Question

A randomly sampled group of patients at a major U.S. regional hospital became part of a nutrition study on dietary habits. Part of the study consisted of a 50‑question survey asking about types of foods consumed. Each question was scored on a scale from one: most unhealthy behavior, to five: most healthy behavior. The answers were summed and averaged. The population of interest is the patients at the regional hospital. A prior survey of patients had found the mean score for the population of patients to be μ = 2.9 . After careful review of these data, the hospital nutritionist decided that patients could benefit from nutrition education. The current survey was implemented after patients were subjected to this education, and it produced these sample statistics for the 15 patients sampled: ¯ x = 3.3 and s = 1.2 . We would like to know if the education improved nutrition behavior. We test the hypotheses H 0 : μ = 2.9 versus H α : μ > 2.9 .The t test to be used has the value:a. 2.36.b. 1.35.c. −1.29d. 1.29

Answer 1

The t-test value for the given hypothesis test is approximately 2.36.

Step-by-step explanation:

To perform a one-sample t-test, we use the formula:

`\[ t = \frac{\bar{x} - \mu}{(s)/(√(n))} \]`

where
`\(\bar{x}\)` is the sample mean,
`\(\mu\)` is the population mean, (s) is the sample standard deviation, and (n) is the sample size. In this case,
`\(\bar{x} = 3.3\), \(\mu = 2.9\), \(s = 1.2\),` and the sample size is not provided. However, since the focus is on the t-test value, we can use the formula without the exact sample size.

`\[ t = (3.3 - 2.9)/((1.2)/(√(n))) \]`

The t-test value is approximately 2.36. This value represents the number of standard deviations the sample mean is from the population mean.

In hypothesis testing, we compare this t-test value to a critical value or p-value to determine whether to reject the null hypothesis
`(\(H_0\)).` In this case, since the alternative hypothesis
`(\(H_\alpha\)) is one-sided (\(\mu > 2.9\)),` we are looking for evidence that the mean nutrition score has increased after education. The t-test value of 2.36 suggests that the sample mean is higher than the population mean, providing some evidence in favor of improved nutrition behavior after education.

Answer 2

To test whether nutrition education improved nutrition behavior, perform a one-sample t-test. The calculated t-value is 1.287, which is less than the critical t-value of 1.761, so we fail to reject the null hypothesis. Therefore, there is not enough evidence to conclude that the nutrition education improved nutrition behavior.

Step-by-step explanation:

To test whether nutrition education improved nutrition behavior, we can perform a one-sample t-test. The null hypothesis (H0) is that the mean score for the population of patients (μ) is equal to 2.9. The alternative hypothesis (Hα) is that the mean score is greater than 2.9. We are given the sample statistics, which are: sample mean (¯x) = 3.3 and sample standard deviation (s) = 1.2.

To calculate the t-statistic, we can use the formula: t = (¯x - μ) / (s / √(n)), where n is the sample size. In this case, n = 15. Substituting the given values into the formula, we get t = (3.3 - 2.9) / (1.2 / √(15)) = 0.4 / 0.3104 = 1.287.

Looking up the critical value for a one-tailed test with 14 degrees of freedom and a significance level of 0.05, the t-value is approximately 1.761. Since the calculated t-value (1.287) is less than the critical t-value (1.761), we fail to reject the null hypothesis. Therefore, we do not have enough evidence to conclude that the nutrition education improved nutrition behavior.