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Chapter 9: Problem 14

Has the consumption of red meat decreased over the last 10 years? A researcher selected hospital nutrition records for 400 subjects surveyed 10 years ago and compared the average amount of beef consumed per year to amounts consumed by an equal number of subjects interviewed this year. The data are given in the table. $$ \begin{array}{lcc} \hline & \text { Ten Years Ago } & \text { This Year } \\ \hline \text { Sample Mean } & 73 & 63 \\ \text { Sample Standard Deviation } & 25 & 28 \\ \hline \end{array} $$ a. Do the data present sufficient evidence to indicate that per-capita beef consumption has decreased over the last 10 years? Test at the \(1 \%\) level of significance. b. Find a \(99 \%\) lower confidence bound for the difference in the average per-capita beef consumptions for the two groups. Does the confidence bound confirm your conclusions in part a? Explain. What additional information does the confidence bound give you?

Short Answer

Expert verified
Based on the given step-by-step solution, we can conclude that there is sufficient evidence to suggest that the per-capita beef consumption has decreased over the last 10 years. The 99% lower confidence bound for the difference in average per-capita beef consumption between the two groups is 6.15 units, indicating that we are 99% confident that the true difference in consumption is at least 6.15 units.

Step by step solution

01

State the null and alternative hypotheses

The null hypothesis (H0) states that there is no significant difference in per-capita beef consumption between the two groups, while the alternative hypothesis (H1) states that the per-capita beef consumption has decreased. \(H_0: \mu_1 - \mu_2 = 0\) \(H_1: \mu_1 - \mu_2 > 0\)
02

Compute the test statistic

To compute the test statistic, we will use the formula for two-sample z-test, where \(\bar{X}_1\) and \(\bar{X}_2\) are the sample means, \(s_1\) and \(s_2\) are the sample standard deviations, n1 and n2 are the sample sizes, and \(\sigma_{\bar{X}_1 - \bar{X}_2}\) is the standard error of the difference between the sample means: $$z = \frac{(\bar{X}_1 - \bar{X}_2) - (\mu_1 - \mu_2)}{\sigma_{\bar{X}_1 - \bar{X}_2}}$$ The standard error \(\sigma_{\bar{X}_1 - \bar{X}_2}\) can be calculated using the following formula: $$\sigma_{\bar{X}_1 - \bar{X}_2} = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}$$ Given the data provided: \(\bar{X}_1 = 73, ~\bar{X}_2 = 63, ~s_1 = 25, ~s_2 = 28, ~n_1 = 400, ~\text{and} ~n_2 = 400\) Plugging in the values, we get: $$\sigma_{\bar{X}_1 - \bar{X}_2} = \sqrt{\frac{25^2}{400} + \frac{28^2}{400}} = \sqrt{\frac{625+784}{400}} \approx 1.67$$ Now, we can compute the test statistic: $$z = \frac{(73 - 63) - 0}{1.67} = \frac{10}{1.67} \approx 5.99$$
03

Determine the critical value and perform the test

At the 1% significance level, the critical value can be found using the z-table. For a one-tailed test, the critical z-value corresponding to a 99% confidence level is 2.33. Now, compare the test statistic with the critical value: Since \(5.99 > 2.33\), we reject the null hypothesis and conclude that there is sufficient evidence to indicate that per-capita beef consumption has decreased over the last 10 years.
04

Find a 99% lower confidence bound for the difference

To find the 99% lower confidence bound for the difference in means, we use the following formula: $$\text{Lower Confidence Limit} = (\bar{X}_1 - \bar{X}_2) - Z_{\alpha} \sigma_{\bar{X}_1 - \bar{X}_2}$$ Using the data and the critical value (\(Z_{\alpha} = 2.33\)) from earlier: $$\text{Lower Confidence Limit} = (73 - 63) - 2.33 \times 1.67 \approx 6.15$$ This means that we are 99% confident that the true difference in average per-capita beef consumption between the two groups is at least 6.15 units. The confidence bound confirms our conclusion in part (a), since the entire confidence interval lies above zero, which is consistent with the alternative hypothesis that per-capita beef consumption has decreased over the last 10 years. The confidence bound also provides additional information about the magnitude of the decrease in per-capita beef consumption.

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Key Concepts

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

Statistical Significance
Statistical significance is a vital concept in hypothesis testing, representing the likelihood that an observed effect in a study is due to chance. This concept is measured against a pre-set significance level, usually denoted as \(\alpha\), which is the threshold at which researchers are willing to reject the null hypothesis. In simple terms, if the probability of observing the study's result (or one more extreme) under the assumption that the null hypothesis is true is less than \(\alpha\), the result is considered statistically significant, and the null hypothesis is rejected.

