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Problem Set 3.5:One-Sample t test in SPSS Criterion: Calculate a one-sample t test in SPSS. Data: Riverbend City online news advertises that it is read
Problem Set 3.5:One-Sample ttest in SPSS
Criterion:Calculate a one-sample t test in SPSS.
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:
- Enter the data from Problem Set 3.5 into SPSS and name the variable as Time.
- In the Toolbar, click Analyze, select Compare Means, and then select One-Sample t Test.
- Select Time, then click Arrow to send it over to the right side of the table. In the box labeled Test Value, enter 8.
- Click OK and copy and paste the output into the Word document.
- State the nondirectional hypothesis. H0-T8, H1 T/=8
- State the critical tfor a = .05(two tails). Tinv(.05.14)= 2.14478668
- Answer the following: Is the length of viewing for Riverbend City online news significantly different than the population mean? Explain.
Standard error is not significant when you compare to the mean
* T-Test [DataSet1] One-Sample Statistics Std. Std. Error N Mean Deviation Mean Time 15 7.3333 5.23268 1.35107 One-Sample Test Test Value = 8 95% Confidence Interval of Sig. (2- Mean the Difference t df tailed) Difference Lower Upper Time -.493 14 .629 -.66667 -3.5644 2.2311 One-Sample Effect Sizes Standardizera Point 95% Confidence Interval Estimate Lower Upper Time Cohen's d 5.23268 -.127 -.633 383 Hedges' correction 5.53554 -.120 -.599 .362 a. The denominator used in estimating the effect sizes. Cohen's d uses the sample standard deviation. Hedges' correction uses the sample standard deviation, plus a correction factorStep by Step Solution
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