If someone will explain how to work these problems out, that'll help me a ton!
A weight loss program claims their patients loose an average of 3 pounds during the first month of the program but a consumer group believes the number is higher than that. A random sample of 8 program participants showed the following weight loss at the end of 1 month:
4.1 3.7 3.1 2.8 3.9 3.2 4.5 4.3
What is the correct decision about the claim at a .05 significance level?
To determine the correct decision about the claim, we need to perform a hypothesis test. Here's how you can work it out step by step:
Step 1: State the hypotheses:
- Null hypothesis (H₀): The average weight loss during the first month of the program is 3 pounds.
- Alternative hypothesis (H₁): The average weight loss during the first month of the program is higher than 3 pounds.
Step 2: Set the significance level (α):
In this case, the significance level is given as 0.05, which means we have a 5% chance of rejecting the null hypothesis even if it is true.
Step 3: Calculate the test statistic:
To compare the sample data with the null hypothesis, we will calculate the test statistic. In this case, we will use a one-sample t-test, as the population standard deviation is not provided.
The formula for the t-test statistic is:
t = (x̄ - μ) / (s / √n)
where x̄ is the sample mean, μ is the population mean, s is the sample standard deviation, and n is the sample size.
Step 4: Find the critical value:
Since we are conducting a one-tailed test (to determine if the average weight loss is higher), we need to find the critical value from the t-distribution corresponding to a 0.05 significance level and degrees of freedom (df) equal to n - 1 (sample size - 1).
You can use a t-table or a statistical software to find the critical value. For df = 7 (8 - 1), the critical value at a 0.05 significance level is approximately 1.895.
Step 5: Calculate the test statistic:
Let's plug the values into the formula:
x̄ = (4.1 + 3.7 + 3.1 + 2.8 + 3.9 + 3.2 + 4.5 + 4.3) / 8 = 35.6 / 8 = 4.45 (rounded to two decimal places)
s = standard deviation of the sample = √[((4.1 - 4.45)² + (3.7 - 4.45)² + ... + (4.3 - 4.45)²) / (n - 1)] = 0.574 (rounded to three decimal places)
n = 8
t = (4.45 - 3) / (0.574 / √8) ≈ 4.8
Step 6: Make a decision:
If the absolute value of the test statistic is greater than the critical value, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
Since 4.8 > 1.895, we reject the null hypothesis.
Step 7: State the conclusion:
Based on the sample data and the hypothesis test, there is enough evidence to conclude that the average weight loss during the first month of the program is higher than 3 pounds at a 0.05 significance level.