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Chapter 7: Q23E (page 294)

The Pew Forum on Religion and Public Life reported on \({\rm{Dec}}{\rm{. 9, 2009}}\), that in a survey of \({\rm{2003}}\) American adults, \({\rm{25\% }}\)said they believed in astrology.

a. Calculate and interpret a confidence interval at the \({\rm{99\% }}\)confidence level for the proportion of all adult Americans who believe in astrology.

b. What sample size would be required for the width of a \({\rm{99\% }}\)CI to be at most .05 irrespective of the value of p?

Short Answer

Expert verified

a) The boundaries of the confidence interval are then\((0.2251,0.2749)\)

b) The sample size is \({\rm{n = 2653}}\)

Step by step solution

01

To Calculate and interpret a confidence interval

Given:

\(\begin{array}{l}{\rm{n = 2003}}\\{\rm{\hat p = 25\% = 0}}{\rm{.25}}\\{\rm{c = 99\% = 0}}{\rm{.99}}\end{array}\)

(a) For confidence level \({\rm{1 - \alpha - 0}}{\rm{.99}}\),determine \({{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ - }}{{\rm{z}}_{{\rm{0}}{\rm{.005}}}}\)

\({{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ - 2}}{\rm{.575}}\)

Note: We take the average of \({\rm{2}}{\rm{.57 and 2}}{\rm{.58}}\)because \(0.005\)lies exactly in the middle between \({\rm{0}}{\rm{.0049 and 0}}{\rm{.0051}}\)

The margin of error is then:

\({\rm{E - }}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\sqrt {\frac{{{\rm{\hat p(1 - \hat p)}}}}{{\rm{n}}}} {\rm{ - 2}}{\rm{.575 \times }}\sqrt {\frac{{{\rm{0}}{\rm{.25(1 - 0}}{\rm{.25)}}}}{{{\rm{2003}}}}} {\rm{\gg 0}}{\rm{.0249}}\)

The boundaries of the confidence interval are then:

\(\begin{array}{l}{\rm{\hat p - E - 0}}{\rm{.25 - 0}}{\rm{.0249 - 0}}{\rm{.2251}}\\{\rm{\hat p + E - 0}}{\rm{.25 + 0}}{\rm{.0249 - 0}}{\rm{.2749}}\end{array}\)

Hence The boundaries of the confidence interval are then\((0.2251,0.2749)\)

02

To find the sample size

(b) Given:

\({\rm{ Width - 0}}{\rm{.05}}\)

The margin of error E is half the width of the confidence interval :

\({\rm{E - }}\frac{{{\rm{ Width }}}}{{\rm{2}}}{\rm{ - }}\frac{{{\rm{0}}{\rm{.05}}}}{{\rm{2}}}{\rm{ - 0}}{\rm{.025}}\)

Formula sample size:

\({\rm{\hat p known: n - }}\frac{{{{\left( {{{\rm{z}}_{{\rm{\alpha /2}}}}} \right)}^{\rm{2}}}{\rm{\hat p\hat q}}}}{{{{\rm{E}}^{\rm{2}}}}}{\rm{ - }}\frac{{{{\left( {{{\rm{z}}_{{\rm{\alpha /2}}}}} \right)}^{\rm{2}}}{\rm{\hat p(1 - \hat p)}}}}{{{{\rm{E}}^{\rm{2}}}}}\)

\({\rm{\hat p}}\)unknown: \({\rm{n - }}\frac{{\left( {{{\left. {{{\rm{z}}_{{\rm{\alpha /2}}}}} \right|}^{\rm{2}}}{\rm{0}}{\rm{.25}}} \right.}}{{{{\rm{E}}^{\rm{2}}}}}\)

For confidence level. \({\rm{1 - \alpha - 0}}{\rm{.99}}\), determine \({{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ - }}{{\rm{z}}_{{\rm{0}}{\rm{.005}}}}\):

\({{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ - 2}}{\rm{.575}}\)

\({\rm{\hat p}}\)is assumed to be unknown (as we are interested in the sample size irrespective of the value of \({\rm{\hat p}}\)), then the sample size is (round up!):

\({\rm{n - }}\frac{{{{\left( {{{\rm{z}}_{{\rm{\alpha /2}}}}} \right)}^{\rm{2}}}{\rm{\hat p(1 - \hat p)}}}}{{{{\rm{E}}^{\rm{2}}}}}{\rm{ - }}\frac{{{\rm{2}}{\rm{.57}}{{\rm{5}}^{\rm{2}}}{\rm{ \times 0}}{\rm{.5(1 - 0}}{\rm{.5)}}}}{{{\rm{0}}{\rm{.02}}{{\rm{5}}^{\rm{2}}}}}{\rm{\gg 2653}}\)

Hence the sample size is \({\rm{n = 2653}}\)

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Most popular questions from this chapter

A state legislator wishes to survey residents of her district to see what proportion of the electorate is aware of her position on using state funds to pay for abortions.

a. What sample size is necessary if the \({\rm{95\% }}\)CI for p is to have a width of at most . \({\rm{10}}\)irrespective of p?

b. If the legislator has strong reason to believe that at least \({\rm{2/3}}\)of the electorate know of her position, how large a sample size would you recommend?

