A random sample of 5805 physicians in Colorado showed that 3332 provided at least some charity care (i.e., treated poor people at no cost). a) Let p represent the proportion of all Colorado physicians who provide some charity care. Find a point estimate for p. (Round your answer to four decimal places.) b) Find a 99% confidence interval for p. (Round your answer to three decimal places.) lower limit displaystyle=? upper limit displaystyle=? c) Is the normal approximation to the binomial justified in this problem? Explain

Question
Confidence intervals
asked 2020-11-27
A random sample of 5805 physicians in Colorado showed that 3332 provided at least some charity care (i.e., treated poor people at no cost).
a)
Let p represent the proportion of all Colorado physicians who provide some charity care. Find a point estimate for p. (Round your answer to four decimal places.)
b)
Find a \(99\%\) confidence interval for p. (Round your answer to three decimal places.)
lower limit \(\displaystyle=?\)
upper limit \(\displaystyle=?\)
c)
Is the normal approximation to the binomial justified in this problem? Explain

Answers (1)

2020-11-28
Step 1
a)
Here, \(\displaystyle{x}={3332}{\quad\text{and}\quad}{n}={5805}.\)
The point estimate of the population proportion is calculated as follows:
\(\displaystyle\hat{{p}}=\frac{x}{{n}}\)
\(\displaystyle=\frac{3332}{{5805}}\)
\(\displaystyle={0.573987}\)
\(\displaystyle\approx{0.5740}\)
Step 2
b)
Since the required confidence interval is \(99\%\), the two-tailed z-critical value is 2.5758 using the Excel function \(\displaystyle“={N}{O}{R}{M}.{S}.{I}{N}{V}{\left({0.995}\right)}\text{"}\).
The \(99\%\) confidence interval for the population proportion is calculated as follows:
\(\displaystyle{C}{I}=\hat{{p}}\pm{z}_{{\frac{\alpha}{{2}}}}\sqrt{{\frac{{\hat{{p}}{\left({1}-\hat{{p}}\right)}}}{{n}}}}\)
\(\displaystyle={0.5740}\pm{\left({2.5758}\right)}\sqrt{{{0.5740}\frac{{{1}-{0.5740}}}{{5805}}}}\)
\(\displaystyle={0.5740}\pm{0.0167}\)
\(\displaystyle={\left({0.5573},{0.5907}\right)}\)
\(\displaystyle\approx{\left({0.557},{0.591}\right)}\)
Lower limit \(\displaystyle={0.557}\)
Upper limit \(\displaystyle={0.591}\)
Step 3
c)
The normal approximation to binomial is possible only if,
\(\displaystyle{n}{p}>{5}\)
\(\displaystyle{n}{q}>{5}\)
Checking the conditions for normality:
\(\displaystyle{n}{p}={5805}\times\frac{3332}{{5805}}\)
\(\displaystyle={3332}\)
\(\displaystyle>{5}\)
\(\displaystyle{n}{q}={5805}\times{\left({1}-\frac{3332}{{5805}}\right)}\)
\(\displaystyle={2473}\)
\(\displaystyle>{5}\)
Thus, normal approximation to binomial is possible.
Yes, \(\displaystyle{n}{p}>{5}{\quad\text{and}\quad}{n}{q}>{5}\)
0

Relevant Questions

asked 2021-02-21
We wish to estimate what percent of adult residents in a certain county are parents. Out of 600 adult residents sampled, 192 had kids. Based on this, plot a \(99\%\) confidence interval for the proportion of adult residents who are parents in a given county.
Express your answer in the form of three inequalities. Give your answers in decimal fractions up to three places \(\displaystyle<{p}<\) Express the same answer using a point estimate and a margin of error. Give your answers as decimals, to three places.
\(\displaystyle{p}=\pm\pm\)
asked 2020-12-12
A machine produces parts with lengths that are normally distributed with \(\displaystyle\sigma={0.66}\). A sample of 20 parts has a mean length of 75.98.
a)
Give a point estimate for \(\displaystyle\mu\). (Give your answer correct to two decimal places.)
b)
Find the \(95\%\) confidence maximum error of estimate for \(\displaystyle\mu\). (Give your answer correct to two decimal places.)
c)
Find the \(95\%\) confidence interval for \(\displaystyle\mu\). (Give your answer correct to two decimal places.)
