Question

The Survey of Study Habits and Attitudes (SSHA) is a psychological test that measures studentsâ study habits and attitude toward school. The survey yields several scores, one of which measures student attitudes toward studying. The mean student attitude score for college students is about 50, and the standard deviation is about 15. A researcher in the Philippines is concerned about the declining performance of college graduates on professional licensure and board exams. She suspects that poor attitudes of students are partly responsible for the decline and that the mean for college seniors who plan to take professional licensure or board exams is less than 50. She gives the SSHA to an SRS of 225 college seniors in the Philippines who plan to take professional licensure or board exams. Suppose we know that the student attitude scores in the population of such students are Normally distributed with standard deviation Ï = 15.Step 1:One sample of 225 students had mean student attitude score x = 48.9. What is the P-value?Give your answer to 4 decimal places.Step 2:Is this outcome statistically significant at the α = 0.05 level? At the α = 0.01 level?a. It is not significant at either level.b. It is significant at the α = 0.05 level but not at the α = 0.01 level.c. It is significant at the α = 0.01 level, hence also at the α = 0.05 level.d. It is significant at the α = 0.01 level but not at the α = 0.05 level.e. It is significant at the α = 0.05 level, hence also at the α = 0.01 level.

Answer 1

Answer:

The test is not significant at either level.

Step-by-step explanation:

Hello!

X: Student attitude toward studying college students.

Mean μ=50

Standard deviation σ=15

X~N(μ;σ²)

The researcher is interested in testing if the poor attitude of students is partly responsible for the decline of the performance of college graduates on professional licensure and board exams.

She suspects that the mean for college seniors who plan to take professional licensure or board exams is less than 50.

1)

The researcher took a sample of 225 students and obtained a mean score of 48.9

If the hypotheses are:

H₀: μ ≥ 50

H₁: μ < 50

$Z_{H_0}= \frac{X[bar]-Mu}{\frac{Sigma}{\sqrt{n} } } ~~N(0;1)$

$Z_{H_0}= \frac{48.9-50}{\frac{15}{\sqrt{225} } } = -1.1$

The p-value is P(Z<-1.1)= 0.1357

2)

The decision rule for the p-value approach is:

If the p-value ≤ α, the decision is to reject the null hypothesis.

If the p-value is > α, the decision is to not reject the null hypothesis.

Using a significance level of α: 0.01 and α: 0.05

Since the calculated p-value is greater than both significance levels, the decision is to not reject the null hypothesis.

The test is not significant at either level.

I hope it helps!