The mean weight of all 20-year old women is 130 pounds. A random sample of 30 women athletes who are 20 years old showed a sample mean of 126 pounds with a standard deviation of 15 pounds. Researchers wanted to determine whether the mean weight for 20-year old women athletes is significantly less than 130, using a significance level of 0.05. (6 pts)
a. Write the null and alternate hypothesis.
b. Calculate the test statistic and write a conclusion for this question.
c. Now suppose a sample of 100 women athletes was taken and the same mean (126) and standard deviation (15) was achieved. Repeat the test.
d. Explain what caused the difference between the outcomes for parts b and c.
Ho: mean1 = mean2
Ha: mean1 > mean2
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√n
If only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.
Repeat with new SD.
I'll leave the explanation up to you.
a. The null hypothesis would be that the mean weight for 20-year old women athletes is not significantly less than 130 pounds. The alternate hypothesis would be that the mean weight for 20-year old women athletes is significantly less than 130 pounds.
Null hypothesis (H0): Mean weight = 130
Alternate hypothesis (H1): Mean weight < 130
b. To calculate the test statistic, we will use the one-sample t-test formula:
t = (sample mean - population mean) / (standard deviation / √sample size)
Given:
Sample mean (x̄) = 126 pounds
Population mean (μ) = 130 pounds
Standard deviation (σ) = 15 pounds
Sample size (n) = 30
Substituting the values into the formula:
t = (126 - 130) / (15 / √30)
t = -4 / (15 / √30)
Using a calculator, we can calculate the value of t to be approximately -1.699.
Next, we compare this calculated t-value with the critical t-value from the t-distribution table at a significance level of 0.05 (one-tailed test).
For a one-tailed test with a significance level of 0.05 and 29 degrees of freedom (n-1), the critical t-value is approximately -1.699.
Since the calculated t-value (-1.699) is less than the critical t-value (-1.699), we reject the null hypothesis.
Therefore, we can conclude that, based on the sample data, the mean weight for 20-year old women athletes is significantly less than 130 pounds at a significance level of 0.05.
c. To repeat the test with a sample size of 100, we need to calculate the new test statistic using the same formula:
t = (sample mean - population mean) / (standard deviation / √sample size)
Given:
Sample mean (x̄) = 126 pounds
Population mean (μ) = 130 pounds
Standard deviation (σ) = 15 pounds
Sample size (n) = 100
Substituting the values into the formula:
t = (126 - 130) / (15 / √100)
t = -4 / (15 / 10)
Using a calculator, we can calculate the value of t to be approximately -8/3 or -2.667.
Next, we compare this calculated t-value with the critical t-value from the t-distribution table at a significance level of 0.05 (one-tailed test).
For a one-tailed test with a significance level of 0.05 and 99 degrees of freedom (n-1), the critical t-value is approximately -1.660.
Since the calculated t-value (-2.667) is less than the critical t-value (-1.660), we reject the null hypothesis.
Therefore, we can conclude that, based on the larger sample data, the mean weight for 20-year old women athletes is still significantly less than 130 pounds at a significance level of 0.05.
d. The difference between the outcomes for parts b and c can be attributed to the increase in the sample size. As the sample size increases from 30 to 100, the degrees of freedom increases from 29 to 99. With a larger sample size, the t-distribution becomes closer to the standard normal distribution, which results in more accurate critical t-values. Consequently, the critical t-value for a given significance level becomes smaller with a larger sample size. Therefore, the calculated t-value (-2.667) in part c is compared to a smaller critical t-value (-1.660) than in part b, leading to a stronger rejection of the null hypothesis in part c compared to part b.