SOLUTION: For many years, TV executives used the guideline that 30 percent of the audience were watching each of the traditional big three prime-time networks and 10 percent were watching ca

Question 658964: For many years, TV executives used the guideline that 30 percent of the audience were watching each of the traditional big three prime-time networks and 10 percent were watching cable stations on a weekday night. A random sample of 550 viewers in the Tampa�St. Petersburg, Florida, area last Monday night showed that 190 homes were tuned in to the ABC affiliate, 100 to the CBS affiliate, 165 to the NBC affiliate, and the remainder were viewing a cable station. At the 0.02 significance level, can we conclude that the guideline is still reasonable? (Round your answers to 3 decimal places.)
H0: The proportions are as stated H1: The proportions are not as stated
Reject H0 if X^2 >____________
X^2 =___________
Reject H0. Proportion of viewers is not as stated.
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
this looks like a chi-square goodness of fit type problem.
i'll calculate through SPSS and then calculate manually so you can see how the formulas work.
we're looking at expected versus actual.
the expected is:
30% watching each of the 3 prime time and 10% watching cable stations on a weekday night.l
actual are:
190 homes watching ABC, 100 CBS, 165 NBC and remainder watching cable.
total viewers = 550.
actual cable viewers = 95
total is 190 + 100 + 165 + 95 = 550
SPSS says that the chi-square value is equal to 58.485 with 3 degrees of freedom and a p-value of .000 which means the results are very significant and extremely likely not to be due by chance.

at the .02 significance level with 3 degrees of freedom (4 categories minus 1 = 3 degrees of freedom), the critical chi-square value is 9.8374. the probability of exceeding that chi-square value is equal to .02.

these results were achieved through SPSS.

if you were to do the calculations manually, you would have to do the following

set up a table of value as shown below:

    cat      actual   expected   (actual - expected)^2 / expected
    abc       190       165                     3.7879
    cbs       100       165                     25.606
    nbc       165       165                     0
    cable     95        55                      29.091

    total     550       550                     58.48485

The formula for calculating the chi-square statistic is:
chi-square = sum(o-e)^2/e
the (o-e)^2 for each of the measurements (observed and expected) is in the 4th column of the table. ths sum is at the bottom.
this value agrees with the SPSS output so it's good.

you can round the answer as required.

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