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March 30, 2017

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In Los Angeles there are three network television stations, each with it own evening news program from 6:00 to 6:30 PM. According to a report in this morning’s local newspaper, a random sample of 150 viewers last night revealed 53 watched the news on KNBC (channel 2), 64 watched KABC (channel 7), and 33 viewed KCBS (channel 2). At the 0.05 significance level, is there a difference in the proportion of viewers watching the three channels?

1. State the null and alternative hypothesis.




2. Identify the level of significance: Given at 0.05

  • statistics - ,

    You can probably use a chi-square for this kind of problem.

    Hypotheses:
    Ho: There was equal distribution of viewers across the channels.
    Ha: There was unequal distribution of viewers across the channels.

    If there was equal distribution across the channels, you would expect 50 viewers watching each channel (150/3 = 50).

    You can use the chi-square formula to calculate the test statistic if you need to do so.

    Once you have the chi-square value calculated, compare this test statistic with a value from the chi-square table using df = 2 (df = number of groups minus 1) and .05 level of significance.

    If the test statistic calculated exceeds the value you find from the table, the null will be rejected (there is a difference). If the test statistic does not exceed the critical value from the table, the null will not be rejected (there is no difference).

    I hope this will help.

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