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Basic Statistics in PHStat

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* Must write out a NULL & HYPOTHESIS and include BUT WHAT DOES THIS MEAN TO A MANAGER??

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Must write out a NULL & HYPOTHESIS
and include BUT WHAT DOES THIS MEAN TO A MANAGER??

Problem #1 For this problem use the following contingency table:

A B Total
1 20 30 50
2 30 45 75
Total 50 75 125

The null hypothesis tested is
H0: There is no association between row and column attributes

The alternative Hypothesis

H1: There is association between row and column attributes

Rejection Criteria: Reject the null hypothesis, if the calculated value of chi square is greater than the critical value of Chi square with 1 d.f at 0.05 significance level

a. Find the expected frequency for each cell.

Expected Frequencies
Column variable
A B Total
1 20 30 50
2 30 45 75
Total 50 75 125

b. Compare the observed and expected frequencies in each cell.

The chi square contribution in each cell is given below

Column variable
A B Total
1 0 0 0
2 0 0 0
Total 0 0 0

c. Compare the X2 statistic. Is it significant at alpha () = .05.

Data
Level of Significance 0.05
Number of Rows 2
Number of Columns 2
Degrees of Freedom 1

Results
Critical Value 3.841459
Chi-Square Test Statistic 0
p-Value 1
Do not reject the null hypothesis

Conclusion: Since the Chi Square test Statistic is less than the critical value , the null hypothesis that there is no association between row variable and column is accepted

Problem #2 A survey for Internet users in the United States, Australia, and Europe studied whether those who downloaded music subsequently purchased a CD or tape of the music ("net Music Inspires Buying." USA Today, January 23, 2001, 1A). The survey reported that 77% of Americans, 78% of Australians, and 54% of Europeans who downloaded music get inspired to purchase the CD or tape of music. Suppose the survey was based on a sample of 500 Americans, 250 ...

Solution Summary

This solution gives the step by step method for solving business statistics problems using PHSTAT

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