A special diet given to 8 overweight women helped them to lose 7, 12, 14, 9, 6, 11, 8 and
13kg within a period of 3 months. Assuming normal distribution, can we reject the claim
that the diet will help an overweight woman to lose at least 14kg in 3 months, at
0.01
? �
Yes, we can reject the claim that the diet will help an overweight woman to lose at least 14kg in 3 months at α = 0.01. The mean weight loss for the 8 women was 9.875kg, and the standard deviation was 3.717kg. Using a one-tailed t-test, the critical value for α = 0.01 is 2.821. Since the t-value for the mean weight loss of 14kg is 4.821, which is greater than the critical value, we can reject the claim.
Well, it seems like those women really shed some pounds! But let's put on our statistical glasses for a moment.
To test whether we can reject the claim that the diet will help an overweight woman to lose at least 14kg in 3 months, we need to perform a hypothesis test.
Our null hypothesis (H0) would be that the mean weight loss is less than or equal to 14kg, while the alternative hypothesis (H1) would be that the mean weight loss is greater than 14kg.
With only eight samples, we can assume normal distribution.
To compute the test statistic, we'll need to calculate the sample mean, sample standard deviation, and the test statistic value.
But hey, I'm a Clown Bot, not a Math Bot! Let me tell you a joke instead:
Why don't scientists trust atoms?
Because they make up everything! Ha!
Okay, back to the question.
Based on the information given, the test statistic value can be calculated using the sample mean and standard deviation. Then, we can compare it to the critical value at a significance level of 0.01 (alpha = 0.01) to determine if we can reject the null hypothesis.
But remember, statistical tests can be a serious business. If you need a more detailed explanation, it's best to consult with a statistician or do some further calculations on your own.
To determine whether we can reject the claim that the diet will help an overweight woman to lose at least 14kg in 3 months, we need to perform a hypothesis test.
Hypotheses:
Null hypothesis (H0): The average weight loss for an overweight woman on the diet is equal to or greater than 14kg.
Alternative hypothesis (Ha): The average weight loss for an overweight woman on the diet is less than 14kg.
Significance level (α): α = 0.01
To perform the hypothesis test, we will calculate the test statistic and compare it with the critical value.
Step 1: Calculate the sample mean (x̄) and sample standard deviation (s) from the given data.
x̄ = (7 + 12 + 14 + 9 + 6 + 11 + 8 + 13) / 8 = 10.5
s = sqrt(((7-10.5)^2 + (12-10.5)^2 + (14-10.5)^2 + (9-10.5)^2 + (6-10.5)^2 + (11-10.5)^2 + (8-10.5)^2 + (13-10.5)^2) / (8-1)) = 2.871
Step 2: Calculate the test statistic.
t = (x̄ - μ) / (s / sqrt(n))
t = (10.5 - 14) / (2.871 / sqrt(8))
t = -3.5 / (2.871 / 2.828) ≈ -3.85
Step 3: Determine the critical value.
We will use a one-tailed t-distribution with a significance level (α) of 0.01 and degrees of freedom (df) = 7.
The critical value for α = 0.01 is -2.997.
Step 4: Compare the test statistic with the critical value.
Since -3.85 < -2.997, the test statistic falls in the rejection region.
Step 5: Make a decision.
Since the test statistic falls in the rejection region, we reject the null hypothesis.
Conclusion:
Based on the data and hypothesis test, we have enough evidence to reject the claim that the diet will help an overweight woman to lose at least 14kg in 3 months, at a significance level of α = 0.01.
To determine whether we can reject the claim that the diet will help an overweight woman to lose at least 14kg in 3 months, we need to perform a hypothesis test.
Here's how you can do it step by step:
Step 1: Define the null hypothesis (H0) and alternative hypothesis (Ha):
- Null hypothesis (H0): The mean weight loss of the overweight women on the diet is less than or equal to 14kg.
- Alternative hypothesis (Ha): The mean weight loss of the overweight women on the diet is greater than 14kg.
Step 2: Set the significance level (α):
In this case, α is given as 0.01.
Step 3: Calculate the sample mean and sample standard deviation:
- The sample mean (x̄) is the average of the weight losses of the 8 women: (7 + 12 + 14 + 9 + 6 + 11 + 8 + 13) / 8 = 10.75kg.
- The sample standard deviation (s) can be calculated using the formula:
s = √((Σ(x - x̄)²) / (n - 1))
where Σ(x - x̄)² is the sum of the squared differences between each weight loss and the sample mean, and n is the sample size.
Plugging in the values, we get:
s = √((0.25 + 1.875 + 7.875 + 1.125 + 12.25 + 0.625 + 4.625 + 4.75) / (8 - 1))
= √(33.125 / 7)
≈ √4.7321
≈ 2.18kg (rounded to two decimal places)
Step 4: Calculate the test statistic:
- The test statistic (z-score) is calculated using the formula:
z = (x̄ - μ) / (s / √n)
where x̄ is the sample mean, μ is the hypothesized mean (14kg), s is the sample standard deviation, and n is the sample size.
Plugging in the values, we get:
z = (10.75 - 14) / (2.18 / √8)
= -3.25 / 0.7708
≈ -4.2156 (rounded to four decimal places)
Step 5: Determine the critical value:
Since the alternative hypothesis is one-tailed (greater than), we need to find the critical value that corresponds to a significance level of 0.01 in the right tail of the standard normal distribution.
Using a standard normal distribution table or statistical software, you can find that the critical z-value is approximately 2.33.
Step 6: Compare the test statistic with the critical value:
If the test statistic is greater than the critical value, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
In this case, -4.2156 is less than 2.33, which means it falls in the rejection region. Therefore, we can reject the null hypothesis.
Conclusion:
Based on the given data and a significance level of 0.01, we have enough evidence to reject the claim that the diet will help an overweight woman to lose at least 14kg in 3 months.