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Testing whether means are different - ANOVA and Chi-Square

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There are three hospitals in the Tulsa, Oklahoma, area. The following data show the number of outpatient surgeries performed at each hospital last week. At the .05 significance level, can we conclude there is a difference in the mean number of surgeries performed by hospital or by day of the week?

Day Number of Surgeries Performed
St. Luke's St. Vincent Mercy
Monday 14 18 24
Tuesday 20 24 14
Wednesday16 22 14
Thursday 18 20 22
Friday 20 28 24

For many years TV executives used the guideline that 30 percent of the audiences were watching each of the prime-time networks and 10 percent were watching cable stations on a weekday night.  A random sample of 500 viewers in the Tampa-St. Petersburg, Florida, and area last Monday night showed that 165 homes were tuned in to the ABC affiliate, 140 to the CBS affiliate, 125 to the NBC affiliate, and the remainders were viewing a cable station.  At the .05 significance level, can we conclude that the guideline is still reasonable?

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Solution Summary

The posting shows how we can test the difference in mean among the groups. Three approaches with the help of two problems are suggested. The approaches discussed are test for proportions, ANOVA and Chi-Square test. All the three tests are done using Excel so that the reader can understand the procedure clearly.

Solution Preview

The problem is solved using two methods - ANOVA and Chi-Square depending upon whether we are making assumption about the distribution of data.
See the attached file for complete solution. The text here may not be copied exactly as some of the symbols / tables may not print. Thanks

Anova: Single Factor

SUMMARY
Groups Count Sum Average Variance
St. Luke's 5 88 17.6 6.8
St. ...

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