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Direct mail advertisers send solicitations​ ("junk mail") to thousands of potential customers in the hope that some will buy the​ company's product. The response rate is usually quite low. Suppose a company wants to test the response to a new flyer and sends it to 1140 people randomly selected from their mailing list of over​ 200,000 people. They get orders from 122 of the recipients. Use this information to complete parts a through d.Create a 95% confidence interval for the percentage of people the company contacts who may buy something. (Show your work. Step by step)

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JeanaShupp

Answer: (0.089, 0.125)

Step-by-step explanation:

Confidence interval for population proportion is given by :-

[tex]\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]

, where n= sample size.

[tex]\hat{p}[/tex] = Sample proportion.

z*= Critical z-value.

Let p be the population proportion of people the company contacts who may buy something.

As per given , sample size : n= 1140

Number of recipients ordered = 122

Then, [tex]\hat{p}=\dfrac{122}{1140}\approx0.107[/tex]

Critical value for 95% confidence interval = z*= 1.96 (By z-table)

So , the 95% confidence interval for the percentage of people the company contacts who may buy something:

[tex]0.107\pm (1.96)\sqrt{\dfrac{0.107(1-0.107)}{1140}}[/tex]

[tex]=0.107\pm (1.96)\sqrt{0.000083817}[/tex]

[tex]=0.107\pm (1.96)(0.00915516)[/tex]

[tex]=0.107\pm 0.018[/tex]

[tex]=(0.107-0.018,\ 0.107+0.018)=(0.089,\ 0.125)[/tex]

Hence, the 95% confidence interval for the percentage of people the company contacts who may buy something = (0.089, 0.125)

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