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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. Undergraduate Enrollment It has been found that \(85.6 \%\) of all enrolled college and university students in the United States are undergraduates. A random sample of 500 enrolled college students. A random sample of 500 enrolled college students in a particular state revealed that 420 of them were undergraduates. Is there sufficient evidence to conclude that the proportion differs from the national percentage? Use \(\alpha=0.05 .\)

Short Answer

Expert verified
The proportion does not differ significantly from the national percentage.

Step by step solution

01

State the Hypotheses and Identify the Claim

First, we need to define the null and alternative hypotheses. - Null Hypothesis \( H_0 \): The proportion of undergraduates in the state is equal to the national proportion, i.e., \( p = 0.856 \).- Alternative Hypothesis \( H_a \): The proportion of undergraduates in the state differs from the national proportion, i.e., \( p eq 0.856 \). The claim is in the alternative hypothesis.
02

Find the Critical Value(s)

We use a significance level of \( \alpha = 0.05 \). Since this is a two-tailed test, we split the significance level in half, giving \( \alpha/2 = 0.025 \) on each side.Using a Z-table, the critical values for a two-tailed test at \( \alpha/2 = 0.025 \) are approximately \( Z = -1.96 \) and \( Z = 1.96 \).
03

Compute the Test Value

The test statistic is calculated using the formula for the standard normal test: \[ Z = \frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}} \]Where:- \( \hat{p} = \frac{420}{500} = 0.84 \) (sample proportion)- \( p = 0.856 \) (population proportion)- \( n = 500 \) (sample size)Substitute the values: \[ Z = \frac{0.84 - 0.856}{\sqrt{\frac{0.856(1-0.856)}{500}}} \] Calculate the values step by step:1. \( 1 - 0.856 = 0.144 \)2. \( 0.856 \times 0.144 = 0.123264 \)3. \( \frac{0.123264}{500} = 0.000246528 \)4. \( \sqrt{0.000246528} = 0.0157 \)5. \( \frac{0.84 - 0.856}{0.0157} = \frac{-0.016}{0.0157} \approx -1.019 \)Thus, the test statistic \( Z \) is approximately \( -1.019 \).
04

Make the Decision

The computed test statistic \( Z = -1.019 \) is between the critical values \( -1.96 \) and \( 1.96 \), falling into the acceptance region.Since \( -1.96 < -1.019 < 1.96 \), we do not reject the null hypothesis.
05

Summarize the Results

There is not sufficient statistical evidence at the \( \alpha = 0.05 \) level to conclude that the proportion of undergraduates in the state differs from the national proportion of 85.6%.

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Key Concepts

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

Null Hypothesis
The null hypothesis, often denoted as \( H_0 \), is a statement used in statistics that suggests there is no significant effect or difference in a particular situation or experiment. It serves as a starting point for statistical testing. In hypothesis testing, it assumes that any kind of difference or significance you see in a test is purely by chance.

For example, in our exercise, the null hypothesis \( H_0 \) states: "The proportion of undergraduates in the state is equal to the national proportion," specifically \( p = 0.856 \). This implies that whatever sample proportion we measure, any variance from the national proportion is due to random sampling error.

The main goal is to test this hypothesis against its counterpart, the alternative hypothesis, to see if we can reject it based on the data we have. Rejection occurs if the statistical evidence suggests that what's observed is unlikely to happen if the null hypothesis is true. Hence, the null hypothesis acts like a baseline that enables comparison.
Alternative Hypothesis
Think of the alternative hypothesis as the mirror image of the null hypothesis. Represented usually by \( H_a \) or \( H_1 \), this hypothesis is what we set out to prove and reflects the researcher's theory of what might be true. If, after testing, the null hypothesis is rejected, the alternative hypothesis is accepted as true.

In the context of our specific exercise, the alternative hypothesis is: "The proportion of undergraduates in the state differs from the national proportion," expressed as \( p eq 0.856 \). Here, this indicates a hypothesis where the state's proportion is different, and not due to chance alone.

To determine if the alternative hypothesis should be accepted or not, statisticians compare the calculated test statistic to the critical value within chosen significance levels. If the evidence aligns more with \( H_a \) than with \( H_0 \), we lean towards its acceptance.
Critical Value
In hypothesis testing, critical values act as the benchmark, a dividing line that helps you make a decision on whether to reject the null hypothesis. These values correspond to the significance level, denoted by \( \alpha \), which is the probability of rejecting the null hypothesis when it is actually true, also known as the Type I error.

To determine these values in a two-tailed test scenario, as in our exercise, the significance level \( \alpha = 0.05 \) is split into two parts: \( \alpha/2 = 0.025 \) on each side. By consulting a Z-table, the critical Z-values are found to be approximately \( Z = -1.96 \) and \( Z = 1.96 \).

The test statistic is then compared to these critical values. If the test statistic is outside of these boundaries, the null hypothesis is rejected. Conversely, if it falls within this range, we continue to accept \( H_0 \), suggesting that any observed difference is not statistically significant at the stated level of \( \alpha \).
Test Statistic
A test statistic is a standardized value that is calculated from sample data during a hypothesis test. It's used to determine whether to reject the null hypothesis. Common types of test statistics include the Z-score, T-score, Chi-square, etc., depending on the data and test being used.

In our given exercise, the Z-score formula is utilized, expressed as: \[ Z = \frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}} \]Where:
  • \( \hat{p} = 0.84 \) (sample proportion)
  • \( p = 0.856 \) (population proportion)
  • \( n = 500 \) (sample size)
Substituting these values into the formula provides the test statistic, \( Z \approx -1.019 \).

The test statistic essentially gauges how far our sample statistic is from the population parameter, in standard deviation units. In conclusion, this calculated Z-score tells us where our sample proportion sits in relation to the hypothesized population proportion, ultimately assisting in making a decision about the null hypothesis.

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Most popular questions from this chapter

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?

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. Television Viewing by Teens Teens are reported to watch the fewest total hours of television per week of all the demographic groups. The average television viewing for teens on Sunday from 1: 00 to 7: 00 P.M. is 58 minutes. A random sample of local teens disclosed the following times for Sunday afternoon television viewing. At \(\alpha=0.01,\) can it be concluded that the average is greater than the national viewing time? (Note: Change all times to minutes.) $$ \begin{array}{llll}{2: 30} & {2: 00} & {1: 30} & {3: 20} \\ {1: 00} & {2: 15} & {1: 50} & {2: 10} \\ {1: 30} & {2: 30} & {}\end{array} $$

Working at Home Workers with a formal arrangement with their employer to be paid for time worked at home worked an average of 19 hours per week. A random sample of 15 mortgage brokers indicated that they worked a mean of 21.3 hours per week at home with a standard deviation of 6.5 hours. At $$\alpha=0.05,$$ is there sufficient evidence to conclude a difference? Construct a $$95 \%$$ confidence interval for the true mean number of paid working hours at home. Compare the results of your confidence interval to the conclusion of your hypothesis test and discuss the implications.

Explain the difference between a one-tailed and a two-tailed test.

Newspaper Reading Times A survey taken several years ago found that the average time a person spent reading the local daily newspaper was 10.8 minutes. The standard deviation of the population was 3 minutes. To see whether the average time had changed since the newspaper's format was revised, the newspaper editor surveyed 36 individuals. The average time that these 36 randomly selected people spent reading the paper was 12.2 minutes. At $$\alpha=0.02,$$ is there a change in the average time an individual spends reading the newspaper? Find the $$98 \%$$ confidence interval of the mean. Do the results agree? Explain.

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