Question

Use the GSS dataset to answer the questions below. All analysis must be completed using your...

Use the GSS dataset to answer the questions below. All analysis must be completed using your own GSS data set. Failure to use your own GSS data set will result in a zero for the question. As above, you will need to read and interpret the question to determine which statistical test is appropriate, as well as whether it is a one-tailed or two-tailed test. Similar to the questions above, you must: 1. State the null and research hypotheses 2. Provide the Z(critical), T(critical), or χ2 (critical) score corresponding to the α thresh- old for your test 3. Explicitly provide your test statistic that is in your SPSS output (the exception is for tests of proportion, as the test statistic is not returned by SPSS) 4. Provide your decision about statistical significance You should also substantively interpret the results of your test (again, keep it short). All of the necessary SPSS output must be pasted into your document.

2. The population mean for a full-time worker is 40 hours per week. On average, do the respondents in the GSS work either significantly more or significantly less than 40 hours a week? Use an α value of 0.01 and the variable HRS1 to answer this question. (10 points)

 One-Sample Statistics N Mean Std. Deviation Std. Error Mean NUMBER OF HOURS WORKED LAST WEEK 858 39.75 15.208 .519
 One-Sample Test Test Value = 0.01 t df Sig. (2-tailed) Mean Difference 95% Confidence Interval of the Difference Lower NUMBER OF HOURS WORKED LAST WEEK 76.538 857 .000 39.738 38.72
 One-Sample Test Test Value = 0.01 95% Confidence Interval of the Difference Upper NUMBER OF HOURS WORKED LAST WEEK 40.76

1)

Ho:X =

Ha:X

2)

Z CRITICAL VALUE IS 2.58 FOR TWO TAIL TEST

T CRITICAL VALUE IS 2.58 FOR TWO TAIL TEST

t-test = (sample mean - population mean)/ standard error

= (39.75-40)/0.518

= -0.481695568

so accept the null hypothesis because of calculated t value is less than critical t (2.58) value so, there is no significant evidence that sample mean and population mean are different .

confidence interval = mean t * se

= 39.75 2.58 * 0.519

= 39.75 1.33902

lower = 38.41098 to upper = 41.08902

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