Question

Confidence Intervals (Proportions)

1

Out of 400 people sampled, 312 preferred Candidate A. Based on this, estimate what proportion of the entire voting population (p) prefers Candidate A.

Use a 99% confidence level, and give your answers as decimals, to three places.

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2 Assume that a sample is used to estimate a population proportion p. Find the 98% confidence interval for a sample of size 348 with 207 successes. Enter your answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places.

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3 Out of 500 people sampled, 260 preferred Candidate A. Based on this, estimate what proportion of the voting population (p) prefers Candidate A.

Use a 99% confidence level, and give your answers as decimals, to three places.

< p

4 Giving a test to a group of students, the grades and gender are summarized below

	A	B	C	Total
Male	9	6	2	17
Female	16	13	12	41
Total	25	19	14	58

Let p represent the proportion of all female students who would receive a grade of B on this test. Use a 98% confidence interval to estimate p to three decimal places.

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Answer 1

1)

Level of Significance,   α =    0.01          
Number of Items of Interest,   x =   312          
Sample Size,   n =    400          
                  
Sample Proportion ,    p̂ = x/n =    0.780          
z -value =   Zα/2 =    2.576   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0207          
margin of error , E = Z*SE =    2.576   *   0.0207   =   0.0534
                  
99%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.780   -   0.0534   =   0.7266
Interval Upper Limit = p̂ + E =   0.780   +   0.0534   =   0.8334
                  
99%   confidence interval is (   0.727 < p <    0.833 )

2)

Level of Significance,   α =    0.02          
Number of Items of Interest,   x =   207          
Sample Size,   n =    348          
                  
Sample Proportion ,    p̂ = x/n =    0.595          
z -value =   Zα/2 =    2.326   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0263          
margin of error , E = Z*SE =    2.326   *   0.0263   =   0.0612
                  
98%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.595   -   0.0612   =   0.5336
Interval Upper Limit = p̂ + E =   0.595   +   0.0612   =   0.6560
                  
98%   confidence interval is (   0.534 < p <    0.656 )

3)

Level of Significance,   α =    0.01          
Number of Items of Interest,   x =   260          
Sample Size,   n =    500          
                  
Sample Proportion ,    p̂ = x/n =    0.520          
z -value =   Zα/2 =    2.576   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0223          
margin of error , E = Z*SE =    2.576   *   0.0223   =   0.0576
                  
99%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.520   -   0.0576   =   0.4624
Interval Upper Limit = p̂ + E =   0.520   +   0.0576   =   0.5776
                  
99%   confidence interval is (   0.462 < p <    0.578 )

4)

Level of Significance,   α =    0.02          
Number of Items of Interest,   x =   13          
Sample Size,   n =    58          
                  
Sample Proportion ,    p̂ = x/n =    0.224          
z -value =   Zα/2 =    2.326   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0548          
margin of error , E = Z*SE =    2.326   *   0.0548   =   0.1274
                  
98%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.224   -   0.1274   =   0.0968
Interval Upper Limit = p̂ + E =   0.224   +   0.1274   =   0.3515
                  
98%   confidence interval is (   0.097 < p <    0.352 )

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