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2 July, 12:21

An article studied differences between expert and consumer ratings by considering medal ratings for wines, which could be gold (G), silver (S), or bronze (B). Three categories were then established.

Rating is the same [ (G, G), (B, B), (S, S) ]

Rating differs by one medal [ (G, S), (S, G), (S, B), (B, S) ]

Rating differs by two medals [ (G, B), (B, G) ]

The observed frequencies for these three categories were 61, 108, and 47, respectively. On the hypothesis of equally likely expert ratings and consumer ratings being assigned completely by chance, each of the nine medal pairs has probability 1/9.

Carry out an appropriate chi-squared test using a significance level of 0.10.

Calculate the test statistic. (Round your answer to two decimal places.)

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  1. 2 July, 13:11
    0
    test statistic = 1.73

    the results are not significant

    Step-by-step explanation:

    Expected probability of all rating = 1/9 or 0.11111

    total number of observations = 61+108+47 = 216

    Observed value for same rating = 61/216 = 0.282407

    Observed value for Rating differing by one medal = 108/216 = 0.5

    Observed value for Rating differing by two medal = 47/216 = 0.2175925926

    Ch - Squared test: Χ² = Σ (O - E) ² / E

    O: observed value

    E: Expected value

    X² = (0.28241-0.11111) ² / 0.1111 + (0.5-0.11111) ² / 0.1111 + (0.21759-0.11111) ² / 0.1111

    X² = 1.727 or 1.73

    degrees of freedom = 216-1=215

    at 0.1 significance and 215 degress of freedom, p-value is 1.0

    The result is not significant as p value is greater than 0.1
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