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Chapter 7: Problem 26

Use the data and confidence level to construct a confidence interval estimate of \(p,\) then address the given question. Before its clinical trials were discontinued, the Genetics \& IVF Institute conducted a clinical trial of the XSORT method designed to increase the probability of conceiving a girl and, among the 945 babies born to parents using the XSORT method, there were 879 girls. The YSORT method was designed to increase the probability of conceiving a boy and, among the 291 babies born to parents using the YSORT method, there were 239 boys. Construct the two \(95 \%\) confidence interval estimates of the percentages of success. Compare the results. What do you conclude?

Short Answer

Expert verified
XSORT confidence interval: (0.912, 0.946)YSORT confidence interval: (0.778, 0.864)XSORT method is more effective.

Step by step solution

01

- Identify the Sample Proportions

For the XSORT method, the sample proportion of girls is given by: \[ \ \hat{p}_1 = \frac{879}{945} \] For the YSORT method, the sample proportion of boys is given by: \[ \ \hat{p}_2 = \frac{239}{291} \]
02

- Calculate the Sample Proportions

Calculate the sample proportions: \[ \hat{p}_1 = \frac{879}{945} \approx 0.929 \] \[ \hat{p}_2 = \frac{239}{291} \approx 0.821 \]
03

- Determine the Standard Errors

Calculate the standard errors for both proportions: \[ SE_1 = \sqrt{ \frac{\hat{p}_1 (1-\hat{p}_1)}{n_1} } = \sqrt{ \frac{0.929(1-0.929)}{945} } \approx 0.0088 \] \[ SE_2 = \sqrt{ \frac{\hat{p}_2 (1-\hat{p}_2)}{n_2} } = \sqrt{ \frac{0.821(1-0.821)}{291} } \approx 0.022 \]
04

- Find the Critical Value

For a 95% confidence interval, the critical value (z-score) from the standard normal distribution is: \[ z = 1.96 \]
05

- Construct the Confidence Intervals

The confidence interval for the XSORT method is: \[ \ \hat{p}_1 \pm z \cdot SE_1 = 0.929 \pm 1.96 \cdot 0.0088 \approx (0.912, 0.946) \] The confidence interval for the YSORT method is: \[ \ \hat{p}_2 \pm z \cdot SE_2 = 0.821 \pm 1.96 \cdot 0.022 \approx (0.778, 0.864) \]
06

- Compare and Conclude

Comparing the two confidence intervals: - XSORT: (0.912, 0.946) - YSORT: (0.778, 0.864) The XSORT method has a higher probability and a narrower confidence interval than the YSORT method, indicating that XSORT is more effective at achieving its intended outcome.

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Key Concepts

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

Sample Proportion
A sample proportion is the fraction of the total sample that possesses a particular attribute of interest. It helps in making inferences about the population proportion based on sample data. In this exercise, we determine the proportion of success in two methods designed to influence the gender of babies. For the XSORT method, the sample proportion of girls is calculated by dividing the number of girls born (879) by the total number of babies (945). Similarly, for the YSORT method, the sample proportion of boys is determined by dividing the number of boys born (239) by the total number of babies (291). The resulting proportions give us a snapshot of the success rate in the sample.
Standard Error
The standard error (SE) measures the variability or dispersion of the sample proportion from the true population proportion. A smaller SE indicates that the sample proportion is a more precise estimate of the population proportion. It's calculated using the formula:
  • For XSORT: \[ SE_1 = \sqrt{ \frac{\hat{p}_1 (1-\hat{p}_1)}{n_1} } \]Plugging in the values, we get:\[ SE_1 \approx 0.0088 \]
  • For YSORT: \[ SE_2 = \sqrt{ \frac{\hat{p}_2 (1-\hat{p}_2)}{n_2} } \]Plugging in the values, we get:\[ SE_2 \approx 0.022 \]
Critical Value
The critical value is a factor used to calculate the margin of error in confidence intervals. It is derived from the standard normal distribution (z-distribution) based on the desired confidence level. For a 95% confidence level, the critical value is 1.96. This means we are 95% confident that the true population parameter lies within the calculated confidence interval. The critical value is essential for constructing confidence intervals, as it scales the standard error to reflect the desired confidence level.
Confidence Level
The confidence level represents the frequency with which the calculated confidence interval would contain the true population parameter if the same population were sampled multiple times. For example, a 95% confidence level implies that in 95 out of 100 samples, the true proportion would fall within the calculated confidence interval. In our exercise, we used a 95% confidence level to build the confidence intervals for both XSORT and YSORT methods, which means we are fairly certain about the range in which the true effect of the methods lies.

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Most popular questions from this chapter

Finding Confidence Intervals. In Exercises \(9-16,\) assume that each sample is a simple random sample obtained from a population with a normal distribution. Speed Dating In a study of speed dating conducted at Columbia University, female subjects were asked to rate the attractiveness of their male dates, and a sample of the results is listed below \((1=\text { not attractive; } 10=\text { extremely attractive). Construct a } 95 \%\) confidence interval estimate of the standard deviation of the population from which the sample was obtained. \(\begin{array}{ccccccccccccccccccccccc}5 & 8 & 3 & 8 & 6 & 10 & 3 & 7 & 9 & 8 & 5 & 5 & 6 & 8 & 8 & 7 & 3 & 5 & 5 & 6 & 8 & 7 & 8 & 8 & 8 & 7\end{array}\)

Construct the confidence interval estimate of the mean. In a test of weight loss programs, 40 adults used the Atkins weight loss program. After 12 months, their mean weight loss was found to be 2.1 lb, with a standard deviation of 4.8 ib. Construct a \(90 \%\) confidence interval estimate of the mean weight loss for all such subjects. Does the Atkins program appear to be effective? Does it appear to be practical?

Find the sample size required to estimate the population mean. The Wechsler IQ test is designed so that the mean is 100 and the standard deviation is 15 for the population of normal adults. Find the sample size necessary to estimate the mean IQ score of college professors. We want to be 99\% confident that our sample mean is within 4 IQ points of the true mean. The mean for this population is clearly greater than \(100 .\) The standard deviation for this population is less than 15 because it is a group with less variation than a group randomly selected from the general population; therefore, if we use \(\sigma=15\) we are being conservative by using a value that will make the sample size at least as large as necessary. Assume then that \(\sigma=15\) and determine the required sample size. Does the sample size appear to be practical?

Use the given data to find the minimum sample size required to estimate a population proportion or percentage. In a study of government financial aid for college students, it becomes necessary to estimate the percentage of full-time college students who earn a bachelor's degree in four years or less. Find the sample size needed to estimate that percentage. Use a 0.05 margin of error, and use a confidence level of \(95 \% .\) a. Assume that nothing is known about the percentage to be estimated. b. Assume that prior studies have shown that about \(40 \%\) of full-time students earn bachelor's degrees in four years or less. c. Does the added knowledge in part (b) have much of an effect on the sample size?

Use the given sample data and confidence level. In each case, (a) find the best point estimate of the population proportion \(p ;(b)\) identify the value of the margin of error \(E ;(c)\) construct the confidence interval; (d) write a statement that correctly interprets the confidence interval. In a study of cell phone use and brain hemispheric dominance, an Intemet survey was e-mailed to 5000 subjects randomly selected from an online group involved with ears. 717 surveys were returned. Construct a \(90 \%\) confidence interval for the proportion of retumed surveys.
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