Solution :
Null and alternative hypotheses :
The null and alternative hypotheses would be as follows:
H0 : The actual follows the professor's prediction in that P(hard copy) = 60%, P(Print) = 25% and P(Web) = 15%.
Ha : The actual prediction does not follow that of the professor's prediction.
Blank #1 :
With the assumption of H0, the expected count of each cell are as follows :
Hard copy E = (126 × 0.60) = 75.6
Print E = (126 × 0.25) = 31.5
Web E = (126 × 0.15) = 18.9
Blank #2 :
Degrees of freedom = (c - 1)
Where, c is number of categories in which prediction has been made.
We have, c = 3
The degrees of freedom for the GOE test are (3 - 1) = 2.
Blank #3 :
To test the hypothesis we shall use chi-square test of goodness of fit. The test statistic is given as follows :
Where, Oi's are observed counts and ei's are expected counts.
From the above table we get,
The value of the test statistic is 2.3201.
Blank #4 :
The p-value is given as follows :
P-value = P(χ2 > value of the test statistic)
P-value = P(χ² > 2.3201)
P-value = 0.3135
The p-value is 0.3135.
Blank #5 :
In our question the significance level is not given. Generally significance level 0.05 or 0.01 is used. We shall use significance level 0.05
Significance level = 0.05
P-value = 0.3135
(0.3135 > 0.05)
Since, p-value is greater than the significance level of 0.05, therefore we shall be fail to reject the null hypothesis (H0) at 0.05 significance level.
We decide to do not reject H0.
Blank #6 :
The evidence does not favor that the actual distribution does not follow that of the professor's prediction.
Please rate the answer. Thank you.
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