Question

A)Assume that a sample is used to estimate a population proportion p. Find the 99.9% confidence interval for a sample of size 177 with 10% successes. Enter your answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places.

< p <

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

	A	B	C	Total
Male	17	10	19	46
Female	13	9	2	24
Total	30	19	21	70

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

< p <

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

	A	B	C	Total
Male	10	7	14	31
Female	13	5	12	30
Total	23	12	26	61

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

Enter your answer as a tri-linear inequality using decimals (not percents).

< p <

State the point estimate for the proportion of all male students who would receive a grade of B on this test:
p≈p≈

State the margin of error for the proportion of all male students who would receive a grade of B on this test:
EBP ≈≈

D) We wish to estimate what percent of adult residents in Ventura County like chocolate. Out of 500 adult residents sampled, 393 like chocolate. Based on this, construct a 95% confidence interval for the proportion (p) of adult residents who like chocolate in Ventura County.

1. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places.

   Confidence interval =

2. 
   
3. Express the same answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places.

   < p <

4. 
   
5. Express the same answer using the point estimate and margin of error. Give your answers as decimals, to three places.

   p =  ±±

Answer 1

A)

Level of Significance,   α =    0.001   
Sample Size,   n =    177          
                  
Sample Proportion ,    p̂ = x/n =    0.1000          
z -value =   Zα/2 =    3.291   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0225          
margin of error , E = Z*SE =    3.291   *   0.0225   =   0.0742
                  
99.9%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.100   -   0.0742   =   0.026
Interval Upper Limit = p̂ + E =   0.100   +   0.0742   =   0.174
                  
99.9%   confidence interval is (   0.026 < p <    0.174 )

B)

Level of Significance,   α =    0.02          
Number of Items of Interest,   x =   10          
Sample Size,   n =    46          
                  
Sample Proportion ,    p̂ = x/n =    0.2174          
z -value =   Zα/2 =    2.326   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0608          
margin of error , E = Z*SE =    2.326   *   0.0608   =   0.1415
                  
98%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.217   -   0.1415   =   0.076
Interval Upper Limit = p̂ + E =   0.217   +   0.1415   =   0.359
                  
98%   confidence interval is (   0.076 < p <    0.359 )

C)

point estimate=0.2174

margin of error , E = Z*SE =    2.326   *   0.0608   =   0.1415