91Ó°ÊÓ

The National Health Statistics Reports dated Oct. 22, 2008 , included the following information on the heights (in.) for non-Hispanic white females: \begin{tabular}{lccc} Age & Sample Size & Sample Mean & Std. Error Mean \\ \hline \(20-39\) & 866 & \(64.9\) & \(.09\) \\ 60 and older & 934 & \(63.1\) & \(.11\) \\ \hline \end{tabular} a. Calculate and interpret a confidence interval at confidence level approximately \(95 \%\) for the difference between population mean height for the younger women and that for the older women. b. Let \(\mu_{1}\) denote the population mean height for those aged \(20-39\) and \(\mu_{2}\) denote the population mean height for those aged 60 and older. Interpret the hypotheses \(H_{0}: \mu_{1}-\) \(\mu_{2}=1\) and \(H_{a}: \mu_{1}-\mu_{2}>1\), and then carry out a test of these hypotheses at significance level .001 using the rejection region approach. c. What is the \(P\)-value for the test you carried out in (b)? Based on this \(P\)-value, would you reject the null hypothesis at any reasonable significance level? Explain. d. What hypotheses would be appropriate if \(\mu_{1}\) referred to the older age group, \(\mu_{2}\) to the younger age group, and you wanted to see if there was compelling evidence for concluding that the population mean height for younger women exceeded that for older women by more than 1 in.?

Step by step solution

01

Identify the Statistical Formulas

To calculate the confidence interval, we will use the formula for the confidence interval for the difference in means: \( (\bar{x}_1 - \bar{x}_2) \pm z^* \times \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} \), where the sample means, standard errors, and sizes are provided.

02

Compute the Standard Error Difference

Using the standard errors provided, we calculate the standard error for the difference in means: \( \sqrt{(0.09)^2 + (0.11)^2} = \sqrt{0.0081 + 0.0121} = \sqrt{0.0202} \approx 0.142 \).

03

Determine the Z-Score for 95% Confidence Level

For a 95% confidence level, the Z-score \( z^* \) is approximately 1.96.

04

Calculate the Confidence Interval

Using the formula, calculate \( (64.9 - 63.1) \pm 1.96 \times 0.142 \). This simplifies to \( 1.8 \pm 0.278 \). The confidence interval is (1.522, 2.078).

05

Hypotheses Interpretation for Part b

The null hypothesis \( H_0: \mu_1 - \mu_2 = 1 \) suggests the difference in means is 1, while the alternative \( H_a: \mu_1 - \mu_2 > 1 \) suggests the difference is greater than 1.

06

Compute the Test Statistic

The test statistic is computed as \( \frac{(\bar{x}_1 - \bar{x}_2) - 1}{\text{SE difference}} = \frac{1.8 - 1}{0.142} \approx 5.63 \).

07

Determine the Rejection Region

For a significance level of \( \alpha = 0.001 \), the critical Z value is approximately 3.291. Since 5.63 > 3.291, we reject \( H_0 \).

08

Calculate the P-Value

Using the Z-table, find the P-value for the test statistic 5.63, which is very close to 0.

09

Interpretation of the P-Value

The P-value is much less than 0.001, so we would reject the null hypothesis at any reasonable significance level.

10

Correct Hypotheses Setup for Part d

If \( \mu_1 \) refers to the older group and \( \mu_2 \) to the younger, the hypotheses for testing if younger is more than 1 inch taller become \( H_0: \mu_2 - \mu_1 = 1 \) and \( H_a: \mu_2 - \mu_1 > 1 \).

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Key Concepts

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

Hypothesis Testing

Hypothesis testing is a critical part of statistics used to make inferences or decisions about a population parameter based on sample data. It involves stating a specific hypothesis about a parameter and using sample data to test whether sufficient evidence exists to support this hypothesis. This typical process begins by formulating two hypotheses: the null hypothesis (denoted as \( H_0 \)) and the alternative hypothesis (denoted as \( H_a \)).

• The null hypothesis usually indicates no effect or no difference, functioning as a baseline for the test. For instance, in the given exercise, \( H_0: \mu_1 - \mu_2 = 1 \) states that the mean height difference between the two age groups is exactly 1 inch.
• The alternative hypothesis reflects the claim we are trying to find evidence for. For example, \( H_a: \mu_1 - \mu_2 > 1 \) suggests the difference exceeds 1 inch, indicating that younger women are taller on average.

Deciding between these hypotheses involves collecting data, calculating a test statistic, and evaluating this statistic to see how compatible the data is with these hypotheses.

Z-Score

The Z-score plays a fundamental role in hypothesis testing and statistics as a whole. It measures the number of standard deviations a data point or sample mean is from the population mean. Specifically, it enables statisticians to understand where a data point lies when compared to a standard normal distribution.
In the context of hypothesis testing, the Z-score helps determine how extreme the observed sample statistic is under the assumption that the null hypothesis is true. For example, computing a Z-score using the formula: \[z = \frac{\bar{x}_1 - \bar{x}_2 - \Delta}{SE_{difference}}\]where \( \Delta \) is the hypothesized difference between population means, gives us a Z-score of 5.63 in this particular exercise. This suggests that the sample result is 5.63 standard deviations away from the hypothesized population mean difference, under the null hypothesis. This extremity in Z-score translates to the probability of observing such a statistic, assuming the null hypothesis is true, which is critical for making informed decisions in hypothesis testing.

Rejection Region

The rejection region is an essential concept in hypothesis testing that refers to a range of values for which the null hypothesis is rejected. The boundary of this region is determined by the significance level \( \alpha \), which represents the probability of rejecting the null hypothesis when it is actually true.

• For instance, at a significance level \( \alpha = 0.001 \), the rejection region for a right-tailed test like the one in the exercise begins at a critical Z-score of approximately 3.291.
• Any test statistic that falls into this region (greater than 3.291 for our test) indicates that the observed data is sufficiently inconsistent with the null hypothesis in favor of the alternative hypothesis. In our example, having a test statistic of 5.63 places it well within the rejection region.

By linking observed data to critical values, the rejection region offers a systematic and objective method to decide whether to support or reject the null hypothesis.

P-Value

A P-value quantifies the likelihood of obtaining a test statistic at least as extreme as the observed one, under the assumption that the null hypothesis is true. It provides a continuous measure of evidence against the null hypothesis: the smaller the P-value, the stronger the evidence against it.
In our exercise, the test statistic (Z = 5.63) yields a very tiny P-value, approaching zero. This indicates a very low probability of observing such an extreme result by random chance if the null hypothesis was true, thereby providing compelling evidence to reject the null.

• A common procedure is to compare the P-value with the chosen significance level \( \alpha \). If the P-value is less than \( \alpha \), we reject the null hypothesis.
• For example, with \( \alpha = 0.001 \), our almost-zero P-value suggests rejecting the null hypothesis at any reasonable significance level, confirming that younger women in the sample are taller by more than 1 inch compared to the older group.

The P-value approach offers precision, especially when detection beyond binary outcomes (reject or do not reject) is desired.