Lower limit \(\displaystyle=?\)
Upper limit \(\displaystyle=?\)
asked 2020-12-07
Would you rather spend more federal taxes on art? Of a random sample of \(n_{1} = 86\) politically conservative voters, \(r_{1} = 18\) responded yes. Another random sample of \(n_{2} = 85\) politically moderate voters showed that \(r_{2} = 21\) responded yes. Does this information indicate that the population proportion of conservative voters inclined to spend more federal tax money on funding the arts is less than the proportion of moderate voters so inclined? Use \(\alpha = 0.05.\) (a) State the null and alternate hypotheses. \(H_0:p_{1} = p_{2}, H_{1}:p_{1} > p_2\)
\(H_0:p_{1} = p_{2}, H_{1}:p_{1} < p_2\)
\(H_0:p_{1} = p_{2}, H_{1}:p_{1} \neq p_2\)
\(H_{0}:p_{1} < p_{2}, H_{1}:p_{1} = p_{2}\) (b) What sampling distribution will you use? What assumptions are you making? The Student's t. The number of trials is sufficiently large. The standard normal. The number of trials is sufficiently large.The standard normal. We assume the population distributions are approximately normal. The Student's t. We assume the population distributions are approximately normal. (c)What is the value of the sample test statistic? (Test the difference \(p_{1} - p_{2}\). Do not use rounded values. Round your final answer to two decimal places.) (d) Find (or estimate) the P-value. (Round your answer to four decimal places.) (e) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level alpha? At the \(\alpha = 0.05\) level, we reject the null hypothesis and conclude the data are statistically significant. At the \(\alpha = 0.05\) level, we fail to reject the null hypothesis and conclude the data are statistically significant. At the \(\alpha = 0.05\) level, we fail to reject the null hypothesis and conclude the data are not statistically significant. At the \(\alpha = 0.05\) level, we reject the null hypothesis and conclude the data are not statistically significant. (f) Interpret your conclusion in the context of the application. Reject the null hypothesis, there is sufficient evidence that the proportion of conservative voters favoring more tax dollars for the arts is less than the proportion of moderate voters. Fail to reject the null hypothesis, there is sufficient evidence that the proportion of conservative voters favoring more tax dollars for the arts is less than the proportion of moderate voters. Fail to reject the null hypothesis, there is insufficient evidence that the proportion of conservative voters favoring more tax dollars for the arts is less than the proportion of moderate voters. Reject the null hypothesis, there is insufficient evidence that the proportion of conservative voters favoring more tax dollars for the arts is less than the proportion of moderate voters.
asked 2021-02-25
A psychologist is interested in constructing a \(99\%\) confidence interval for the proportion of people who accept the theory that a person's spirit is no more than the complicated network of neurons in the brain. 68 of the 702 randomly selected people who were surveyed agreed with this theory. Round answers to 4 decimal places where possible.
a)
With \(99\%\) confidence the proportion of all people who accept the theory that a person's spirit is no more than the complicated network of neurons in the brain is between ____ and ____.
b)
If many groups of 702 randomly selected people are surveyed, then a different confidence interval would be produced from each group. About ____ percent of these confidence intervals will contain the true population proportion of all people who accept the theory that a person’s spirit is no more than the complicated network of neurons in the brain and about ____ percent will not contain the true population proportion.
asked 2021-02-12
What is the optimal time for a scuba diver to be on the bottom of the ocean? That depends on the depth of the dive. The U.S. Navy has done a lot of research on this topic. The Navy defines the "optimal time" to be the time at each depth for the best balance between length of work period and decompression time after surfacing. Let \(\displaystyle{x}=\) depth of dive in meters, and let \(\displaystyle{y}=\) optimal time in hours. A random sample of divers gave the following data.
\(\begin{array}{|c|c|} \hline x & 13.1 & 23.3 & 31.2 & 38.3 & 51.3 &20.5 & 22.7 \\ \hline y & 2.78 & 2.18 & 1.48 & 1.03 & 0.75 & 2.38 & 2.20 \\ \hline \end{array}\)
(a)
Find \(\displaystyleΣ{x},Σ{y},Σ{x}^{2},Σ{y}^{2},Σ{x}{y},{\quad\text{and}\quad}{r}\). (Round r to three decimal places.)
\(\displaystyleΣ{x}=\)
\(\displaystyleΣ{y}=\)
\(\displaystyleΣ{x}^{2}=\)
\(\displaystyleΣ{y}^{2}=\)
\(\displaystyleΣ{x}{y}=\)
\(\displaystyle{r}=\)
(b)
Use a \(1\%\) level of significance to test the claim that \(\displaystyle\rho<{0}\). (Round your answers to two decimal places.)
\(\displaystyle{t}=\)
critical \(\displaystyle{t}=\)
Conclusion
Reject the null hypothesis. There is sufficient evidence that \(\displaystyle\rho<{0}\).Reject the null hypothesis. There is insufficient evidence that \(\displaystyle\rho<{0}\).
Fail to reject the null hypothesis. There is sufficient evidence that \(\displaystyle\rho<{0}\).Fail to reject the null hypothesis. There is insufficient evidence that \(\displaystyle\rho<{0}.\)
(c)
Find \(\displaystyle{S}_{{e}},{a},{\quad\text{and}\quad}{b}\). (Round your answers to four decimal places.)
