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Perform these steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. It seems that people are choosing or finding it necessary to work later in life. Random samples of 200 men and 200 women age 65 or older were selected, and 80 men and 59 women were found to be working. At \(\alpha=0.01,\) can it be concluded that the proportions are different?

Short Answer

Expert verified
The test statistic is approximately 2.21, which does not exceed the critical value 2.58, so we fail to reject the null hypothesis. Thus, there is not enough evidence to conclude that the proportions are different.

Step by step solution

01

State the Hypotheses

We need to state the null hypothesis and the alternative hypothesis. The null hypothesis, denoted as \(H_0\), is that there is no difference in the proportion of men and women aged 65 or older who are working. This can be stated as \(H_0: p_1 = p_2\). The alternative hypothesis \(H_a\) is that the proportions are different, \(H_a: p_1 eq p_2\). The claim is in the alternative hypothesis.
02

Find the Critical Value(s)

Since we are conducting a two-tailed test at \(\alpha = 0.01\), we need to find the critical values for a standard normal distribution. Looking at a Z-table, the critical values for a two-tailed test at \(\alpha = 0.01\) are approximately \(-2.58\) and \(2.58\). These are the values that the test statistic must exceed in either direction to reject the null hypothesis.
03

Compute the Test Value

We compute the test value using the formula for the test statistic for the difference in two proportions. The formula is: \[ z = \frac{(\hat{p}_1 - \hat{p}_2)}{\sqrt{p(1-p)\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}} \] where \(\hat{p}_1\) and \(\hat{p}_2\) are the sample proportions, \(n_1\) and \(n_2\) are the sample sizes, and \(p\) is the pooled sample proportion.First, calculate the sample proportions: \[ \hat{p}_1 = \frac{80}{200} = 0.4, \quad \hat{p}_2 = \frac{59}{200} = 0.295 \]Next, calculate the pooled proportion: \[ p = \frac{80 + 59}{200 + 200} = \frac{139}{400} = 0.3475 \]Now, compute the test statistic: \[ z = \frac{0.4 - 0.295}{\sqrt{0.3475(1-0.3475)\left(\frac{1}{200} + \frac{1}{200}\right)}} = \frac{0.105}{\sqrt{0.3475 \times 0.6525 \times 0.01}} = \frac{0.105}{\sqrt{0.0022639}} = \frac{0.105}{0.0476} \approx 2.21 \]
04

Make the Decision

Compare the computed test statistic \(z \approx 2.21\) with the critical values \(-2.58\) and \(2.58\). Since \(2.21\) does not exceed \(2.58\) in absolute value, we fail to reject the null hypothesis. Therefore, there is not enough statistical evidence at the \(0.01\) level of significance to conclude that the proportions are different.
05

Summarize the Results

At the \(0.01\) level of significance, based on our hypothesis test, we cannot conclude there is a difference in the proportions of men and women aged 65 or older who are working. The data does not provide sufficient evidence to support the claim that the proportions are different.

