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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. A random survey of 80 women who were victims of violence found that 24 were attacked by relatives. A random survey of 50 men found that 6 were attacked by relatives. At \(\alpha=0.10,\) can it be shown that the percentage of women who were attacked by relatives is greater than the percentage of men who were attacked by relatives?

Short Answer

Expert verified
The claim is supported; women are more likely attacked by relatives than men.

Step by step solution

01

State the Hypotheses and Identify the Claim

We are testing to see if the proportion of women who were attacked by relatives is greater than the proportion for men. Let \( p_1 \) be the proportion of women attacked by relatives and \( p_2 \) be the proportion of men attacked by relatives. The null hypothesis is \( H_0: p_1 \leq p_2 \). The alternative hypothesis is \( H_1: p_1 > p_2 \). This is a right-tailed test. The claim is that the percentage of women attacked by relatives is greater.
02

Find the Critical Value

Using \( \alpha = 0.10 \), look up the critical value for a right-tailed test from the standard normal distribution table. The critical value \( z_c \) at \( \alpha = 0.10 \) is approximately 1.28.
03

Compute the Test Value

Calculate the sample proportions: \( \hat{p}_1 = \frac{24}{80} = 0.3 \) and \( \hat{p}_2 = \frac{6}{50} = 0.12 \). Then, compute the pooled sample proportion: \( \hat{p} = \frac{24 + 6}{80 + 50} = \frac{30}{130} \approx 0.231 \). Now, calculate the standard error: \[ SE = \sqrt{\hat{p}(1 - \hat{p}) \left(\frac{1}{80} + \frac{1}{50}\right)} = \sqrt{0.231(1 - 0.231) \left(\frac{1}{80} + \frac{1}{50}\right)} \approx 0.068 \]. Finally, compute the test statistic: \[ z = \frac{\hat{p}_1 - \hat{p}_2}{SE} = \frac{0.3 - 0.12}{0.068} \approx 2.65 \].
04

Make the Decision

Compare the test statistic to the critical value. Since \( 2.65 > 1.28 \), we reject the null hypothesis.
05

Summarize the Results

There is sufficient evidence at the \( \alpha = 0.10 \) level to support the claim that a greater percentage of women than men were attacked by relatives.

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Key Concepts

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

Null Hypothesis
When starting a hypothesis test, one of the first things we need to establish is the null hypothesis. The null hypothesis, denoted as \(H_0\), serves as a statement of no effect or no difference. It is assumed to be true until evidence indicates otherwise.

In our example, the null hypothesis is expressing that the proportion of women attacked by relatives is less than or equal to the proportion of men attacked by relatives. Mathematically, we write this as \(H_0: p_1 \leq p_2\). This hypothesis represents our starting belief, suggesting that any observation contrary to this is due to random chance rather than a difference between groups.

Understanding the null hypothesis's role is crucial, as it sets the base we compare our observations against. Only when the evidence against the null hypothesis is clear and strong (like in our example) do we consider rejecting it.
Alternative Hypothesis
The alternative hypothesis, denoted as \(H_1\), is the statement you aim to support with your data. In essence, it reflects a new effect or a specific difference that may exist due to some factor.

In the given scenario, the alternative hypothesis posits that the proportion of women attacked by relatives is indeed greater than the proportion of men attacked by relatives. This is stated mathematically as \(H_1: p_1 > p_2\). The alternative hypothesis aligns with what we suspect might be true based on prior knowledge or theory.

Choosing the correct form of the alternative hypothesis is vital, as it determines the direction of the test. Here, it's a right-tailed test since we're interested in showing that one proportion is specifically greater than the other. Thus, it defines how we evaluate our collected data and draw conclusions from it.
Critical Value
The critical value is a threshold that demarcates whether you reject the null hypothesis. It is determined based on the level of significance \(\alpha\), which in this example is set at 0.10.

By referring to a standard normal distribution table, we find that for a right-tailed test, the critical value \(z_c\) is approximately 1.28 when \(\alpha = 0.10\). This means that if our test statistic exceeds this value, we have enough evidence to reject the null hypothesis in favor of the alternative hypothesis.

The critical value ensures that results lying beyond this point are rare under the null hypothesis, providing justification for its rejection. Hence, it's a pivotal part of hypothesis testing, helping convert observed data into logical decisions.
Test Statistic
The test statistic is a calculated value used to compare against the critical value to decide on the null hypothesis. It's derived from our sample data, in this case using the difference between sample proportions.

First, the sample proportions for women and men are calculated as \(\hat{p}_1 = 0.3\) and \(\hat{p}_2 = 0.12\), respectively. Then, we determine a pooled sample proportion \(\hat{p} = 0.231\) to reflect the overall attack rate.

The standard error is then computed to measure variability, and finally, our test statistic \(z\) is calculated: \[ z = \frac{\hat{p}_1 - \hat{p}_2}{SE} = \frac{0.3 - 0.12}{0.068} \approx 2.65 \]The obtained test statistic of 2.65 is compared to the critical value. Because it exceeds 1.28, it suggests that the null hypothesis doesn’t hold, allowing us to support the alternative claim. This method of using a test statistic ensures our conclusions are grounded in statistical analysis.

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Most popular questions from this chapter

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. The percentages of ontime arrivals for major U.S. airlines range from 68.6 to \(91.1 .\) Two regional airlines were surveyed with the following results. At \(\alpha=0.01\), is there a difference in proportions? $$ \begin{array}{lcc} & \text { Airline A } & \text { Airline B } \\ \hline \text { No. of flights } & 300 & 250 \\ \text { No. of on-time flights } & 213 & 185 \end{array} $$

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. In today's economy, everyone has become savings savvy. It is still believed, though, that a higher percentage of women than men clip coupons. A random survey of 180 female shoppers indicated that 132 clipped coupons while 56 out of 100 men did so. At \(\alpha=0.01\), is there sufficient evidence that the proportion of couponing women is higher than the proportion of couponing men? Use the \(P\) -value method.

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. Cholesterol Levels A medical researcher wishes to see if he can lower the cholesterol levels through diet in 6 people by showing a film about the effects of high cholesterol levels. The data are shown. At \(\alpha=0.05,\) did the cholesterol level decrease on average? $$ \begin{array}{lrrrrrr} \text { Patient } & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \text { Before } & 243 & 216 & 214 & 222 & 206 & 219 \\ \hline \text { After } & 215 & 202 & 198 & 195 & 204 & 213 \end{array} $$

A random sample of monthly gasoline prices was taken from 2011 and from \(2015 .\) The samples are shown. Using \(\alpha=0.01,\) can it be concluded that gasoline cost more in 2015 ? Use the \(P\) -value method. $$\begin{array}{l|lllllll}\mathbf{2 0 1 1} & 2.02 & 2.47 & 2.50 & 2.70 & 3.13 & 2.56 & \\\\\hline \mathbf{2 0 1 5} & 2.36 & 2.46 & 2.63 & 2.76 & 3.00 & 2.85 & 2.77\end{array}$$

The number of grams of carbohydrates contained in 1 -ounce servings of randomly selected chocolate and nonchocolate candy is listed here. Is there sufficient evidence to conclude that the difference in the means is statistically significant? Use \(\alpha=0.10\) $$ \begin{array}{lllllllll}\text { Chocolate: } & 29 & 25 & 17 & 36 & 41 & 25 & 32 & 29 \\\& 38 & 34 & 24 & 27 & 29 & & & \\\\\text { Nonchocolate: } & 41 & 41 & 37 & 29 & 30 & 38 & 39 & 10 \\ & 29 & 55 & 29 & & & & &\end{array}$$

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