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For Exercises 5 through \(20,\) 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. Runaways A corrections officer read that \(58 \%\) of runaways are female. He believes that the percentage is higher than \(58 .\) He selected a random sample of 90 runaways and found that 63 were female. At \(\alpha=0.05,\) can you conclude that his belief is correct?

Short Answer

Expert verified
The officer's belief is correct; more than 58% of runaways are female.

Step by step solution

01

State the Hypotheses

We start by setting up our null and alternative hypotheses. The null hypothesis \(H_0\) states that the proportion of female runaways is \(p = 0.58\). The alternative hypothesis \(H_1\) claims that the proportion is greater than \(0.58\). Thus, our hypotheses are:\[H_0: p = 0.58 \H_1: p > 0.58\]The claim is that more than 58% of runaways are female.
02

Find the Critical Value

For this problem, we need to determine the critical value for a right-tailed test at \(\alpha = 0.05\). We use a standard normal distribution (Z-distribution) since the sample size is large. The critical value for \(\alpha = 0.05\) in a right-tailed test is approximately 1.645. This means if our test statistic (Z) is greater than 1.645, we will reject the null hypothesis.
03

Compute the Test Value

Next, we calculate the test statistic using the formula for a proportion:\[Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}\]where \(\hat{p} = \frac{63}{90} = 0.7\) is the sample proportion, \(p_0 = 0.58\) is the hypothesized population proportion, and \(n = 90\) is the sample size.\[Z = \frac{0.7 - 0.58}{\sqrt{\frac{0.58 \times 0.42}{90}}} = \frac{0.12}{\sqrt{0.0027}} \approx \frac{0.12}{0.052} \approx 2.31\]
04

Make the Decision

Since the calculated test statistic \(Z = 2.31\) is greater than the critical value of 1.645, we reject the null hypothesis \(H_0\). This suggests that the sample data provides sufficient evidence to support the claim that more than 58% of runaways are female.
05

Summarize the Results

Based on our hypothesis test, there is significant evidence at the \(\alpha = 0.05\) level to conclude that the proportion of female runaways is indeed greater than 58%. The officer's belief that the percentage of female runaways is higher is supported.

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Key Concepts

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

Understanding the Null Hypothesis
In hypothesis testing, the null hypothesis acts as a baseline statement of "no effect" or "no change." It is denoted by \(H_0\). In terms of proportion testing, it represents an initial hypothesis about the population proportion. In our exercise, the null hypothesis is that the proportion \(p\) of female runaways is \(58\%\), expressed as \(H_0: p = 0.58\).

It serves as a statement to be tested. Unless we have strong evidence, we assume the null hypothesis is true. It is like the "default setting" in our experiment. Only if our test provides significant results that go against the null, do we consider alternative views. In this case, the alternative hypothesis \(H_1: p > 0.58\) suggests a different scenario where the percentage of female runaways is more than 58%. Always frame the null and alternative hypotheses clearly, as they form the foundation of hypothesis testing.
Exploring Critical Values
Once the hypotheses are set, the next step is to locate the critical value. But what exactly is a critical value? Simply put, a critical value is a threshold that determines the boundary for decision-making in hypothesis testing. Depending on where the test statistic falls relative to this critical value, we decide to either reject or fail to reject the null hypothesis.

In our context, for a right-tailed test with a significance level \(\alpha = 0.05\), we find the critical value using a Z-distribution. The critical value for this specific setup is approximately 1.645. This means if the calculated Z-test statistic surpasses 1.645, it lies in the rejection region, indicating we have enough evidence to reject the null hypothesis.

Think of it like setting a mark on a ruler: if your measurement extends beyond the mark, it indicates change or difference from what is assumed under the null hypothesis.
Decoding the Z-Distribution
The Z-distribution is a important concept in statistics, particularly for hypothesis testing involving proportions. It is often synonymous with the standard normal distribution, which is a bell-shaped curve that's symmetrical around the mean of zero.

