The 90 % confidence interval for the difference μ 1 − μ 2 .

Question

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Answer 1

Question

Chapter 10, Problem 14CQ

To determine

To find: The 90% confidence interval for the difference μ1−μ2 .

Expert Solution & Answer

Answer to Problem 14CQ

The 90% confidence interval for the difference μ1−μ2 is (1832.7054,3253.2946) .

Explanation of Solution

Given information:

The sample is 76 , the mean annual income is $34214 and the standard deviation is $2171 . The sample in neighboring town is 88 , the mean income is $31671 with standard of deviation of $3279 .

Concept used:

Excel formula is used.

Calculation:

The point estimate for the difference is,

μ1−μ2=x1¯−x2¯=34214−31671=$2543

The formula for degree of freedom is,

dof= ( s 1 2 n 1 + s 2 2 n 2 )2 ( s 1 2 n 1 ) 2 n 1 −1+ ( s 2 2 n 2 ) 2 n 2 −1= ( ( 2171 ) 2 76 + ( 3279 ) 2 88 )2 ( ( 2171 ) 2 76 ) 2 76−1+ ( ( 3279 ) 2 88 ) 2 76−1=152.2363=152

Assume level of significance as 0.10 .

For 152 degree of freedom and 0.10 level of significance, the critical value is,

tc=[=TINV(probability,deg_freedom)]=[=TINV(0.1,152)]=1.6549=1.655

The formula for margin of error is,

E=tc s 1 2 n 1 + s 2 2 n 2 =1.655 ( 2171 ) 2 76+ ( 3279 ) 2 88=710.294573=710.3

The confidence interval is given as,

CI=( x 1 ¯− x 2 ¯)±tc s 1 2 n 1 + s 2 2 n 2 =( x 1 ¯− x 2 ¯)±E=2543±710.294573=1832.7054,3253.2946

Therefore, the 90% confidence interval for the difference μ1−μ2 is (1832.7054,3253.2946) .

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Answer 2

Question

Chapter 10, Problem 14CQ

To determine

To find: The 90% confidence interval for the difference μ1−μ2 .

Expert Solution & Answer

Answer to Problem 14CQ

The 90% confidence interval for the difference μ1−μ2 is (1832.7054,3253.2946) .

Explanation of Solution

Given information:

The sample is 76 , the mean annual income is $34214 and the standard deviation is $2171 . The sample in neighboring town is 88 , the mean income is $31671 with standard of deviation of $3279 .

Concept used:

Excel formula is used.

Calculation:

The point estimate for the difference is,

μ1−μ2=x1¯−x2¯=34214−31671=$2543

The formula for degree of freedom is,

dof= ( s 1 2 n 1 + s 2 2 n 2 )2 ( s 1 2 n 1 ) 2 n 1 −1+ ( s 2 2 n 2 ) 2 n 2 −1= ( ( 2171 ) 2 76 + ( 3279 ) 2 88 )2 ( ( 2171 ) 2 76 ) 2 76−1+ ( ( 3279 ) 2 88 ) 2 76−1=152.2363=152

Assume level of significance as 0.10 .

For 152 degree of freedom and 0.10 level of significance, the critical value is,

tc=[=TINV(probability,deg_freedom)]=[=TINV(0.1,152)]=1.6549=1.655

The formula for margin of error is,

E=tc s 1 2 n 1 + s 2 2 n 2 =1.655 ( 2171 ) 2 76+ ( 3279 ) 2 88=710.294573=710.3

The confidence interval is given as,

CI=( x 1 ¯− x 2 ¯)±tc s 1 2 n 1 + s 2 2 n 2 =( x 1 ¯− x 2 ¯)±E=2543±710.294573=1832.7054,3253.2946

Therefore, the 90% confidence interval for the difference μ1−μ2 is (1832.7054,3253.2946) .

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Answer 3

Question

Chapter 10, Problem 14CQ

To determine

To find: The 90% confidence interval for the difference μ1−μ2 .

Expert Solution & Answer

Answer to Problem 14CQ

The 90% confidence interval for the difference μ1−μ2 is (1832.7054,3253.2946) .

