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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.)

Respuesta :

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Answer:

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

As the p-value is greater than 0.1 so, the result is not significant and this can be determined by using the given data.

Given :

  • 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 a probability of 1/9.

First, determine the total number of observations:

[tex]\rm Total \; Observations = 61+108+47=216[/tex]

Now, according to the Ch-squared test:

[tex]\rm X^2 = \sum \dfrac{(O-E)^2}{E}[/tex]

where O represents the observed value and E represents the expected value.

[tex]\rm X^2 = \dfrac{\dfrac{61}{216}-\dfrac{1}{9}}{\dfrac{1}{9}}+\dfrac{\dfrac{108}{216}-\dfrac{1}{9}}{\dfrac{1}{9}}+\dfrac{\dfrac{47}{216}-\dfrac{1}{9}}{\dfrac{1}{9}}[/tex]

Now, simplify the above expression in order to determine the value of [tex]\rm X^2[/tex].

[tex]\rm X^2 = 1.73[/tex]

Now, the degree of freedom is given by:

= 216 - 1 = 215

So, the p-value at 0.1 significance level and 215 degree of freedom is 1.

As the p-value is greater than 0.1 so, the result is not significant.

For more information, refer to the link given below:

https://brainly.com/question/23044118

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