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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

The amount of lateral expansion (mils) was determined for a sample of n=\({\rm{9}}\)pulsed-power gas metal arc welds used in LNG ship containment tanks. The resulting sample standard deviation was s=\({\rm{2}}{\rm{.81}}\)mils. Assuming normality, derive a \({\rm{95\% }}\)CI for s2 and for s.

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 \({\rm{95\% }}\)

presented a sample of \({\rm{n = 153}}\)observations on ultimate tensile strength, the previous section gave summary quantities and requested a large-sample confidence interval. Because the sample size is large, no assumptions about the population distribution are required for the validity of the CI.

a. Is any assumption about the tensile-strength distribution required prior to calculating a lower prediction bound for the tensile strength of the next specimen selected using the method described in this section? Explain.

b. Use a statistical software package to investigate the plausibility of a normal population distribution.

c. Calculate a lower prediction bound with a prediction level of \({\rm{95\% }}\)for the ultimate tensile strength of the next specimen selected.

a. Under the same conditions as those leading to the interval\({\rm{(7}}{\rm{.5),p((}}\overline {\rm{X}} {\rm{ - \mu )/(\sigma /}}\sqrt {\rm{n}} {\rm{) < 1}}{\rm{.645 = }}{\rm{.95}}{\rm{.}}\)Use this to derive a one-sided interval for\({\rm{\mu }}\)that has infinite width and provides a lower confidence bound on m. What is this interval for the data in Exercise 5(a)?

b. Generalize the result of part (a) to obtain a lower bound with confidence level\({\rm{100(1 - \alpha )\% }}\)

c. What is an analogous interval to that of part (b) that provides an upper bound on\({\rm{\mu }}\)? Compute this 99% interval for the data of Exercise 4(a).

Let X1, X2,鈥�, Xn be a random sample from a uniform distribution on the interval (0,胃), so that

\({\rm{f(x) = }}\left\{ \begin{aligned}{l}\frac{{\rm{1}}}{{\rm{\theta }}}{\rm{ 0}} \le {\rm{x}} \le {\rm{\theta }}\\{\rm{0 otherwise}}\end{aligned} \right.\)

Then if Y= max(Xi ), it can be shown that the rv U=Y/胃 has density function

\({{\rm{f}}_U}{\rm{(u) = }}\left\{ \begin{aligned}{l}{\rm{n}}{{\rm{u}}^{n - 1}}{\rm{ 0}} \le u \le 1\\{\rm{0 otherwise}}\end{aligned} \right.\)

a. Use\({{\rm{f}}_U}{\rm{(u)}}\) to verify that

\({\rm{P}}\left( {{{(\alpha /2)}^{1/n}} < \frac{Y}{\theta } \le {{(1 - \alpha /2)}^{1/n}}} \right) = 1 - \alpha \)

and use this derive a \(100(1 - \alpha )\% \) CI for 胃.

b. Verify that \({\rm{P}}\left( {{\alpha ^{1/n}} < Y/\theta \le 1} \right) = 1 - \alpha \), and derive a 100(1 - 伪)% CI for 胃 based on this probability statement.

c.Which of the two intervals derived previously is shorter? If my waiting time for a morning bus is uniformly distributed and observed waiting times are x1 = 4.2, x2 = 3.5, x颅3 = 1.7, x4 = 1.2, and x5 = 2.4, derive a 95% CI for 胃 by using the shorter of the two intervals.
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