\(\displaystyle{S}_{{e}}=\)
\(\displaystyle{a}=\)
\(\displaystyle{b}=\)
asked 2020-10-23
A random sample of \(\displaystyle{n}_{{1}}={16}\) communities in western Kansas gave the following information for people under 25 years of age.
\(\displaystyle{X}_{{1}}:\) Rate of hay fever per 1000 population for people under 25
\(\begin{array}{|c|c|} \hline 97 & 91 & 121 & 129 & 94 & 123 & 112 &93\\ \hline 125 & 95 & 125 & 117 & 97 & 122 & 127 & 88 \\ \hline \end{array}\)
A random sample of \(\displaystyle{n}_{{2}}={14}\) regions in western Kansas gave the following information for people over 50 years old.
\(\displaystyle{X}_{{2}}:\) Rate of hay fever per 1000 population for people over 50
\(\begin{array}{|c|c|} \hline 94 & 109 & 99 & 95 & 113 & 88 & 110\\ \hline 79 & 115 & 100 & 89 & 114 & 85 & 96\\ \hline \end{array}\)
(i) Use a calculator to calculate \(\displaystyle\overline{{x}}_{{1}},{s}_{{1}},\overline{{x}}_{{2}},{\quad\text{and}\quad}{s}_{{2}}.\) (Round your answers to two decimal places.)
(ii) Assume that the hay fever rate in each age group has an approximately normal distribution. Do the data indicate that the age group over 50 has a lower rate of hay fever? Use \(\displaystyle\alpha={0.05}.\)
(a) What is the level of significance?
State the null and alternate hypotheses.
\(\displaystyle{H}_{{0}}:\mu_{{1}}=\mu_{{2}},{H}_{{1}}:\mu_{{1}}<\mu_{{2}}\)
\(\displaystyle{H}_{{0}}:\mu_{{1}}=\mu_{{2}},{H}_{{1}}:\mu_{{1}}>\mu_{{2}}\)
\(\displaystyle{H}_{{0}}:\mu_{{1}}=\mu_{{2}},{H}_{{1}}:\mu_{{1}}\ne\mu_{{2}}\)
\(\displaystyle{H}_{{0}}:\mu_{{1}}>\mu_{{2}},{H}_{{1}}:\mu_{{1}}=\mu_{{12}}\)
(b) What sampling distribution will you use? What assumptions are you making?
The standard normal. We assume that both population distributions are approximately normal with known standard deviations.
The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations,
The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations,
The Student's t. We assume that both population distributions are approximately normal with known standard deviations,
What is the value of the sample test statistic? (Test the difference \(\displaystyle\mu_{{1}}-\mu_{{2}}\). Round your answer to three decimalplaces.)
What is the value of the sample test statistic? (Test the difference \(\displaystyle\mu_{{1}}-\mu_{{2}}\). Round your answer to three decimal places.)
(c) Find (or estimate) the P-value.
P-value \(\displaystyle>{0.250}\)
\(\displaystyle{0.125}<{P}-\text{value}<{0},{250}\)
\(\displaystyle{0},{050}<{P}-\text{value}<{0},{125}\)
\(\displaystyle{0},{025}<{P}-\text{value}<{0},{050}\)
\(\displaystyle{0},{005}<{P}-\text{value}<{0},{025}\)
P-value \(\displaystyle<{0.005}\)
Sketch the sampling distribution and show the area corresponding to the P-value.
P.vaiue Pevgiue
P-value f P-value
asked 2020-12-29
The presidential election is coming. Five survey companies (A, B, C, D, and E) are doing survey to forecast whether or not the Republican candidate will win the election. Each company randomly selects a sample size between 1000 and 1500 people. All of these five companies interview people over the phone during Tuesday and Wednesday. The interviewee will be asked if he or she is 18 years old or above and U.S. citizen who are registered to vote. If yes, the interviewee will be further asked: will you vote for the Republican candidate? On Thursday morning, these five companies announce their survey sample and results at the same time on the newspapers. The results show that a% (from A), b% (from B), c% (from C), d% (from D), and e% (from E) will support the Republican candidate. The margin of error is plus/minus 3% for all results. Suppose that \(\displaystyle{c}{>}{a}{>}{d}{>}{e}{>}{b}\). When you see these results from the newspapers, can you exactly identify which result(s) is (are) not reliable and not accurate? That is, can you identify which estimation interval(s) does (do) not include the true population proportion? If you can, explain why you can, if no, explain why you cannot and what information you need to identify. Discuss and explain your reasons. You must provide your statistical analysis and reasons.
asked 2020-11-20
In 2014, the Centers for Disearse reported the percentage of people 18 years of age and older who smoke. Suppose that a study designed to collect new data on smokers and nonsmokers uses a preliminary estimate of the proportion who smoke of 0.30.
