During the first 13 weeks of the television season, the Saturday evening 8:00 P.M. to 9:00 P.M. audience proportions were recorded as ABC 29

Question

During the first 13 weeks of the television season, the Saturday evening 8:00 P.M. to 9:00 P.M. audience proportions were recorded as ABC 29%, CBS 28%, NBC 25%, and independents 18%. A sample of 300 homes two weeks after a Saturday night schedule revision yielded the following viewing audience data: ABC 95 homes, CBS 65 homes, NBC 87 homes, and independents 53 homes. Test with = .05 to determine whether the viewing audience proportions changed.

Required:
Compute the value of the 2 test statistic (to 2 decimals). The p value is Has a significant change occurred in viewing audience proportions?

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Vodka 3 years 2021-08-21T02:04:44+00:00 1 Answers 114 views 0

Answers ( )

  1. thugiang
    0
    2021-08-21T02:06:24+00:00
    Reply

    Answer:

    – The value of the 2 test statistic is 6.98

    – There is no sufficient evidence to reject the claim of the specific distribution

    Step-by-step explanation:

    Given

    Proportion (p)

    ABC = 29\% = 0.29\\CBS = 28\% = 0.28\\NBC = 25\% = 0.25\\Others = 18\% = 0.18\\

    Samples

    ABC = 95\\CBD = 65\\NBC = 87\\Others = 53\\\alpha = 0.05

    First, we formulate a table for chi square totals:

    x^2 = \frac{(O - E)^2}{E}

    Where

    O = Observed Frequency and E = Expected Frequency

    So, we have:

    \begin{array}{cccccc}{\ } & {O} & {E = np} & {O - E} & {(O -E)^2} & {x^2 = \frac{(O - E)^2}{E}} \ \\ \\ {0.29} & {95} & {87} & {8} & {64} & {0.736} &{0.28} & {65} & {84} & {-19} & {361} & {4.300} & {0.25} & {87} & {75} & {12} & {144} & {1.920} & {0.18} & {53} & {54} & {-1} & {1} & {0.019}\ \end{array}

    For ABC

    E = np = 300 * 0.29= 87

    O-E =95 - 87 = 8

    (O-E)^2 = 8^2 = 64

    x^2 = \frac{(O - E)^2}{E} = \frac{64}{87} = 0.736

    For CBS

    E =np = 300 * 0.28 = 84

    O -E = 65 - 84 = -19

    (O -E)^2 = (-19)^2 = 361

    x^2 = \frac{(O - E)^2}{E} = \frac{361}{84} = 4.300

    For NBC

    E =np = 300 * 0.25 = 75

    O -E = 87 - 75 = 12

    (O -E)^2 = (12)^2 = 144

    x^2 = \frac{(O - E)^2}{E} = \frac{144}{75} = 1.920

    For others:

    E = np = 300 * 0.18 =54

    O - E = 53 - 54 = -1

    (O - E)^2 = (-1)^2= 1

    x^2 = \frac{(O - E)^2}{E} = \frac{1}{54} = 0.019

    \sum x^2 = 0.736 + 4.300 + 1.920 + 0.019

    \sum x^2 = 6.975

    \sum x^2 = 6.98 — approximated

    The value of the 2 test statistic is 6.98

    Next, is to determine the p value:

    Calculate the degree of freedom

    df = n - 1

    df = 4 - 1

    df = 3

    Using the x^2 table in the 3rd row,

    \sum x^2 = 6.98 corresponds to:

    p = 0.072699

    The value of p is:

    0.05 < P < 0.10

    This implies that, we fail to reject H o

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