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Key Concepts

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

Null Hypothesis
The concept of the null hypothesis is foundational in statistics, especially when performing hypothesis testing. A null hypothesis, often represented as \(H_0\), is a statement that implies no effect or no difference exists between groups or variables.
In the context of the exercise, the null hypothesis (\(H_0: p_1 = p_2\)) asserts that there is no difference in the proportion of men and women aged 65 or older who are working. This is a default position suggesting that any observed discrepancies are due to sampling error or chance, rather than a true difference.
  • The null hypothesis is always tested with the aim to possibly reject it.
  • Failing to reject \(H_0\) implies that there is no statistical evidence to support a difference in the populations in question.
By testing the null hypothesis, we aim to determine if there is enough statistical evidence in favor of a significant effect or difference. If the evidence is insufficient, the null hypothesis remains valid.
Alternative Hypothesis
The alternative hypothesis is the statement that opposes the null hypothesis and suggests that there is a significant difference or effect. It is typically denoted by \(H_a\) or \(H_1\).
In this problem, the alternative hypothesis (\(H_a: p_1 eq p_2\)) suggests that the proportions of men and women working at age 65 or older are different.
  • The alternative hypothesis is what the researcher aims to support or prove.
  • It is where the claim usually lies if one suspects that a noticeable difference exists. In this exercise, the claim is linked to the alternative hypothesis.
When the test results provide enough evidence against the null hypothesis, the alternative hypothesis is considered more likely. Acceptance of \(H_a\) implies acknowledging the presence of a difference or effect, based on the statistical evaluation performed.
Sample Proportions
Sample proportions are statistical measures that provide an estimate of the population proportions based on sample data.
The sample proportion is calculated by dividing the number of occurrences of an event by the total number of samples. In this exercise, the sample proportions are calculated as follows:
  • For men: \( \hat{p}_1 = \frac{80}{200} = 0.4\)
  • For women: \( \hat{p}_2 = \frac{59}{200} = 0.295\)
Sample proportions serve as the observed values used in hypothesis tests, forming the basis of calculations needed to compare observed differences against expected ones under the null hypothesis. They help determine if observed differences are statistically significant or if they likely occurred by chance.
Critical Value
Critical values are thresholds that help decide whether to reject the null hypothesis. For hypothesis tests, these values mark the cutoff points of the acceptance region of \(H_0\) in a standard normal distribution or other relevant distribution.In this exercise, since a significance level \(\alpha = 0.01\) is used for a two-tailed test, the critical values are:
  • \(-2.58\)
  • \(2.58\)
Any test statistic beyond these values (either smaller than \(-2.58\) or larger than \(2.58\)) would lead to the rejection of \(H_0\). Critical values act as the boundaries of decision making in hypothesis testing and must be carefully calculated based on the significance level and the nature of the test (one-tailed or two-tailed).If the test statistic falls within these critical values, it suggests that there is not enough evidence to reject the null hypothesis, thus maintaining its validity.
Pooled Proportion
The pooled proportion is a combined estimate of a proportion calculated from two independent samples, especially when testing for differences between two proportions. It is crucial when comparing proportions as it provides a weighted average estimate used in the standard error computation.In the context of the exercise, the pooled proportion \(p\) is calculated as:\[ p = \frac{80 + 59}{200 + 200} = \frac{139}{400} = 0.3475 \]
  • Pooled proportion takes into account the total number of successful outcomes and total sample size from both groups.
  • It is important for calculating the standard error accurately when the null hypothesis assumes equal proportions.
A correct calculation of the pooled proportion ensures that subsequent hypothesis test statistics are accurate and decisions on \(H_0\) can be confidently made based on those statistics.

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Most popular questions from this chapter

Whiting, Indiana, leads the "Top 100 Cities with the Oldest Houses" list with the average age of houses being 66.4 years. Farther down the list resides Franklin, Pennsylvania, with an average house age of 59.4 years. Researchers selected a random sample of 20 houses in each city and obtained the following statistics. At \(\alpha=0.05,\) can it be concluded that the houses in Whiting are older? Use the \(P\) -value method. $$\begin{array}{lrr} & \text { Whiting } & \text { Franklin } \\\\\hline \text { Mean age } & 62.1 \text { years } & 55.6 \text { years } \\\\\text { Standard deviation } & 5.4 \text { years } & 3.9 \text { years }\end{array}$$

Find the \(95 \%\) confidence interval of the difference in the distance that day students travel to school and the distance evening students travel to school. Two random samples of 40 students are taken, and the data are shown. Find the \(95 \%\) confidence interval of the difference in the means. $$ \begin{array}{lccc} & \bar{X} & \sigma & n \\ \hline \text { Day students } & 4.7 & 1.5 & 40 \\ \text { Evening Students } & 6.2 & 1.7 & 40 \end{array} $$

Perform these steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. A recent random survey of 100 individuals in Michigan found that 80 drove to work alone. A similar survey of 120 commuters in New York found that 62 drivers drove alone to work. Find the \(95 \%\) confidence interval for the difference in proportions.

Perform each of the following steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. A researcher wishes to see if there is a difference between the mean number of hours per week that a family with no children participates in recreational activities and a family with children participates in recreational activities. She selects two random samples and the data are shown. At \(\alpha=0.10\), is there a difference between the means? $$ \begin{array}{lrcl} & \bar{X} & \sigma & n \\ \hline \text { No children } & 8.6 & 2.1 & 36 \\ \text { Children } & 10.6 & 2.7 & 36 \end{array} $$

Perform each of these steps. Assume that all variables are normally or approximately normally distributed a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Overweight Dogs A veterinary nutritionist developed a diet for overweight dogs. The total volume of food consumed remains the same, but one-half of the dog food is replaced with a low-calorie "filler" such as canned green beans. Six overweight dogs were randomly selected from her practice and were put on this program. Their initial weights were recorded, and they were weighed again after 4 weeks. At the 0.05 level of significance, can it be concluded that the dogs lost weight? $$ \begin{array}{l|llllll} \text { Before } & 42 & 53 & 48 & 65 & 40 & 52 \\ \hline \text { After } & 39 & 45 & 40 & 58 & 42 & 47 \end{array} $$

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