91Ó°ÊÓ

Please provide the following information for Problems. (a) What is the level of significance? State the null and alternate hypotheses. (b) Check Requirements What sampling distribution will you use? What assumptions are you making? Compute the sample test statistic and corresponding \(z\) or \(t\) value as appropriate. (c) Find (or estimate) the \(P\) -value. Sketch the sampling distribution and show the area corresponding to the \(P\) -value. (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level \(\alpha ?\) (e) Interpret your conclusion in the context of the application. Note: For degrees of freedom \(d . f .\) not in the Student's \(t\) table, use the closest \(d . f .\) that is smaller. In some situations, this choice of \(d . f .\) may increase the \(P\) -value a small amount and therefore produce a slightly more "conservative" answer. Sociology: High School Dropouts This problem is based on information taken from Life in America's Fifty States by G. S. Thomas. A random sample of \(n_{1}=153\) people ages 16 to 19 was taken from the island of Oahu, Hawaii, and 12 were found to be high school dropouts. Another random sample of \(n_{2}=128\) people ages 16 to 19 was taken from Sweetwater County, Wyoming, and 7 were found to be high school dropouts. Do these data indicate that the population proportion of high school dropouts on Oahu is different (either way) from that of Sweetwater County? Use a \(1 \%\) level of significance.

Step by step solution

01

Define Level of Significance and Hypotheses

The level of significance is \( \alpha = 0.01 \).The null hypothesis \( H_0 \) states that the population proportions of high school dropouts on Oahu and Sweetwater County are equal: \( H_0: p_1 = p_2 \).The alternative hypothesis \( H_a \) states that the population proportions are not equal: \( H_a: p_1 eq p_2 \).

02

Check Requirements and Determine Distribution

For proportions comparison, we will use the normal approximation to the binomial distribution (z-test) since both sample sizes are large enough for normal approximation:* Oahu: \( n_1 = 153 \), \( x_1 = 12 \)* Sweetwater: \( n_2 = 128 \), \( x_2 = 7 \)Assumptions: The samples are random, independent, and the sample sizes are considered large enough under the binomial assumption to apply a normal approximation (\( np \) and \( n(1-p) \) are greater than 5 for both samples).

03

Compute the Sample Test Statistic

Compute the sample proportions: \( \hat{p}_1 = \frac{12}{153} \approx 0.0784 \) and \( \hat{p}_2 = \frac{7}{128} \approx 0.0547 \).The pooled sample proportion is:\[ \hat{p} = \frac{x_1 + x_2}{n_1 + n_2} = \frac{12 + 7}{153 + 128} \approx 0.067 \]The test statistic (z-score) is:\[ z = \frac{\hat{p}_1 - \hat{p}_2}{\sqrt{\hat{p}(1-\hat{p})\left(\frac{1}{n_1} + \frac{1}{n_2}\right)}} \]\[ z = \frac{0.0784 - 0.0547}{\sqrt{0.067 \times (1-0.067) \left(\frac{1}{153} + \frac{1}{128}\right)}} \approx 0.971 \]

04

Calculate the P-value

The test statistic \( z \approx 0.971 \) is used to determine the \( P \)-value, considering both tails of the standard normal distribution due to the two-tailed test.The \( P \)-value is approximately \( 2 \times P(Z > 0.971) \approx 0.331 \). This value can be found using a standard normal distribution table or calculator.

05

Decision to Reject or Fail to Reject Null Hypothesis

Since the \( P \)-value of \( 0.331 \) is greater than our significance level \( \alpha = 0.01 \), we fail to reject the null hypothesis \( H_0 \).

06

Interpretation of Conclusion

The conclusion is that there is not enough statistical evidence at the 1% significance level to suggest that the proportion of high school dropouts on Oahu is different from that of Sweetwater County.

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Key Concepts

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

Level of Significance

In hypothesis testing, the level of significance, denoted as \( \alpha \), is a threshold set before conducting a test. It determines how much risk of error you are willing to accept when drawing a conclusion. Essentially, it's the probability of making a Type I error, which occurs when you incorrectly reject a true null hypothesis. For example, a 1% level of significance means you accept a 1% risk of concluding that a relationship or effect exists when it actually does not.

• A smaller \( \alpha \) level signifies a more stringent criterion, reducing the chance of committing a Type I error but increasing the risk of a Type II error (failing to reject a false null hypothesis).
• In the given problem, \( \alpha = 0.01 \), indicating that strong evidence is required to reject the null hypothesis.

Null and Alternative Hypotheses

The null hypothesis \( H_0 \) and the alternative hypothesis \( H_a \) are foundational components of hypothesis testing. The null hypothesis posits that there is no effect or difference, serving as a default or starting assumption that will only be rejected if evidence strongly contradicts it.
The alternative hypothesis represents the conclusion you might reach if the evidence allows you to reject the null hypothesis. It suggests there is a meaningful effect or difference.

• In the exercise, the null hypothesis is \( H_0: p_1 = p_2 \), suggesting that the proportions of high school dropouts in Oahu and Sweetwater County are equal.
• The alternative hypothesis is \( H_a: p_1 eq p_2 \), indicating that the proportions are different.

Essentially, you are testing whether the data provide sufficient evidence to support the claim of a difference in dropout proportions between the two regions.

P-value

The \( P \)-value is a crucial aspect of hypothesis testing. It measures the probability of observing the test results, or something more extreme, assuming that the null hypothesis is true.

A smaller \( P \)-value indicates stronger evidence against the null hypothesis. In technical terms, if the \( P \)-value is less than or equal to \( \alpha \), the null hypothesis is rejected. However, if it is greater, the null hypothesis fails to be rejected.

• From the problem, the calculated \( P \)-value is approximately \( 0.331 \).
• Since this \( P \)-value exceeds the \( 1\% \) level of significance, we do not have enough evidence to reject the null hypothesis.

Z-test

The Z-test is a statistical procedure used to decide whether to accept or reject the null hypothesis when comparing sample proportions or means. For proportion comparisons, like in our scenario, the Z-test helps determine if any observed difference in sample proportions is statistically significant.

• It is based on the standard normal distribution, which is why it requires the calculation of a Z-score - a value representing how many standard deviations an element is from the mean.
• In our exercise, we used a Z-test due to the sufficiently large sample sizes, allowing for a normal approximation of the binomial distribution.

The calculated Z-score from the problem is approximately \( 0.971 \). This value is then used to find the corresponding \( P \)-value, which ultimately helps in making a decision about the null hypothesis.