Data:Riverbend City online news advertises that it is read longer than the national news. The mean for national news is 8 hours per week. The following sample of the Riverbend City online news readers is: 5, 7, 6, 2, 4, 8, 5, 4, 18, 21, 8, 7, 4, 5, 6.

Instructions: Complete the following:

1. Enter the data from Problem Set 3.5 into SPSS and name the variable as Time.
2. In the Toolbar, click Analyze, select Compare Means, and then select One-Sample t Test.
3. Select Time, then click Arrow to send it over to the right side of the table. In the box labeled Test Value, enter 8.
4. Click OK and copy and paste the output into the Word document.

One-Sample Statistics
	N	Mean	Std. Deviation	Std. Error Mean
Time	15	7.3333	5.23268	1.35107

One-Sample Test
	Test Value = 8
t	df	Sig. (2-tailed)	Mean Difference	95% Confidence Interval of the Difference
Lower	Upper
Time	-.493	14	.629	-.66667	-3.5644	2.2311

1) State the nondirectional hypothesis.

H0: = 8

H1: 8

2) State the critical tfor a = .05(two tails).

Critical t for a = .05 is 2.145

3) Answer the following: Is the length of viewing for Riverbend City online news significantly different than the population mean? Explain.

P= 0.629, if P > .05 do not reject null hypothesis. Therefore, there is not enough evidence to determine that the viewing for Riverbend City online news is significantly different than the population mean.

4)Identify the IV and DV in the study.

5) State the null hypothesis and the directional (one-tailed) alternative hypothesis.

6) Can you reject the null hypothesis at = .05? Explain why or why not.