For instance, in our exercise where the level of significance is set at 1% (\(\alpha = 0.01\)), it means there's only a 1% chance of concluding that beef consumption has decreased when, in reality, it has not (a Type I error). The calculated z-value (5.99) in the provided solution greatly exceeds the critical z-value (2.33) at this significance level, leading us to reject the null hypothesis confidently, thereby asserting the observed decrease in beef consumption is statistically significant and unlikely due to random variation alone.
Confidence Interval
A confidence interval is a range of values, derived from sample statistics, that is likely to contain the true population parameter with a specified degree of confidence. It provides not only an estimate of the unknown parameter but also the precision and reliability of the estimate. Confidence intervals are often used in conjunction with hypothesis tests and can provide a richer understanding of the data.

For the exercise's part (b), a 99% lower confidence bound gives us a range below which the true difference in average per-capita beef consumptions likely falls. The result, a lower bound of 6.15, suggests that we can be 99% confident that the true decrease in average beef consumption over the last 10 years is at least 6.15 units. This interval does not include zero, which further reinforces our findings of a statistically significant decrease and provides insightful context to the magnitude of the change.
Two-Sample Z-Test
The two-sample z-test is a statistical method used to determine whether there is a significant difference between the means of two independent groups. When the population variances are known and when sample sizes are large enough to satisfy the conditions of the Central Limit Theorem, this test is a valid approach to compare the means.

In the step-by-step solution provided, we apply a two-sample z-test to compare per-capita beef consumption from the records of two different years. We calculate the test statistic by determining the difference between the sample means and comparing it with the hypothesized difference, considering the variability of the data. The formula incorporates standard deviations and sample sizes to account for the dispersions and sizes of the sample groups. As our calculated z-value is significantly greater than the critical value from the z-table, the test suggests a significant difference in beef consumption, hence allowing us to reject the null hypothesis with confidence. By employing the two-sample z-test, researchers can make informed decisions regarding the population means based on sample data.

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Most popular questions from this chapter

Independent random samples of 140 observations were randomly selected from binomial populations 1 and 2 , respectively. Sample 1 had 74 successes and sample 2 had 81 successes. Use this information to answer the questions. Suppose that, for practical reasons, you know that \(p\), cannot be larger than \(p_{2}\). Test the appropriate hypothesis using \(\alpha=.10\).

How do states stack up against each other in SAT scores? To compare California and Massachusetts scores, random samples of 100 students from each state were selected and their SAT scores recorded with the following results: $$ \begin{array}{lccc} \hline & & \text { Sample } & \text { Standard } \\ \text { State } & \text { Mean } & \text { Size } & \text { Deviation } \\ \hline \text { Massachusetts } & 1122 & 100 & 194 \\ \text { California } & 1048 & 100 & 165 \end{array} $$ a. Use the critical value approach to test for a significant difference in the average SAT scores for these two states at the \(5 \%\) level of significance. b. Use the \(p\) -value approach to test for a significant difference in the average SAT scores for these two states. If you were writing a research report, how would you report your results?

As a group, students majoring in the engineering disciplines have the highest salary expectations, followed by those studying the computer science fields, according to a Michigan State University study. \({ }^{8}\) To compare the starting salaries of college graduates majoring in electrical engineering and computer science, random samples of 50 recent college graduates in each major were selected and the following information was obtained: $$ \begin{array}{lcc} \hline \text { Major } & \text { Mean (\$) } & \text { SD } \\ \hline \text { Electrical Engineering } & 62,428 & 12,500 \\ \text { Computer Science } & 57,762 & 13,330 \\ \hline \end{array} $$ a. Do the data provide sufficient evidence to indicate a difference in average starting salaries for college graduates who majored in electrical engineering and computer science? Test using \(\alpha=.05\). b. Calculate a \(95 \%\) confidence interval for the difference in the two population means. Does this confirm your conclusion in part a? Explain.

Does a baby's sleeping position affect the development of motor skills? A study in the Archives of Pediatric Adolescent Medicine examined 343 full-term infants at their 4 -month checkups for various milestones, such as rolling over, grasping a rattle, reaching for an object, and so on. \({ }^{21}\) The baby's favored sleep position - either on the stomach or on the back or side- was reported by the parents of 320 of the children, with the sample results shown here. $$ \begin{array}{lcc} \hline & \text { Stomach } & \text { Back or Side } \\ \hline \text { Number of Infants } & 121 & 199 \\ \text { Number That Roll Over } & 93 & 119 \\ \hline \end{array} $$ The researcher reported that infants who slept in the side or back position were less likely to roll over at the 4 -month checkup than infants who slept primarily in the stomach position \((P<.001) .\) Use a large-sample test of hypothesis to confirm or refute the researcher's conclusion.

The braking ability was compared for two 2018 automobile models. Random samples of 64 automobiles were tested for each type. The recorded measurement was the distance (in meters) required to stop when the brakes were applied at 80 kilometers per hour. These are the computed sample means and variances: \begin{tabular}{ll} \hline Model I & Model II \\ \hline \(\bar{x}_{1}=36.0\) & \(\bar{x}_{2}=33.2\) \\ \(s_{1}^{2}=9.48\) & \(s_{2}^{2}=8.09\) \\ \hline \end{tabular} Do the data provide sufficient evidence to indicate a difference between the mean stopping distances for the two models?
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