Math  /  Data & Statistics

QuestionSuppose the weights of seventh-graders at a certain school vary according to a Normal distribution, with a mean of 100 pounds and a standard deviation of 7.5 pounds. A researcher believes the average weight has decreased since the implementation of a new breakfast and lunch program at the school. She finds, in a random sample of 35 students, an average weight of 98 pounds.
What is the PP-value for an appropriate hypothesis test of the researcher's claim? 1.578-1.578 0.057 0.115 0.943

Studdy Solution

STEP 1

1. The weights of seventh-graders are normally distributed with a mean (μ\mu) of 100 pounds and a standard deviation (σ\sigma) of 7.5 pounds.
2. The sample size is 35 students, with a sample mean (xˉ\bar{x}) of 98 pounds.
3. We are conducting a one-sample t-test for the mean.
4. The null hypothesis (H0H_0) is that the population mean is 100 pounds.
5. The alternative hypothesis (HaH_a) is that the population mean is less than 100 pounds.

STEP 2

1. Set up the hypotheses.
2. Calculate the test statistic.
3. Determine the PP-value.
4. Compare the PP-value to the significance level.

STEP 3

Set up the hypotheses:
- Null hypothesis (H0H_0): μ=100\mu = 100 - Alternative hypothesis (HaH_a): μ<100\mu < 100

STEP 4

Calculate the test statistic using the formula for the t-statistic:
t=xˉμσnt = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}
where: - xˉ=98\bar{x} = 98 - μ=100\mu = 100 - σ=7.5\sigma = 7.5 - n=35n = 35
Substitute the values:
t=981007.535t = \frac{98 - 100}{\frac{7.5}{\sqrt{35}}}
Calculate the denominator:
7.5351.267\frac{7.5}{\sqrt{35}} \approx 1.267
Calculate the t-statistic:
t=21.2671.578t = \frac{-2}{1.267} \approx -1.578

STEP 5

Determine the PP-value for the calculated t-statistic (1.578-1.578) using a t-distribution with n1=34n-1 = 34 degrees of freedom.
The PP-value is approximately 0.057.

STEP 6

Compare the PP-value to a common significance level (e.g., α=0.05\alpha = 0.05):
- If PP-value <α< \alpha, reject H0H_0. - If PP-value α\geq \alpha, do not reject H0H_0.
Since the PP-value is 0.057, which is slightly greater than 0.05, we do not reject H0H_0.
The PP-value is:
0.057 \boxed{0.057}

Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord