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Reika [66]
1 year ago
13

Type the correct answer in each box. use numerals instead of words. in a survey conducted by a retail store, 58% of the sample r

espondents said they prefer to shop at places with loyalty cards. if the margin of error is 3.4%, the expected population proportion that prefers to shop at places with loyalty cards is between % and %.
Mathematics
1 answer:
solmaris [256]1 year ago
3 0

The required expected population proportion that prefers to shop at places with loyalty cards is between 54.6% and 61.4%.

Given That,
In a survey conducted by a retail store, 58% of the sample respondents said they prefer to shop at places with loyalty cards. if the margin of error is 3.4%.

<h3>What is population proportion?</h3>

A population proportion is defined as the percentage of a population that belongs to a individual category. Certainty gaps are used to evaluate population proportions.

Here, the margin of error is 3.4%.
expected population proportion = 58% ± 3.4%
expected population proportion interval = (58% - 3.4%, 58% + 3.4%)
expected population proportion interval = (54.6%, 61.4%).


Thus, The required expected population proportion that prefers to shop at places with loyalty cards is between 54.6% and 61.4%.

Learn more about population proportion here.
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The 2003 Statistical Abstract of the United States reported the percentage of people years of age and older who smoke. Suppose t
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Complete Question

The 2003 Statistical Abstract of the United States reported the percentage of people 18 years of age and older who smoke. Suppose that a study designed to collect new data on smokers and non smokers uses a preliminary estimate of the proportion who smoke of 0.30.

a. How large a sample should be taken to estimate the proportion of smokers in the population with a margin of error of 0.02? use 95% confidence.

b. Assume that the study uses your sample size recommendation in part (a) and finds 520 smokers. What is the point estimate of the proportion of smokers in the population?

c. What is the 95% confidence interval for the proportion of smokers in the population?

Answer:

a

n = 2017

b

\^ p = 0.2578

c

0.238  <  0.278

Step-by-step explanation:

Considering question a

From the question we are told that

    The margin of error is  E = 0.02

     The preliminary estimate of the proportion who smoke is  p = 0.30

From the question we are told the confidence level is  95% , hence the level of significance is    

      \alpha = (100 - 95 ) \%

=>   \alpha = 0.05

Generally from the normal distribution table the critical value  of  \frac{\alpha }{2} is  

   Z_{\frac{\alpha }{2} } =  1.96

Generally the sample size is mathematically  represented as  

      n = [ \frac{Z_{\frac{\alpha }{2} }  }{E} ] ^2  *  p(1 - p )

=>   n = [ \frac{1.96  }{0.02} ] ^2  *  0.30(1 -  0.30  )

=>   n = 2017

Considering question b

 From the question we are told that

   The number of smokers  is  k =  520  

Generally the  point estimate of the proportion of smokers in the population is mathematically represented as

        \^ p =  \frac{520 }{2017 }

=>    \^ p = 0.2578

Considering  question c

Generally 95% confidence interval is mathematically represented as  

      \^ p -E <  p <  \^ p +E

=>   0.2578 - 0.02 <  0.2578 + 0.02

=>   0.238  <  0.278

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To find out how much it is in cups, we must know how many cups 1 quart is equal to. 

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