A study of the ability of individuals to walk in a straight line reported the accompanying data on cadence (strides per second) for a sample of n =\({\rm{20}}\) randomly selected healthy men.

\({\rm{.95 }}{\rm{.85 }}{\rm{.92 }}{\rm{.95 }}{\rm{.93 }}{\rm{.86 1}}{\rm{.00 }}{\rm{.92 }}{\rm{.85 }}{\rm{.81 }}{\rm{.78 }}{\rm{.93 }}{\rm{.93 1}}{\rm{.05 }}{\rm{.93 1}}{\rm{.06 1}}{\rm{.06 }}{\rm{.96 }}{\rm{.81 }}{\rm{.96}}\)

A normal probability plot gives substantial support to the assumption that the population distribution of cadence is approximately normal. A descriptive summary of the data from Minitab follows:

Variable N Mean Median TrMean StDev SEMean cadence

\({\rm{20 0}}{\rm{.9255 0}}{\rm{.9300 0}}{\rm{.9261 0}}{\rm{.0809 0}}{\rm{.0181}}\)

Variable Min Max Q1 Q3 cadence

\({\rm{0}}{\rm{.7800 1}}{\rm{.0600 0}}{\rm{.8525 0}}{\rm{.9600}}\)

a. Calculate and interpret a \({\rm{95\% }}\) confidence interval for population mean cadence.

b.Calculate and interpret a \({\rm{95\% }}\)prediction interval for the cadence of a single individual randomly selected from this population.

c. Calculate an interval that includes at least \({\rm{99\% }}\)of the cadences in the population distribution using a confidence level of \({\rm{95\% }}\)

On the basis of extensive tests, the yield point of a particular type of mild steel-reinforcing bar is known to be normally distributed with\({\rm{\sigma = 100,}}\)The composition of bars has been slightly modified, but the modification is not believed to have affected either the normality or the value of\({\rm{\sigma }}\).

a. Assuming this to be the case, if a sample of 25 modified bars resulted in a sample average yield point of 8439 lb, compute a 90% CI for the true average yield point of the modified bar.

b. How would you modify the interval in part (a) to obtain a confidence level of 92%?

The article 鈥淕as Cooking, Kitchen Ventilation, and Exposure to Combustion Products鈥� reported that for a sample of 50 kitchens with gas cooking appliances monitored during a one week period, the sample mean CO2 level (ppm) was \({\rm{654}}{\rm{.16}}\), and the sample standard deviation was \({\rm{164}}{\rm{.43}}{\rm{.}}\)

a. Calculate and interpret a \({\rm{95\% }}\) (two-sided) confidence interval for true average CO2 level in the population of all homes from which the sample was selected.

b. Suppose the investigators had made a rough guess of \({\rm{175}}\) for the value of s before collecting data. What sample size would be necessary to obtain an interval width of ppm for a confidence level of \({\rm{95\% }}\)

The following observations are lifetimes (days) subsequent to diagnosis for individuals suffering from blood cancer (鈥淎 Goodness of Fit Approach to the Class of Life Distributions with Unknown Age,鈥� Quality and Reliability Engr. Intl., \({\rm{2012: 761--766):}}\)

\(\begin{array}{*{20}{l}}{{\rm{115 181 255 418 441 461 516 739 743 789 807}}}\\{{\rm{865 924 983 1025 1062 1063 1165 1191 1222 1222 1251}}}\\{{\rm{1277 1290 1357 1369 1408 1455 1478 1519 1578 1578 1599}}}\\{{\rm{1603 1605 1696 1735 1799 1815 1852 1899 1925 1965}}}\end{array}\)

a. Can a confidence interval for true average lifetime be calculated without assuming anything about the nature of the lifetime distribution? Explain your reasoning. (Note: A normal probability plot of the data exhibits a reasonably linear pattern.)

b. Calculate and interpret a confidence interval with a \({\rm{99\% }}\)confidence level for true average lifetime.
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