Prob. 32ECh. 8.1 - Prob. 33ECh. 8.1 - Prob. 34ECh. 8.1 - Prob. 35ECh. 8.1 - Prob. 36ECh. 8.1 - Prob. 37ECh. 8.1 - Prob. 38ECh. 8.1 - Prob. 39ECh. 8.1 - Prob. 40ECh. 8.1 - Prob. 41ECh. 8.1 - Prob. 42ECh. 8.1 - Prob. 43ECh. 8.1 - Prob. 44ECh. 8.1 - Prob. 45ECh. 8.1 - Prob. 46ECh. 8.2 - Prob. 47UYKCh. 8.2 - Prob. 48UYKCh. 8.2 - Prob. 49UYKCh. 8.2 - Prob. 50UYKCh. 8.2 - Prob. 51UYKCh. 8.2 - Prob. 52UYKCh. 8.2 - Prob. 53UYKCh. 8.2 - Prob. 54UYKCh. 8.2 - Prob. 55UYKCh. 8.2 - Prob. 56ECh. 8.2 - Prob. 57ECh. 8.2 - Prob. 58ECh. 8.2 - Prob. 59ECh. 8.2 - Prob. 60ECh. 8.2 - Prob. 61ECh. 8.2 - Prob. 62ECh. 8.2 - Prob. 63ECh. 8.2 - Prob. 64ECh. 8.2 - Prob. 65ECh. 8.2 - Prob. 66ECh. 8.2 - Prob. 67ECh. 8.2 - Prob. 68ECh. 8.2 - Prob. 69ECh. 8.2 - Prob. 70ECh. 8.2 - Prob. 71ECh. 8.2 - Prob. 72ECh. 8.2 - Prob. 73ECh. 8 - Prob. 74ECh. 8 - Prob. 75ECh. 8 - Prob. 76ECh. 8 - Prob. 77ECh. 8 - Prob. 78ECh. 8 - Prob. 79ECh. 8 - Prob. 80ECh. 8 - Prob. 81ECh. 8 - Prob. 82ECh. 8 - Prob. 83ECh. 8 - Prob. 84ECh. 8 - Prob. 85ECh. 8 - Prob. 86ECh. 8 - Prob. 87ECh. 8 - Prob. 88ECh. 8 - Prob. 89ECh. 8 - Prob. 90ECh. 8 - Prob. 91ECh. 8 - Prob. 92ECh. 8 - Prob. 93ECh. 8 - Prob. 94ECh. 8 - Prob. 95ECh. 8 - Prob. 96ECh. 8 - Prob. 97E
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