The Survey of Study Habits and Attitudes (SSHA) is a psychological test that measures students' study habits and attitudes toward school. Scores range from 0 to 200. The mean score for college students is about 119, and the standard deviation is about 27.8. A teacher suspects that the mean μ for older students is higher than 119. She gives the SSHA to an SRS of 28 students who are at least 30 years old. Suppose we know that scores in the population of older students are Normally distributed with standard deviation σ = 27.8. We seek evidence against the claim that μ = 119.



If the study shows that = 127.9 g/dl, what is the P-value (±0.0001) of this outcome?

To find the P-value for this outcome, we can conduct a one-sample t-test.

1. State the null hypothesis (H0) and alternative hypothesis (Ha):
- H0: μ = 119 (the mean score for older students is equal to 119)
- Ha: μ > 119 (the mean score for older students is higher than 119)

2. Calculate the t-score. The formula for calculating the t-score is:
t = (x - μ) / (s / sqrt(n))
where x is the sample mean, μ is the population mean, s is the standard deviation, and n is the sample size.

Given the information in the question, x = 127.9, μ = 119, s = 27.8, and n = 28.
Plugging in the values, we get:
t = (127.9 - 119) / (27.8 / sqrt(28))
t = 8.9 / (27.8 / 5.29)
t ≈ 8.9 / 5.27
t ≈ 1.687

3. Determine the degrees of freedom (df). In this case, it is the sample size minus 1:
df = n - 1 = 28 - 1 = 27

4. Calculate the P-value. We will use a t-distribution table or a statistical software to find the P-value associated with the calculated t-score. Since the alternate hypothesis is μ > 119, we are interested in the upper tail of the t-distribution.

Using the t-distribution table or software, with 27 degrees of freedom, the P-value for t = 1.687 will be found to be a small value, significantly less than 0.0001.

Therefore, the P-value is less than 0.0001, which means that there is significant evidence against the null hypothesis.