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Statistical Significance versus Practical Significance The manufacturer of a daily dietary supplement claims that its product will help people lose weight. The company obtains a random sample of 950 adult males aged 20 to 74 who take the supplement and finds their mean weight loss after 8 weeks to be 0.9 pound with standard deviation weight loss of 7.2 pounds. (a) State the null and alternative hypotheses. (b) Test the hypothesis at the \(\alpha=0.1\) level of significance. Is a mean weight loss of 0.9 pound significant? (c) Do you think that a mean weight loss of 0.9 pound is worth the expense and commitment of a daily dietary supplement? In other words, does the weight loss have any practical significance? (d) Test the hypothesis at the \(\alpha=0.1\) level of significance with \(n=40\) subjects. Assume the same sample statistics. Is a sample mean weight loss of 0.9 pound significantly more than 0 pound? What do you conclude about the impact of large samples on the hypothesis test?

Step by step solution

01

- State the Null and Alternative Hypotheses

The null hypothesis (\text{H}_0) and alternative hypothesis (\text{H}_1) can be stated as follows: \(\text{H}_0: \mu = 0\) (There is no mean weight loss) \(\text{H}_1: \mu > 0\) (There is a mean weight loss, with \mu\ being the mean weight loss)

02

- Determine the Test Statistic

Use the sample mean, population mean under the null hypothesis, sample standard deviation, and sample size to calculate the test statistic (t). The formula is: \[ t = \frac{\bar{x} - \mu}{s / \sqrt{n}} \] Substituting the given values: \( \bar{x} = 0.9, \mu = 0, s = 7.2, n = 950 \) \[ t = \frac{0.9 - 0}{7.2 / \sqrt{950}} \]

03

- Calculate the Test Statistic

Using the given data: \[ t = \frac{0.9}{7.2 / \sqrt{950}} \approx 2.331 \]

04

- Determine the Critical Value and Make a Decision

At the \( \alpha = 0.1 \) significance level for a one-tailed test, the critical value (t_c) from the t-distribution table with \( n-1=949 \) degrees of freedom is approximately 1.282. Since the calculated t-value 2.331 is greater than the critical value 1.282, reject the null hypothesis.

05

- Assess Practical Significance

Although the statistical test indicates significance, a mean weight loss of 0.9 pound may not be practically significant. Consider whether this small mean weight loss is worth the expense and commitment required for a daily dietary supplement.

06

- Test Hypothesis with Smaller Sample Size

Repeat the hypothesis test with \( n = 40 \). The test statistic formula again is: \[ t = \frac{0.9}{7.2 / \sqrt{40}} \approx 0.787 \] The critical value (t_c) from the t-distribution table with \( 39 \) degrees of freedom at the \( \alpha = 0.1 \) level for a one-tailed test is approximately 1.304. Since the calculated t-value 0.787 is less than the critical value 1.304, do not reject the null hypothesis.

07

- Conclusion on Impact of Large Samples

Large sample sizes can make it easier to detect small differences, leading to statistical significance that may not imply practical significance. In a smaller sample, the same level of effect might not show statistical significance.

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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 (\text{H}_0) represents a default position that there is no effect or no difference. In the context of our dietary supplement study, the null hypothesis states that taking the supplement does not lead to a mean weight loss. This can be mathematically expressed as \text{H}_0: \( \mu = 0 \). The null hypothesis is essentially a statement to be tested against the alternative hypothesis.

alternative hypothesis

On the flip side, the alternative hypothesis (\text{H}_1) proposes that there is an effect or a difference. For our study, this means the dietary supplement does lead to a mean weight loss. The alternative hypothesis can be written as \text{H}_1: \( \mu > 0 \).
This hypothesis is considered when the evidence from the data shows that the null hypothesis is unlikely.
Assessing the alternative hypothesis involves comparing the computed test statistic to a critical value to determine if the observed mean weight loss is statistically significant.

t-test

A t-test is a statistical test used to determine if there is a significant difference between the means of two groups, which may be related in certain features. In this study, we use a one-sample t-test to check if the mean weight loss from the supplement is different from zero.
We calculate the t-value using the formula:
\[ t = \frac{\bar{x} - \mu}{s / \sqrt{n}} \], where \( \bar{x} \) is the sample mean, \( \mu \) is the population mean under the null hypothesis, \( s \) is the sample standard deviation, and \( n \) is the sample size.
Determination of the t-value allows us to make decisions regarding the null hypothesis based on the critical value from the t-distribution table.

practical significance

Although statistical significance is important, it's equally crucial to consider practical significance. This asks whether the observed effect, in our case a mean weight loss of 0.9 pounds, has real-world relevance or utility.
For example, even if the weight loss is statistically significant, it might not be enough to justify the cost, effort, and potential side effects of taking a daily dietary supplement.
Practical significance requires evaluating the effect size and considering whether the result has substantial and meaningful implications in the real world.