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91Ó°ÊÓ

In a random sample of 40 brown-haired individuals, 22 indicated that they use hair coloring. In another random sample of 40 blond individuals, 26 indicated that they use hair coloring. Use a \(92 \%\) confidence interval to estimate the difference in the population proportions of brunettes and blondes that use hair coloring.

Short Answer

Expert verified
Using the steps described above and the provided sample data, you should be able to calculate the 92% confidence interval for difference in the population proportions of brunettes and blondes who use hair coloring.

Step by step solution

01

Calculate the Sample Proportions

First, find the sample proportions for each group. For brown-haired individuals, 22 of 40 use hair coloring, so the sample proportion, \(p_B\), is \(22/40\). For blond individuals, 26 of 40 use hair coloring, so the sample proportion, \(p_A\), is \(26/40\).
02

Calculate the Difference in Sample Proportions

Subtract the sample proportion of brown-haired individuals from the sample proportion of blond individuals to get the difference in sample proportions, \(d = p_A - p_B\).
03

Calculate the Standard Error

The standard error, \(SE\), for the difference in sample proportions is calculated using the formula \(SE = \sqrt{ [ p_A (1 - p_A) / n_A ] + [ p_B (1 - p_B) / n_B ] }\). Here \(n_A\) and \(n_B\) represent the number of individuals in the A and B groups, respectively.
04

Determine the Z-Value

For a confidence interval of 92%, look up or calculate the Z value from a standard normal distribution. Let's denote it as \(Z_{0.04}\) because 4% (100% - 92%) of the probability density would be split equally into two tails of the distribution.
05

Calculate the Confidence Interval

The formula to calculate a confidence interval for a difference in population proportions is \(CI = d \pm Z_{0.04} \times SE\). Plug in the calculated values to obtain the confidence interval.

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Key Concepts

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

Sample Proportions
Understanding sample proportions is fundamental when it comes to statistical analysis of categorical data. A sample proportion represents the fraction of individuals in a sample with a certain characteristic. For example, if you have a group of people and you want to know the proportion of them that have blue eyes, you simply divide the number of blue-eyed individuals by the total number of people in the group.

In the context of the given exercise, the sample proportions are calculated for two different groups of individuals based on their hair color and whether they use hair coloring. The formula used is \( p = \frac{x}{n} \) where \(x\) is the number of individuals with the characteristic of interest (using hair coloring) and \(n\) is the total number of individuals in the sample. It's the starting point for estimating the population proportions using confidence intervals.
Standard Error
The concept of standard error plays a pivotal role in quantifying the precision of an estimated sample statistic, such as a sample mean or sample proportion. The standard error measures the variability of the sampling distribution and is essentially the standard deviation of this distribution.

To put it simply, smaller standard errors suggest that the sample statistic is likely closer to the actual population parameter. In the scenario of our exercise, the standard error is applied to the difference in sample proportions. It incorporates both sample proportions and the size of the samples. The formula given in the exercise \( SE = \sqrt{ [ p_A (1 - p_A) / n_A ] + [ p_B (1 - p_B) / n_B ] } \) takes into account the variability within each group and gives an estimate of uncertainty around the difference in population proportions.
Population Proportions
Population proportions reflect the true percentage of individuals in an entire population who possess a particular attribute. It’s what we are trying to estimate with our sample data. Unlike sample proportions, which are derived from a subset and can vary from sample to sample, the population proportion remains constant—it is what it is, regardless of what our sample data shows.

When we work with samples, we rarely know the actual population proportion, so we use our sample proportion as an estimate. This estimate can be improved by calculating a confidence interval, which gives a range within which we can be a certain percentage confident that the true population proportion lies. In our example, the confidence interval will give us this range for the difference between the population proportions of brown-haired and blond individuals who use hair coloring.
Normal Distribution
The normal distribution, often called the bell curve, is a common and very important probability distribution in statistics. It is characterized by its symmetric shape and is defined by two parameters: the mean and the standard deviation. Most values remain close to the mean and the likelihood of extreme values drops off rapidly.

In terms of confidence intervals, when we assume that the difference in sample proportions is normally distributed, we can use the properties of the normal distribution to make inferences about the population. The Z-value in the confidence interval formula corresponds to the desired level of confidence and is found using the normal distribution. For a 92% confidence interval, we seek the Z-value that captures the middle 92% of the distribution, leaving 4% in the tails—hence the term \(Z_{0.04}\). This Z-value is then used to build the confidence interval around our sample estimate, providing a range that is likely to contain the true difference in population proportions.

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Most popular questions from this chapter

Use a computer to demonstrate the truth of the theory presented in this section. a. The underlying assumptions are "the populations are normally distributed," and while conducting a hypothesis test for the equality of two standard deviations, it is assumed that the standard deviations are equal. Generate very large samples of two theoretical populations: \(N(100,20)\) and \(N(120,20)\) Find graphic and numerical evidence that the populations satisfy the assumptions. b. Randomly select 100 samples, each of size \(8,\) from both populations and find the standard deviation of each sample. c. Using the first sample drawn from each population as a pair, calculate the \(F\) t-statistic. Repeat for all samples. Describe the sampling distribution of the \(100 F \star\) -values using both graphic and numerical statistics. d. Generate the probability distribution for \(F(7,7),\) and compare it with the observed distribution of \(F \star .\) Do the two graphs agree? Explain.

A test that measures math anxiety was given to 50 male and 50 female students. The results were as follows: $$\begin{aligned}&\text { Males: } \quad \bar{x}=70.5, s=13.2\\\&\text { Females: } \bar{x}=75.7, s=13.6\end{aligned}$$ Construct a \(95 \%\) confidence interval for the difference between the mean anxiety scores.

Determine the \(p\) -value for each hypothesis test for the mean difference. a. \(\quad H_{o}: \mu_{d}=0\) and \(H_{a}: \mu_{d}>0,\) with \(n=20\) and \(t \star=1.86\) b. \(\quad H_{o}: \mu_{d}=0\) and \(H_{a}: \mu_{d} \neq 0,\) with \(n=20\) and \(t \star =-1.86\) c. \(\quad H_{o}: \mu_{d}=0\) and \(H_{a}: \mu_{d}<0,\) with \(n=29\) and \(t \star =-2.63\) d. \(\quad H_{o}: \mu_{d}=0.75\) and \(H_{a}: \mu_{d}>0.75,\) with \(n=10\) and \(t \star =3.57\)

10.80 One reason for being conservative when determining the number of degrees of freedom to use with the \(t\) distribution is the possibility that the population variances might be unequal. Extremely different values cause a lowering in the number of df used. Repeat Exercise 10.79 using theoretical normal distributions of \(N(100,9)\) and \(N(120,27)\) and both sample sizes of 8 Check all three properties of the sampling distribution: normality, its mean value, and its standard error. Describe in detail what you discover. Do you think we should be concerned about the choice of df? Explain.

The Soap and Detergent Association issued it fifth annual Clean Hands Report Card survey for 2009 From the answers to a series of hygiene-related question posed to American adults, it was found that \(62 \%\) of 44 women washed their hands more than 10 times per day while \(37 \%\) of 446 men did the same. Find the \(95 \%\) confidence interval for the difference in proportions of womer and men that washed their hands more than 10 times day.

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