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Consider the following hypothesis test: \\[ \begin{array}{l} H_{0}: p=.20 \\ H_{\mathrm{a}}: p \neq .20 \end{array} \\] A sample of 400 provided a sample proportion \(\bar{p}=.175\) a. Compute the value of the test statistic. b. What is the \(p\) -value? c. \(\quad\) At \(\alpha=.05,\) what is your conclusion? d. What is the rejection rule using the critical value? What is your conclusion?

Short Answer

Expert verified
Fail to reject \( H_0 \); the sample proportion is not statistically different from 0.20.

Step by step solution

01

Understanding the Hypothesis

We have the null hypothesis (\( H_0 \)) that the population proportion \( p = 0.20 \), and the alternative hypothesis (\( H_a \)) that \( p eq 0.20 \). This is a two-tailed test.
02

Compute the Test Statistic

The formula for the test statistic \( z \) for a proportion is \[ z = \frac{\bar{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}} \] where \( \bar{p} = 0.175 \), \( p_0 = 0.20 \), and \( n = 400 \). Substitute these values into the formula: \[ z = \frac{0.175 - 0.20}{\sqrt{\frac{0.20 \times 0.80}{400}}} = \frac{-0.025}{\sqrt{\frac{0.16}{400}}} = \frac{-0.025}{0.02} = -1.25 \].
03

Calculate the p-value

For a two-tailed test, the p-value is the probability that \( z \leq -1.25 \) or \( z \geq 1.25 \). Using a standard normal distribution table, find that \( P(z \leq -1.25) \approx 0.1056 \). Since this is a two-tailed test, double the probability: \( p\text{-value} = 2 \times 0.1056 = 0.2112 \).
04

Draw a Conclusion at \( \alpha = 0.05 \)

Compare the p-value to \( \alpha = 0.05 \). Since \( 0.2112 > 0.05 \), we fail to reject the null hypothesis \( H_0 \).
05

Rejection Rule Using Critical Value

For a two-tailed test with \( \alpha = 0.05 \), the critical values are \( z = \pm 1.96 \). The test statistic \( z = -1.25 \) does not fall beyond these critical values. Thus, as it is not in the rejection region, we fail to reject \( H_0 \). Our conclusion remains that there is not enough evidence to suggest \( p eq 0.20 \).

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Key Concepts

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

Test Statistic
The test statistic is an essential numerical measure that helps us perform hypothesis testing. It allows us to evaluate whether the observed sample data deviates significantly from what was expected under the null hypothesis. In our case, the sample proportion was 0.175, different from the hypothesized population proportion of 0.20. To compute the test statistic for a population proportion, we use the formula:\[z = \frac{\bar{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}} \]In this formula:
  • \( \bar{p} \) is the sample proportion.
  • \( p_0 \) is the hypothesized population proportion.
  • \( n \) is the sample size.
For our example:
  • \( \bar{p} = 0.175 \)
  • \( p_0 = 0.20 \)
  • \( n = 400 \)
By substituting these values into the formula, the calculation yields a test statistic \( z = -1.25 \). The negative sign indicates that the sample proportion is less than the hypothesized proportion. This number tells us how many standard deviations the sample proportion is from the hypothesized proportion.
Population Proportion
In statistics, the "population proportion" refers to the fraction or percentage of the total population that exhibits a particular characteristic. It's an unknown parameter that we estimate using sample data. Knowing the population proportion helps us make informed decisions about the whole population based on a smaller sample.In hypothesis testing, particularly when assessing proportions, we typically start with a hypothesized or known population proportion. In our problem, the population proportion we are testing against is 0.20, or 20%.
This number acts as a benchmark for evaluating our sample proportion.Understanding the population proportion is vital because it serves as the null hypothesis value against which the sample's behavior, represented by \( \bar{p} \), is compared. This comparison allows us to determine whether there is significant evidence to reject the null hypothesis in favor of an alternative one.
Two-tailed Test
A two-tailed test is a method in hypothesis testing used when we are interested in determining if there is any statistically significant difference between a sample estimate and a hypothesized population value, in either direction. This means the observed sample value could be significantly greater or less than the population value.Our test involves the null hypothesis stating the population proportion \( p = 0.20 \) and the alternative hypothesis \( p eq 0.20 \). The "not equal to" in the alternative hypothesis indicates a two-tailed approach.
In practice, this two-tailed test analyzes both ends of the distribution curve:
  • To capture if the proportion is significantly lower than 0.20
  • To check if it's significantly higher than 0.20
A two-tailed test is ideal when we have no prior direction or expectation of the difference. We are solely interested in detecting any deviation, whether positive or negative, from the hypothesized value.
p-value Calculation
The p-value is a crucial concept in hypothesis testing that quantifies the evidence against the null hypothesis. It represents the probability that the test statistic would be at least as extreme as the actually observed statistic, under the assumption that the null hypothesis is true.In a two-tailed test like ours, we calculate the p-value by determining the probability of the test statistic being either less than or greater than a specified value. For a standard normal distribution, these correspond to the probabilities of being below \( z = -1.25 \) or above \( z = 1.25 \).Using a standard normal distribution table:
  • The probability \( P(z \leq -1.25) \approx 0.1056 \)
Since this is a two-tailed test, we double this probability:\[ p\text{-value} = 2 \times 0.1056 = 0.2112 \]A higher p-value like 0.2112 means the sample result is not extremely unusual if the null hypothesis is true. Therefore, it indicates that we do not have sufficient evidence to reject the null hypothesis at the 0.05 significance level.

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Most popular questions from this chapter

The Employment and Training Administration reported that the U.S. mean unemployment insurance benefit was \(\$ 238\) per week (The World Almanac, 2003 ). A researcher in the state of Virginia anticipated that sample data would show evidence that the mean weekly unemployment insurance benefit in Virginia was below the national average. a. Develop appropriate hypotheses such that rejection of \(H_{0}\) will support the researcher's contention. b. For a sample of 100 individuals, the sample mean weekly unemployment insurance benefit was \(\$ 231\) with a sample standard deviation of \(\$ 80 .\) What is the \(p\) -value? c. \(\quad\) At \(a=.05,\) what is your conclusion? d. Repeat the preceding hypothesis test using the critical value approach.

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