Question

In a medical study, 76 objects were placed on a low fat diet. After 12 months, there sample mean weight loss was x̅=2.2 kilograms with a sample standard deviation of 6.1 kilograms. We will perform a statistical analysis (with the significance level α=0.05 to answer the question if the mean weight loss is greater than 0.To determine if we reject or fail to reject the null hypothesis we need to compare the p-value with the following number:

Answer 1

In hypothesis testing, to determine whether to reject or fail to reject the null hypothesis, we compare the p-value to the significance level \(\alpha\). In your case, the significance level is \(\alpha = 0.05\).

Here are the steps involved in the hypothesis test:

1. **State the Hypotheses**
- Null Hypothesis (\(H_0\)): \(\mu \leq 0\) (the mean weight loss is less than or equal to 0)
- Alternative Hypothesis (\(H_a\)): \(\mu > 0\) (the mean weight loss is greater than 0)

2. **Calculate the Test Statistic**
The test statistic for a one-sample t-test can be calculated using the formula:
\[
t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}
\]
where:
- \(\bar{x} = 2.2\) (sample mean)
- \(\mu_0 = 0\) (hypothesized mean under the null hypothesis)
- \(s = 6.1\) (sample standard deviation)
- \(n = 76\) (sample size)

Plugging the values into the formula:
\[
t = \frac{2.2 - 0}{6.1 / \sqrt{76}} = \frac{2.2}{6.1 / \sqrt{76}}
\]
First, calculate \(6.1 / \sqrt{76}\):
\[
\sqrt{76} \approx 8.7178
\]
\[
6.1 / \sqrt{76} \approx 0.699
\]
Now, calculate \(t\):
\[
t = \frac{2.2}{0.699} \approx 3.15
\]

3. **Determine the degrees of freedom**
The degrees of freedom for this test is:
\[
df = n - 1 = 76 - 1 = 75
\]

4. **Find the p-value**
To find the p-value associated with the calculated t-statistic of approximately 3.15 with 75 degrees of freedom, you can use a t-distribution table or an online calculator.

5. **Comparison**
If the p-value is less than or equal to \(0.05\), you will reject the null hypothesis. If the p-value is greater than \(0.05\), you will fail to reject the null hypothesis.

The specific number to compare the p-value to in this case is \(0.05\).

To summarize, you compare the p-value you calculate or obtain from statistical software with \(0.05\) to determine whether to reject or fail to reject the null hypothesis.