When testing proportions, we use the Z-distribution to calculate the test statistic, known as the Z-score. This score helps us understand how far our sample result is from the hypothesized population proportion under the null hypothesis. In our exercise, we computed a Z-score using the formula: \[Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}\]where \(\hat{p} = 0.7\) is the sample proportion, and \(p_0 = 0.58\) is the hypothesized proportion.

Understanding the Z-distribution is crucial because it provides a standardized way of measuring probabilities and making informed decisions based on sample data.
Grasping Proportion Testing
Proportion testing is a type of hypothesis test used to determine whether an observed sample proportion is significantly different from a claimed population proportion. In other words, it's about understanding whether the segment of a population exhibiting a specific characteristic diverges from a known proportion under certain conditions.

In the exercise provided, proportion testing helps us evaluate whether the percentage of female runaways (sample proportion of 0.7) significantly exceeds a claimed proportion of 0.58. We essentially compare these proportions via a calculated test statistic, which in our case is evaluated using the Z-test for proportions.

Ultimately, proportion testing provides insights into how sample findings align (or don't align) with a specified population hypothesis, making it a powerful tool in research and real-world decision-making scenarios. By testing the proportions, we gain valuable information about the validity of our assumptions regarding a population.

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Most popular questions from this chapter

For Exercises I through 25, 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 diagrams to show the critical region (or regions), and use the traditional method of hypothesis testing unless otherwise specified. Sick Days A manager states that in his factory, the average number of days per year missed by the employees due to illness is less than the national average of \(10 .\) The following data show the number of days missed by 40 randomly selected employees last year. Is there sufficient evidence to believe the manager's statement at \(\alpha=0.05 ? \sigma=3.63 .\) Use the \(P\) -value method. $$ \begin{array}{cccccccc}{0} & {6} & {12} & {3} & {3} & {5} & {4} & {1} \\ {3} & {9} & {6} & {0} & {7} & {6} & {3} & {4} \\ {7} & {4} & {7} & {1} & {0} & {8} & {12} & {3} \\ {2} & {5} & {10} & {5} & {15} & {3} & {2} & {5} \\ {3} & {11} & {8} & {2} & {2} & {4} & {1} & {9}\end{array} $$

What is meant by a type I error? A type II error? How are they related?

For Exercises I through 25, 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 diagrams to show the critical region (or regions), and use the traditional method of hypothesis testing unless otherwise specified. Stopping Distances A study found that the average stopping distance of a school bus traveling 50 miles per hour was 264 feet. A group of automotive engineers decided to conduct a study of its school buses and found that for 20 randomly selected buses, the average stopping distance of buses traveling 50 miles per hour was 262.3 feet. The standard deviation of the population was 3 feet. Test the claim that the average stopping distance of the company's buses is actually less than 264 feet. Find the \(P\) -value. On the basis of the \(P\) -value, should the null hypothesis be rejected at \(\alpha=0.01 ?\) Assume that the variable is normally distributed.

For Exercises 7 through \(23,\) perform each of the following steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Find the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Assume that the population is approximately normally distributed. Sleep Time A person read that the average number of hours an adult sleeps on Friday night to Saturday morning was 7.2 hours. The researcher feels that college students do not sleep 7.2 hours on average. The researcher randomly selected 15 students and found that on average they slept 8.3 hours. The standard deviation of the sample is 1.2 hours. At \(\alpha=0.05,\) is there enough evidence to say that college students do not sleep 7.2 hours on average?

For Exercises 5 through \(20,\) assume that the variables are normally or approximately normally distributed. Use the traditional method of hypothesis testing unless otherwise specified. Manufactured Machine Parts A manufacturing process produces machine parts with measurements the standard deviation of which must be no more than \(0.52 \mathrm{mm}\). A random sample of 20 parts in a given lot revealed a standard deviation in measurement of \(0.568 \mathrm{mm}\). Is there sufficient evidence at \(\alpha=0.05\) to conclude that the standard deviation of the parts is outside the required guidelines?

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