Question

Harris Interactive® conducted a poll of American adults in August of 2011 to study the use of online medical information. Of the 1,019 randomly chosen adults, 60% had used the Internet within the past month to obtain medical information. Use the results of this survey to create an approximate 95% confidence interval estimate for the percentage of all American adults who have used the Internet to obtain medical information in the past month. Calculate the 90% confidence interval. Which confidence interval has the smaller margin of error? Why does this make sense? Which confidence interval is more likely to contain the true percentage of all American adults who have used the Internet to obtain medical information in the past month? Why do you think so?

Answer 1

Answer:

The 95% confidence interval $0.57 < p <0.63$

The 90% confidence interval  $0.5748 < p <0.6252$

The 90% confidence interval has the smaller margin of error.

This makes sense because the 90% confidence interval is narrower and more precise(i.e a decreased error band  ) than the 95% confidence interval

95% confidence interval is more likely to contain the true percentage

This is because 95% confidence interval signify's a higher confidence  of containing the true percentage than the 90% confidence interval  

Step-by-step explanation:

From the question we are told that

   The sample size is  n =  1019

    The  sample proportion is  $\r p = 0.60$

Given from the question that the confidence level is 95% then the level of significance is mathematically represented as

         $\alpha = (100 - 95)\%$

        $\alpha = 0.05$

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

    $Z_{\frac{\alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

         $E = Z_{\frac{\alpha }{2} } * \sqrt{\frac{\r p (1 - \r p )}{n} }$

=>     $E = 1.96 * \sqrt{\frac{0.60 (1 - 0.60 )}{1019} }$

=>    $E = 0.030$

Generally the 95% confidence interval is mathematically represented as

         $\r p - E < p < \r p +E$

        $0.60 - 0.030 < p <0.60 + 0.030$

        $0.57 < p <0.63$

This interval means that there is 95% chance that the true percentage will be in this interval

Given from the question that the confidence level is 90% then the level of significance is mathematically represented as

         $\alpha = (100 - 90)\%$

        $\alpha = 0.10$

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

    $Z_{\frac{\alpha }{2} } = 1.645$

Generally the margin of error is mathematically represented as

         $E = Z_{\frac{\alpha }{2} } * \sqrt{\frac{\r p (1 - \r p )}{n} }$

=>     $E = 1.645 * \sqrt{\frac{0.60 (1 - 0.60 )}{1019} }$

=>    $E = 0.0252$

Generally the 90% confidence interval is mathematically represented as

         $\r p - E < p < \r p +E$

        $0.60 - 0.0252 < p <0.60 + 0.0252$

        $0.5748 < p <0.6252$

This interval means that there is 90% chance that the true percentage will be in this interval

From the calculation the 90% confidence interval has the smaller margin of error.

This makes sense because the 90% confidence interval is narrower and more precise(i.e a decreased error band  ) than the 95% confidence interval

Generally the confidence interval that is more likely to contain the true percentage  of all American adults who have used the Internet to obtain medical information in the past month is the 95% confidence interval ,

This is because it signify's a higher chance of containing the true percentage than the 90% confidence