The Survey of Study Habits and Attitudes (SSHA) is a psychological test that mea

Question

The Survey of Study Habits and Attitudes (SSHA) is a psychological test that measures students’ 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 ? = 15.

Step 1: One sample of 225 students had mean student attitude score x = 48.9. Enter this x, along with the other required information, into the P-Value of a Test of Significance Applet What is the P-value? Give your answer to 4 decimal places. Fill in the blank: Step 2: OIt is significant at the a 0.05 level but not at the a 0.01 level. O It is significant at the a 0.01 level but not at the a 0.05 level. It is significant at the a0.05 level, hence also at the a 0.01 level. O It is significant at the a 0.01 level, hence also at the a 0.05 level. O It is not significant at either level.

Explanation / Answer

Solution:-

1)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u > 50
Alternative hypothesis: u < 50

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample z-test.

Analyze sample data. Using sample data, we compute the standard error (SE), z statistic test statistic (z).

SE = s / sqrt(n)

S.E = 1.0
z = (x - u) / SE

z = - 1.10

where s is the standard deviation of the sample, x is the sample mean, u is the hypothesized population mean, and n is the sample size.

The observed sample mean produced a z statistic test statistic of - 1.10.

Thus the P-value in this analysis is 0.136.

Interpret results. Since the P-value (0.136) is greater than the significance level (0.05), we cannot reject the null hypothesis.

Step 2 - It is not significant at neither level.

Step3

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u > 50
Alternative hypothesis: u < 50

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample z-test.

Analyze sample data. Using sample data, we compute the standard error (SE), z statistic test statistic (z).

SE = s / sqrt(n)

S.E = 1.0
z = (x - u) / SE

z = - 2.60

where s is the standard deviation of the sample, x is the sample mean, u is the hypothesized population mean, and n is the sample size.

The observed sample mean produced a z statistic test statistic of - 2.60

Thus the P-value in this analysis is 0.009

Interpret results. Since the P-value (0.009) is less than the significance level (0.05), we have to reject the null hypothesis.

Step 4

This is significant ar both the a = 0.01 level and at the a = 0.05 level.

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