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A recent report claims that college non-graduates get married at an earlier age than college graduates. To support the claim, random samples of size 100 were selected from each group, and the mean age at the time of marriage was recorded. The mean and standard deviation of the college non-graduates were \(22.5\) years and \(1.4\) years respectively, while the mean and standard deviation of the college graduates were 23 years and \(1.8\) years. Test the claims of the report at the \(.05\) level of sionificance.

Short Answer

Expert verified
We test the claim that college non-graduates get married at an earlier age than college graduates using a two-sample t-test with a significance level of \(0.05\). Our null hypothesis (H0) is that there is no significant difference in the mean age at the time of marriage between the two groups, and our alternative hypothesis (H1) is that college non-graduates get married at an earlier age. After calculating the t-test statistic and comparing it to the critical t-value, we make a decision whether to reject the null hypothesis or not. Based on this decision, we determine if there is enough evidence to support the claim.

Step by step solution

01

State the null and alternative hypotheses

We will be testing the following hypotheses: Null hypothesis (H0): There is no significant difference in the mean age at the time of marriage between college non-graduates and college graduates: \(μ1 = μ2\). Alternative hypothesis (H1): College non-graduates get married at an earlier age than college graduates: \(μ1 < μ2\).
02

Calculate the test statistic

To perform a two-sample t-test, we need to calculate the t-test statistic. The formula for the test statistic is: \[t = \frac{(\bar{X}_1 - \bar{X}_2) - (μ1 - μ2)}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}\] Where \(\bar{X}_1\) and \(\bar{X}_2\) are the sample means of the college non-graduates and college graduates respectively, \(s_1\) and \(s_2\) are the sample standard deviations, and \(n_1\) and \(n_2\) are the sample sizes. Plugging the given values: \[\bar{X}_1 = 22.5\] \[\bar{X}_2 = 23\] \[s_1 = 1.4\] \[s_2 = 1.8\] \[n_1 = n_2 = 100\] Now, calculating the t-test statistic: \[t = \frac{(22.5 - 23) - 0}{\sqrt{\frac{1.4^2}{100} + \frac{1.8^2}{100}}}\]
03

Calculate the degrees of freedom

Degrees of freedom (df) for the two-sample t-test can be approximated using the following formula: \[df \approx \frac{(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2})^2}{\frac{(s_1^2/n_1)^2}{n_1-1} + \frac{(s_2^2/n_2)^2}{n_2-1}}\] Plugging the given values: \[df \approx \frac{(\frac{1.4^2}{100} + \frac{1.8^2}{100})^2}{\frac{(1.4^2/100)^2}{100-1} + \frac{(1.8^2/100)^2}{100-1}}\] Calculate the df value.
04

Find the critical t-value and make a decision

We will now find the critical t-value using the given level of significance \(\alpha = 0.05\) and calculated degrees of freedom. Since it's a one-tailed test (\(μ1 < μ2\)), we will look for the critical value associated with \(\alpha = 0.05\) and the calculated df. Next, compare the calculated test statistic t with the critical t-value: - If the calculated value is less than the critical t-value, reject the null hypothesis. - If the calculated value is greater than or equal to the critical t-value, fail to reject the null hypothesis. Make a decision based on the comparison.
05

Interpret the results

Based on the decision made in step 4, we can draw one of the following conclusions: - If we rejected the null hypothesis, it means that there is enough evidence to support the claim that college non-graduates get married at an earlier age than college graduates. - If we failed to reject the null hypothesis, it means the observed difference could be due to chance, and we do not have enough evidence to support the claim that college non-graduates get married at an earlier age than college graduates.

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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 represented as \( H_0 \), serves as the starting assumption for statistical testing. In this exercise, the null hypothesis suggests that there is no significant difference in the mean age at marriage between college non-graduates and college graduates. More formally, it is stated as \( \mu_1 = \mu_2 \), where \( \mu_1 \) is the mean age of marriage for non-graduates, and \( \mu_2 \) is the mean age for graduates.

The null hypothesis is essential because it provides a baseline or a point of comparison. It effectively states the assumption that any observed difference in the sample means is due to random chance, rather than a genuine effect. When testing hypotheses, we aim to gather enough evidence to either support or reject the null hypothesis.

  • If the test supports the null hypothesis, the data does not provide enough evidence to suggest a difference between the groups.
  • If the null hypothesis is rejected, it points towards a statistically significant difference.
Alternative Hypothesis
While the null hypothesis implies no difference, the alternative hypothesis presents what we consider when the null is rejected. Denoted as \( H_1 \) in hypothesis testing, it aligns with the research question or claim.

In this scenario, the alternative hypothesis argues that college non-graduates get married at an earlier age than college graduates, formally expressed as \( \mu_1 < \mu_2 \). The alternative hypothesis is crucial because it captures the essence of the research being conducted; it is what the researchers suspect may be true based on preliminary observations, theories, or reports.

The goal of hypothesis testing is to collect enough evidence to either accept this alternative view or to continue supporting the status quo as reflected by the null hypothesis.

  • A successful test result for \( H_1 \) provides statistical evidence that the researchers' initial suspicion might be true.
  • If \( H_1 \) is not supported, it indicates the lack of sufficient evidence to make this claim.
Level of Significance
The level of significance, denoted by \( \alpha \), is a threshold used to determine when we should reject the null hypothesis. In this exercise, the level of significance is set at 0.05, which is a common choice in hypothesis testing.

This value represents the probability of rejecting the null hypothesis when it is actually true, known as a Type I error. Choosing \( \alpha = 0.05 \) indicates a 5% risk of making this type of error.

  • A smaller \( \alpha \) would mean a more stringent test, minimizing the risk of Type I errors but increasing the chance of Type II errors (failing to reject a false null hypothesis).
  • A larger \( \alpha \) increases the chance of committing a Type I error, making the test more lenient.

Understanding this concept is fundamental because it helps researchers balance the risk of committing errors in their conclusions. It also sets the critical value or cutoff for the test statistic, against which the computed statistic is compared to make decisions.
Degrees of Freedom
Degrees of freedom are an important factor in statistical tests like the t-test. They represent the number of values in a calculation that are free to vary. In a two-sample t-test, the degrees of freedom typically depend on the sizes of the two samples and their variances.

In this exercise, the degrees of freedom are calculated using an approximation formula. This accounts for the sample sizes and standard deviations given:

\[ df \approx \frac{(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2})^2}{(\frac{s_1^2/n_1)^2}{n_1-1} + \frac{(s_2^2/n_2)^2}{n_2-1}} \]

The degrees of freedom affect the shape of the t-distribution used for hypothesis testing and have a direct impact on the critical t-value.

  • More degrees of freedom typically imply a t-distribution that approximates a normal distribution more closely.
  • Fewer degrees of freedom mean a t-distribution with heavier tails, which affects the critical values needed for a given level of significance.
Understanding how degrees of freedom influence the outcome of statistical tests is crucial for accurate interpretation and decision-making.

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Most popular questions from this chapter

Suppose a coin is assumed to be biased and it is believed that the probability, p, of obtaining a head on one toss is \(1 / 10\). Suppose the coin is tossed twice. Is there any value for \(\mathrm{k}\), the number of heads; for which we can reject the null hypothesis \(\mathrm{p}=1 / 10\) in favor of the alternate hypothesis \(\mathrm{p}>1 / 10\) at the \(1 \%\) level of significance? Suppose the null hypothesis is \(p=1 / 2\). Is there any value, \(k\), for which we can reject this null hypothesis in favor of the alternate hypothesis \(p<1 / 2\) at the \(1 \%\) level of significance when the coin is tossed twice?

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