Explanation of Solution

Given information:

The sample is 76 , the mean annual income is $34214 and the standard deviation is $2171 . The sample in neighboring town is 88 , the mean income is $31671 with standard of deviation of $3279 .

Concept used:

Excel formula is used.

Calculation:

The point estimate for the difference is,

μ1−μ2=x1¯−x2¯=34214−31671=$2543

The formula for degree of freedom is,

dof= ( s 1 2 n 1 + s 2 2 n 2 )2 ( s 1 2 n 1 ) 2 n 1 −1+ ( s 2 2 n 2 ) 2 n 2 −1= ( ( 2171 ) 2 76 + ( 3279 ) 2 88 )2 ( ( 2171 ) 2 76 ) 2 76−1+ ( ( 3279 ) 2 88 ) 2 76−1=152.2363=152

Assume level of significance as 0.10 .

For 152 degree of freedom and 0.10 level of significance, the critical value is,

tc=[=TINV(probability,deg_freedom)]=[=TINV(0.1,152)]=1.6549=1.655

The formula for margin of error is,

E=tc s 1 2 n 1 + s 2 2 n 2 =1.655 ( 2171 ) 2 76+ ( 3279 ) 2 88=710.294573=710.3

The confidence interval is given as,

CI=( x 1 ¯− x 2 ¯)±tc s 1 2 n 1 + s 2 2 n 2 =( x 1 ¯− x 2 ¯)±E=2543±710.294573=1832.7054,3253.2946

Therefore, the 90% confidence interval for the difference μ1−μ2 is (1832.7054,3253.2946) .

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Answer 4

Question

Chapter 10, Problem 14CQ

To determine

To find: The 90% confidence interval for the difference μ1−μ2 .

Expert Solution & Answer

Answer to Problem 14CQ

The 90% confidence interval for the difference μ1−μ2 is (1832.7054,3253.2946) .

Explanation of Solution

Given information:

The sample is 76 , the mean annual income is $34214 and the standard deviation is $2171 . The sample in neighboring town is 88 , the mean income is $31671 with standard of deviation of $3279 .

Concept used:

Excel formula is used.

Calculation:

The point estimate for the difference is,

μ1−μ2=x1¯−x2¯=34214−31671=$2543

The formula for degree of freedom is,

dof= ( s 1 2 n 1 + s 2 2 n 2 )2 ( s 1 2 n 1 ) 2 n 1 −1+ ( s 2 2 n 2 ) 2 n 2 −1= ( ( 2171 ) 2 76 + ( 3279 ) 2 88 )2 ( ( 2171 ) 2 76 ) 2 76−1+ ( ( 3279 ) 2 88 ) 2 76−1=152.2363=152

Assume level of significance as 0.10 .

For 152 degree of freedom and 0.10 level of significance, the critical value is,

tc=[=TINV(probability,deg_freedom)]=[=TINV(0.1,152)]=1.6549=1.655

The formula for margin of error is,

E=tc s 1 2 n 1 + s 2 2 n 2 =1.655 ( 2171 ) 2 76+ ( 3279 ) 2 88=710.294573=710.3

The confidence interval is given as,

CI=( x 1 ¯− x 2 ¯)±tc s 1 2 n 1 + s 2 2 n 2 =( x 1 ¯− x 2 ¯)±E=2543±710.294573=1832.7054,3253.2946

Therefore, the 90% confidence interval for the difference μ1−μ2 is (1832.7054,3253.2946) .

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Answer 5

Chapter 10, Problem 14CQ

To determine

To find: The 90% confidence interval for the difference μ1−μ2 .

Expert Solution & Answer

Answer to Problem 14CQ

The 90% confidence interval for the difference μ1−μ2 is (1832.7054,3253.2946) .

Explanation of Solution

Given information:

The sample is 76 , the mean annual income is $34214 and the standard deviation is $2171 . The sample in neighboring town is 88 , the mean income is $31671 with standard of deviation of $3279 .

Concept used:

Excel formula is used.

