Question

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 Roses.
2. In the Toolbar, click Analyze, select Compare Means, and then select One-Sample t Test.
3. Select Roses, 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.

5. State the nondirectional hypothesis.
6. State the critical t for a = .05 (two tails).
7. Answer the following: Is the length of viewing for Riverbend City online news significantly different than the population mean? Explain.

Answer 1

The SPSS output is:

The hypothesis being tested is:

H₀: µ = 8

H_a: µ ≠ 8

The test statistic is -0.493.

The critical t-value is 2.145.

The p-value is 0.629.

Since the p-value (0.629) is greater than the significance level (0.05), we fail to reject the null hypothesis.

Therefore, we cannot conclude that the length of viewing for Riverbend City's online news is significantly different than the population mean.

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