Question

In: Statistics and Probability

An apple orchard farmer wants to know the true population proportion of all apples harvested at...

An apple orchard farmer wants to know the true population proportion of all apples harvested at their farm which have the less valuable condition of being blemished, or in other words, are cider apples.

This true population proportion is a parameter value represented by the symbol p. As a population parameter, this value can never be known with certainty. It can however, be estimated with varying levels of confidence.

To estimate this parameter p, the farmer draws a simple random sample (SRS) of 524 apples from the orchard. Within this sample, the farmer finds that exactly 47 of the apples are cider apples.

The farmer then computes a 96%-level confidence interval estimate for the parameter p.

In percentage form, and rounded to four digits past the decimal point: What is the approximate value of the Lower Limit of this confidence interval?

Include a percentage symbol at the end of your numerical answer (with no spaces).

In percentage form, and rounded to four digits past the decimal point: What is the approximate value of the Upper Limit of this confidence interval?

Include a percentage symbol at the end of your numerical answer (with no spaces).

In percentage form, and rounded to four digits past the decimal point: What is the approximate value of the Margin of Error for this confidence interval?

Include a percentage symbol at the end of your numerical answer (with no spaces).

Suppose the farmer now wishes that they had instead built a confidence interval with a 99% confidence level, and a margin of error no greater than 2%.

To achieve these new specifications: The farmer intends to draw another SRS of apples in the future, and build another confidence interval.

What minimum sample size of apples will the farmer have to draw, in order to achieve these new confidence interval specifications?

(Note: Your answer should be a whole number here.)

Solutions

Expert Solution

Level of Significance,   α =    0.04          
Number of Items of Interest,   x =   47          
Sample Size,   n =    524          
                  
Sample Proportion ,    p̂ = x/n =    0.090          
z -value =   Zα/2 =    2.054   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0125          
margin of error , E = Z*SE =    2.054   *   0.0125   =   0.0256

= 2.56%
                  
96%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.090   -   0.0256   =   0.0641

6.41%


Interval Upper Limit = p̂ + E =   0.090   +   0.0256   =   0.1153

= 11.53%

--------------------------------

sample proportion ,   p̂ =    0.089694656                          
sampling error ,    E =   0.02                          
Confidence Level ,   CL=   0.99                          
                                  
alpha =   1-CL =   0.01                          
Z value =    Zα/2 =    2.576   [excel formula =normsinv(α/2)]                      
                                  
Sample Size,n = (Z / E)² * p̂ * (1-p̂) = (   2.576   /   0.02   ) ² *   0.09   * ( 1 -   0.09   ) =    1354.3
                                  
                                  
so,Sample Size required=       1355             

Thanks in advance!

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