a) How large a sample should be taken to estimate the proportion of smokers in the population with a margin of error of 0.02? Use 95% confidence.(Round your answer up to the nearest integer.)
b)Assume that the study uses your sample size recommendation in part (a) and finds 470 smokers. What is the point estimate of the proportion of smokers in the population? (Round your answer to four decimal places.)
c) What is the 95% confidence interval for the proportion of smokers in the population? (Round your answer to four decimal places.)
asked 2021-02-11
Several models have been proposed to explain the diversification of life during geological periods. According to Benton (1997), The diversification of marine families in the past 600 million years (Myr) appears to have followed two or three logistic curves, with equilibrium levels that lasted for up to 200 Myr. In contrast, continental organisms clearly show an exponential pattern of diversification, and although it is not clear whether the empirical diversification patterns are real or are artifacts of a poor fossil record, the latter explanation seems unlikely. In this problem, we will investigate three models fordiversification. They are analogous to models for populationgrowth, however, the quantities involved have a differentinterpretation. We denote by N(t) the diversification function,which counts the number of taxa as a function of time, and by rthe intrinsic rate of diversification.
(a) (Exponential Model) This model is described by \(\displaystyle{\frac{{{d}{N}}}{{{\left.{d}{t}\right.}}}}={r}_{{{e}}}{N}\ {\left({8.86}\right)}.\) Solve (8.86) with the initial condition N(0) at time 0, and show that \(\displaystyle{r}_{{{e}}}\) can be estimated from \(\displaystyle{r}_{{{e}}}={\frac{{{1}}}{{{t}}}}\ {\ln{\ }}{\left[{\frac{{{N}{\left({t}\right)}}}{{{N}{\left({0}\right)}}}}\right]}\ {\left({8.87}\right)}\)
(b) (Logistic Growth) This model is described by \(\displaystyle{\frac{{{d}{N}}}{{{\left.{d}{t}\right.}}}}={r}_{{{l}}}{N}\ {\left({1}\ -\ {\frac{{{N}}}{{{K}}}}\right)}\ {\left({8.88}\right)}\) where K is the equilibrium value. Solve (8.88) with the initial condition N(0) at time 0, and show that \(\displaystyle{r}_{{{l}}}\) can be estimated from \(\displaystyle{r}_{{{l}}}={\frac{{{1}}}{{{t}}}}\ {\ln{\ }}{\left[{\frac{{{K}\ -\ {N}{\left({0}\right)}}}{{{N}{\left({0}\right)}}}}\right]}\ +\ {\frac{{{1}}}{{{t}}}}\ {\ln{\ }}{\left[{\frac{{{N}{\left({t}\right)}}}{{{K}\ -\ {N}{\left({t}\right)}}}}\right]}\ {\left({8.89}\right)}\) for \(\displaystyle{N}{\left({t}\right)}\ {<}\ {K}.\)
(c) Assume that \(\displaystyle{N}{\left({0}\right)}={1}\) and \(\displaystyle{N}{\left({10}\right)}={1000}.\) Estimate \(\displaystyle{r}_{{{e}}}\) and \(\displaystyle{r}_{{{l}}}\) for both \(\displaystyle{K}={1001}\) and \(\displaystyle{K}={10000}.\)
(d) Use your answer in (c) to explain the following quote from Stanley (1979): There must be a general tendency for calculated values of \(\displaystyle{\left[{r}\right]}\) to represent underestimates of exponential rates,because some radiation will have followed distinctly sigmoid paths during the interval evaluated.
(e) Explain why the exponential model is a good approximation to the logistic model when \(\displaystyle\frac{{N}}{{K}}\) is small compared with 1.
asked 2020-12-28
Is statistical inference intuitive to babies? In other words, are babies able to generalize from sample to population? In this study,1 8-month-old infants watched someone draw a sample of five balls from an opaque box. Each sample consisted of four balls of one color (red or white) and one ball of the other color. After observing the sample, the side of the box was lifted so the infants could see all of the balls inside (the population). Some boxes had an “expected” population, with balls in the same color proportions as the sample, while other boxes had an “unexpected” population, with balls in the opposite color proportion from the sample. Babies looked at the unexpected populations for an average of 9.9 seconds (sd = 4.5 seconds) and the expected populations for an average of 7.5 seconds (sd = 4.2 seconds). The sample size in each group was 20, and you may assume the data in each group are reasonably normally distributed. Is this convincing evidence that babies look longer at the unexpected population, suggesting that they make inferences about the population from the sample? Let group 1 and group 2 be the time spent looking at the unexpected and expected populations, respectively. A) Calculate the relevant sample statistic. Enter the exact answer. Sample statistic: _____ B) Calculate the t-statistic. Round your answer to two decimal places. t-statistic = ___________ C) Find the p-value. Round your answer to three decimal places. p-value =
...