Calculation:

The point estimate for the difference is,

μ1−μ2=x1¯−x2¯=34214−31671=$2543

The formula for degree of freedom is,

dof= ( s 1 2 n 1 + s 2 2 n 2 )2 ( s 1 2 n 1 ) 2 n 1 −1+ ( s 2 2 n 2 ) 2 n 2 −1= ( ( 2171 ) 2 76 + ( 3279 ) 2 88 )2 ( ( 2171 ) 2 76 ) 2 76−1+ ( ( 3279 ) 2 88 ) 2 76−1=152.2363=152

Assume level of significance as 0.10 .

For 152 degree of freedom and 0.10 level of significance, the critical value is,

tc=[=TINV(probability,deg_freedom)]=[=TINV(0.1,152)]=1.6549=1.655

The formula for margin of error is,

E=tc s 1 2 n 1 + s 2 2 n 2 =1.655 ( 2171 ) 2 76+ ( 3279 ) 2 88=710.294573=710.3

The confidence interval is given as,

CI=( x 1 ¯− x 2 ¯)±tc s 1 2 n 1 + s 2 2 n 2 =( x 1 ¯− x 2 ¯)±E=2543±710.294573=1832.7054,3253.2946

Therefore, the 90% confidence interval for the difference μ1−μ2 is (1832.7054,3253.2946) .

Answer 6

Chapter 10, Problem 14CQ

To determine

To find: The 90% confidence interval for the difference μ1−μ2 .

Expert Solution & Answer

Answer to Problem 14CQ

The 90% confidence interval for the difference μ1−μ2 is (1832.7054,3253.2946) .

Explanation of Solution

Given information:

The sample is 76 , the mean annual income is $34214 and the standard deviation is $2171 . The sample in neighboring town is 88 , the mean income is $31671 with standard of deviation of $3279 .

Concept used:

Excel formula is used.

Calculation:

The point estimate for the difference is,

μ1−μ2=x1¯−x2¯=34214−31671=$2543

The formula for degree of freedom is,

dof= ( s 1 2 n 1 + s 2 2 n 2 )2 ( s 1 2 n 1 ) 2 n 1 −1+ ( s 2 2 n 2 ) 2 n 2 −1= ( ( 2171 ) 2 76 + ( 3279 ) 2 88 )2 ( ( 2171 ) 2 76 ) 2 76−1+ ( ( 3279 ) 2 88 ) 2 76−1=152.2363=152

Assume level of significance as 0.10 .

For 152 degree of freedom and 0.10 level of significance, the critical value is,

tc=[=TINV(probability,deg_freedom)]=[=TINV(0.1,152)]=1.6549=1.655

The formula for margin of error is,

E=tc s 1 2 n 1 + s 2 2 n 2 =1.655 ( 2171 ) 2 76+ ( 3279 ) 2 88=710.294573=710.3

The confidence interval is given as,

CI=( x 1 ¯− x 2 ¯)±tc s 1 2 n 1 + s 2 2 n 2 =( x 1 ¯− x 2 ¯)±E=2543±710.294573=1832.7054,3253.2946

Therefore, the 90% confidence interval for the difference μ1−μ2 is (1832.7054,3253.2946) .

Answer 7

Chapter 10, Problem 14CQ

To determine

To find: The 90% confidence interval for the difference μ1−μ2 .

Answer 8

Expert Solution & Answer

Answer to Problem 14CQ

The 90% confidence interval for the difference μ1−μ2 is (1832.7054,3253.2946) .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given information:

The sample is 76 , the mean annual income is $34214 and the standard deviation is $2171 . The sample in neighboring town is 88 , the mean income is $31671 with standard of deviation of $3279 .

Concept used:

Excel formula is used.

Calculation:

The point estimate for the difference is,

μ1−μ2=x1¯−x2¯=34214−31671=$2543

The formula for degree of freedom is,

dof= ( s 1 2 n 1 + s 2 2 n 2 )2 ( s 1 2 n 1 ) 2 n 1 −1+ ( s 2 2 n 2 ) 2 n 2 −1= ( ( 2171 ) 2 76 + ( 3279 ) 2 88 )2 ( ( 2171 ) 2 76 ) 2 76−1+ ( ( 3279 ) 2 88 ) 2 76−1=152.2363=152