The National Assessment of Educational Progress administers standardized achievement tests to nationwide samples of 17-year-olds in school. One year, the tests covered history and literature. You may assume that a simple random sample of size 6,000 was taken. Only 36.1% of the students in the sample knew that Chaucer wrote The Canterbury Tales, but 95.2% knew that Edison invented the light bulb.11 (a) If possible, find a 95%-confidence interval for the percentage of all 17-year-olds in school who knew that Chaucer wrote The Canterbury Tales. If this is not possible, why not? (b) If possible, find a 95%-confidence interval for the percentage of all 17- year-olds in school who knew that Edison invented the light bulb. If this is not possible, why not?

Answer:a) Is possible see the conditions below.And the 95% confidence interval would be given (0.349;0.373). b) Is possible see the conditions below.And the 95% confidence interval would be given (0.947;0.957). Step-by-step explanation:A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  The margin of error is the range of values below and above the sample statistic in a confidence interval.  Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  The population proportion have the following distribution [tex]p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})[/tex] The confidence interval would be given by this formula [tex]\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex] For the 95% confidence interval the value of [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2=0.025[/tex], with that value we can find the quantile required for the interval in the normal standard distribution. [tex]z_{\alpha/2}=1.96[/tex](a) If possible, find a 95%-confidence interval for the percentage of all 17-year-olds in school who knew that Chaucer wrote The Canterbury Tales. If this is not possible, why not?np=6000*0.361=2166>10n(1-p)=6000(1-0.361)=3834>10Conditions satisfied. And replacing into the confidence interval formula we got: [tex]0.361 - 1.96 \sqrt{\frac{0.361(1-0.361)}{6000}}=0.349[/tex] [tex]0.361 + 1.96 \sqrt{\frac{0.361(1-0.361)}{6000}}=0.373[/tex] And the 95% confidence interval would be given (0.349;0.373). We are confident a 95% that about 34.9% to 37.3% of all 17-year-olds in school who knew that Chaucer wrote The Canterbury Tales(b) If possible, find a 95%-confidence interval for the percentage of all 17- year-olds in school who knew that Edison invented the light bulb. If this is not possible, why not? np=6000*0.952=5712>10n(1-p)=6000(1-0.952)=288>10Conditions satisfied. And replacing into the confidence interval formula we got: [tex]0.952 - 1.96 \sqrt{\frac{0.952(1-0.952)}{6000}}=0.947[/tex] [tex]0.952 + 1.96 \sqrt{\frac{0.952(1-0.952)}{6000}}=0.957[/tex] And the 95% confidence interval would be given (0.947;0.957). We are confident a 95% that about 94.7% to 95.7% of all 17- year-olds in school who knew that Edison invented the light bulb