A company claims that its new diet pill reduces the weight by more than 5 pounds on average in a week. A sample of 30 people are tested and the mean weight loss is found to be 5.3 pounds with a standard deviation of 1.2 pounds. Using a significance level of 0.05, is there sufficient evidence to support the company's claim? The null hypothesis is that the mean weight loss is 5 pounds.
Answer
Step 4: Compare the calculated t-value with the critical t-value (the t-value corresponding to a significance level of 0.05 with 29 degrees of freedom) to decide whether to reject the null hypothesis.
Step 1 ：Step 1: State the null hypothesis (\(H_0\)) and alternate hypothesis (\(H_a\)). Here, \(H_0: \mu = 5\) and \(H_a: \mu > 5\)
Step 2 ：Step 2: Determine the test statistic. We use the formula for a one-sample t-test: \(t = \frac{\bar{x} - \mu}{s/\sqrt{n}}\), 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. Substituting our values, we get: \(t = \frac{5.3 - 5}{1.2/\sqrt{30}}\)
Step 3 ：Step 3: Calculate the t-value using the above formula.
Step 4 ：Step 4: Compare the calculated t-value with the critical t-value (the t-value corresponding to a significance level of 0.05 with 29 degrees of freedom) to decide whether to reject the null hypothesis.
