A personality psychologist asked college students how practical they are on a sc

Question

A personality psychologist asked college students how practical they are on a scale of 1 (not at all) to 10 (extremely). Assume that the distribution of college students’ “practicality” score is normal (M = 7.15, SD = 1.71). I randomly selected the “practical” score of 9 students from this year’s SOC 390 (that was also one of the questions I asked you in the beginning of the semester), and the mean of the 9 students’ “practical” responses is 8.6. Are these 9 students different from the general population of college students?

2a) What is the null hypothesis? What is the research hypothesis?

2b) Determine the characteristics of the comparison distribution. What is the mean of the comparison distribution? What is the standard error?

2c) Determine the cutoff score on the comparison distribution (a = 0.05).

2d) Determine your sample’s score on the comparison distribution.

2e) Decide whether to reject the null hypothesis.

2f) Calculate the 95% and 99% confidence intervals.

Explanation / Answer

Solution:-

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

Null hypothesis: = 7.15
Alternative hypothesis: 7.15

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

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

b) Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = s / sqrt(n)

S.E = 0.57

DF = n - 1 = 9 - 1

D.F = 8

c)

tcritical = 2.31

d)

t = (x - ) / SE

t = 2.544

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

Since we have a two-tailed test, the P-value is the probability that the t statistic having 8 degrees of freedom is less than - 2.544 or greater than 2.544.

e) Interpret results. Since the t-value (2.544) is greater than the t critical (2.31), hence we have to reject the null hypothesis.

Thus, the P-value = 0.0345

Interpret results. Since the P-value (0.0345) is less than the significance level (0.05), we have to reject the null hypothesis.

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