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Chapter 7: Problem 17

Use the data and confidence level to construct a confidence interval estimate of \(p,\) then address the given question. A random sample of 860 births in New York State included 426 boys. Construct a \(95 \%\) confidence interval estimate of the proportion of boys in all births. It is believed that among all births, the proportion of boys is \(0.512 .\) Do these sample results provide strong evidence against that belief?

Short Answer

Expert verified
The 95% confidence interval is (0.462, 0.528). The sample results do not provide strong evidence against the belief that the proportion of boys is 0.512.

Step by step solution

01

- Identify the Sample Proportion

Calculate the sample proportion (\( \hat{p} \)). Divide the number of boys (426) by the total number of births (860).\( \hat{p} = \frac{426}{860} \approx 0.4953 \)
02

- Determine the Standard Error

Calculate the standard error (SE) for the sample proportion using the formula: \( SE = \sqrt{ \frac{\hat{p}(1 - \hat{p})}{n} } \) where \( n \) is the sample size (860).\( SE = \sqrt{ \frac{0.4953(1 - 0.4953)}{860} } \approx 0.017 \)
03

- Find the Z-Score for the Confidence Level

For a \(95\%\) confidence interval, the Z-score is 1.96 (from standard normal distribution tables).
04

- Calculate the Confidence Interval

Use the formula for the confidence interval: \( \hat{p} \pm Z \times SE \). Plugging in the values: \( 0.4953 \pm 1.96 \times 0.017 \). This gives the interval: \( (0.462, 0.528) \).
05

- Compare to the Believed Proportion

The believed proportion of boys is 0.512. Check if this value falls within the confidence interval \( (0.462, 0.528) \). Since 0.512 is within the interval, the sample results do not provide strong evidence against the belief.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sample Proportion
The sample proportion is a key starting point in confidence interval estimation. It represents the ratio of a particular characteristic (in this case, the number of boys) within a sample. To calculate it, divide the number of boys (426) by the total number of observed births (860). This gives a sample proportion (\( \hat{p} \)) of approximately 0.4953. This simple calculation sets the stage for further statistical analysis in hypothesis testing and confidence interval estimation.
Standard Error
The standard error (SE) measures the variability of the sample proportion. It helps to understand how much the sample proportion is expected to fluctuate if you were to take multiple random samples from the same population.
Calculate the standard error with the formula: \( SE = \sqrt{ \frac{\hat{p}(1 - \hat{p})}{n} } \), where \( n \) is the sample size (860). For our example: \( SE = \sqrt{ \frac{0.4953(1 - 0.4953)}{860} } \);approximately results in 0.017. Knowing the SE is crucial because it will be used to create the confidence interval.
Z-score
A Z-score indicates how many standard errors a data point is from the mean in a standard normal distribution. In the context of confidence intervals, the Z-score translates the confidence level into a standardized value. A higher Z-score corresponds to a higher confidence level. For a 95% confidence interval, the Z-score is 1.96.
This value is obtained from standard normal distribution tables and is used to calculate the range within which the true population proportion lies with a certain level of confidence.
Hypothesis Testing
Hypothesis testing involves making a claim (hypothesis) and then using sample data to assess the claim's validity. In our example, the hypothesis is that the proportion of boys in all births is 0.512. After calculating the confidence interval (\( 0.462, 0.528 \)) for the sample proportion, we check whether the hypothesized population proportion (0.512) falls within this interval.
Since 0.512 is within the interval, we do not have strong evidence to reject the hypothesis.
However, if the value were outside this range, it would mean our sample provides strong evidence against the hypothesized proportion, leading us to reconsider the belief.

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Most popular questions from this chapter

Use the data and confidence level to construct a confidence interval estimate of \(p,\) then address the given question. One of Mendel's famous genetics experiments yielded 580 peas, with 428 of them green and 152 yellow. a. Find a \(99 \%\) confidence interval estimate of the percentage of green peas. b. Based on his theory of genetics, Mendel expected that \(75 \%\) of the offspring peas would be green. Given that the percentage of offspring green peas is not \(75 \%,\) do the results contradict Mendel's theory? Why or why not?

Find the sample size required to estimate the population mean. Data Set 1 "Body Data" in Appendix B includes pulse rates of 153 randomly selected adult males, and those pulse rates vary from a low of 40 bpm to a high of 104 bpm. Find the minimum sample size required to estimate the mean pulse rate of adult males. Assume that we want \(99 \%\) confidence that the sample mean is within 2 bpm of the population mean. a. Find the sample size using the range rule of thumb to estimate \(\sigma .\) b. Assume that \(\sigma=11.3\) bpm, based on the value of \(s=11.3\) bpm for the sample of 153 male pulse rates. c. Compare the results from parts (a) and (b). Which result is likely to be better?

Construct the confidence interval estimate of the mean. Listed below are arrival delays (minutes) of randomly selected American Airlines flights from New York (JFK) to Los Angeles (LAX). Negative numbers correspond to flights that arrived before the scheduled arrival time. Use a \(95 \%\) confidence interval. How good is the on-time performance? $$\begin{array}{llllllllllll} -5 & -32 & -13 & -9 & -19 & 49 & -30 & -23 & 14 & -21 & -32 & 11 \end{array}$$

Finding Sample Size Instead of using Table \(7-2\) for determining the sample size required to estimate a population standard deviation \(\sigma,\) the following formula can be used $$n=\frac{1}{2}\left(\frac{z_{\alpha / 2}}{d}\right)^{2}$$ where \(z_{\alpha / 2}\) corresponds to the confidence level and \(d\) is the decimal form of the percentage error. For example, to be \(95 \%\) confident that \(s\) is within \(15 \%\) of the value of \(\sigma,\) use \(z_{\alpha / 2}=1.96\) and \(d=0.15\) to get a sample size of \(n=86 .\) Find the sample size required to estimate \(\sigma,\) assuming that we want \(98 \%\) confidence that \(s\) is within \(15 \%\) of \(\sigma .\)

Construct the confidence interval estimate of the mean. Listed below are amounts of arsenic \((\mu \mathrm{g},\) or micrograms, per serving) in samples of brown rice from California (based on data from the Food and Drug Administration). Use a \(90 \%\) confidence level. The Food and Drug Administration also measured amounts of arsenic in samples of brown rice from Arkansas. Can the confidence interval be used to describe arsenic levels in Arkansas? $$\begin{array}{cccccccccc} 5.4 & 5.6 & 8.4 & 7.3 & 4.5 & 7.5 & 1.5 & 5.5 & 9.1 & 8.7 \end{array}$$
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