Question

A student conducts a simple random sample of students from her high school and finds that 21 out of 100 students in her sample regularly walk to school. Give a point estimate for the proportion of all students at her high school who regularly walk to school. For each combination of sample size and sample proportion, find the approximate margin of error for the 95% confidence level. (Round the answers to three decimal places.) In a sample of 16 students, 15 were right-handed. Can we construct a 95% confidence interval for the proportion of all students who are right-handed? A study reported that 49% of Internet users have searched for information about themselves online. The 49% figure was based on a random sample of Internet users. Suppose that the sample size was n = 400. In a randomly selected sample of 400 registered voters in a community, 260 individuals say that they plan to vote for Candidate Y in the upcoming election. A certain set of plants were constantly dying in the dry environment that was provided. The plants were moved to a more humid environment where life would improve.

Answer 1

Level of Significance,   α =    0.05          
Number of Items of Interest,   x =   21          
Sample Size,   n =    100          
                  
Sample Proportion ,    p̂ = x/n =    0.210          
z -value =   Zα/2 =    1.960   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.04          
margin of error , E = Z*SE =    1.960   *   0.0407   =   0.08

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Level of Significance,   α =    0.05          
Number of Items of Interest,   x =   15          
Sample Size,   n =    16          
                  
Sample Proportion ,    p̂ = x/n =    0.938          
z -value =   Zα/2 =    1.960   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.06          
margin of error , E = Z*SE =    1.960   *   0.0605   =   0.12
                  
95%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.938   -   0.1186   =   0.8189
Interval Upper Limit = p̂ + E =   0.938   +   0.1186   =   1.0561
                  
95%   confidence interval is (   0.82   < p <    1.06   )

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