Question

In: Statistics and Probability

A random sample of 42 taxpayers claimed an average of ​$9,857 in medical expenses for the...

A random sample of 42 taxpayers claimed an average of ​$9,857 in medical expenses for the year. Assume the population standard deviation for these deductions was ​$2,409. Construct confidence intervals to estimate the average deduction for the population with the levels of significance shown below.

a)1%

b)2%

c)20%

Solutions

Expert Solution

1)

99% Confidence Interval :-   
X̅ ± Z( α /2) σ / √ ( n )  
Z(α/2) = Z (0.01 /2) = 2.576  
9857 ± 2.576 * 2409/√(42)  
Lower Limit = 9857 - 2.576 * 2409/√(42)  
Lower Limit = 8899.46
Upper Limit = 9857 + 2.576 * 2409/√(42)  
Upper Limit = 10814.54
99% Confidence interval is ( 8899.46 , 10814.54 )   

2)

98% Confidence Interval :-
X̅ ± Z( α /2) σ / √ ( n )
Z(α/2) = Z (0.02 /2) = 2.326
9857 ± 2.326* 2409/√(42)
Lower Limit = 9857 - 2.326 * 2409/√(42)
Lower Limit = 8992.39
Upper Limit = 9857 + 2.326 * 2409/√(42)
Upper Limit = 10721.61
98% Confidence interval is ( 8992.39 , 10721.61)

3)

Confidence Interval :-
X̅ ± Z( α /2) σ / √ ( n )
Z(α/2) = Z (0.2 /2) = 1.282
9857 ± 1.282 * 2409/√(42)
Lower Limit = 9857 - 1.282 * 2409/√(42)
Lower Limit = 9380.46
Upper Limit = 9857 + 1.282 * 2409/√(42)
Upper Limit = 10333.54
80% Confidence interval is ( 9380.46 , 10333.54 )


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