Pls help this is due ASAP I’ll brainlest for the right answer
1 answer:
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Solution :
In this case we have to use for the goodness of fit.
Null hypothesis is : Data follows the given distribution
Alternate hypothesis is : Data do not follow the given distribution
The level of significance,
The test statistics is given by :
Chi square =
Here O is the observed frequencies
E is the expected frequencies
So we have, N = number of the categories = 4
df = degrees o freedom = N - 1
= 4 - 1 = 3
Critical value =
Calculating the table for the test statistics are given as :
Category prop. O E
A 0.2 14 20 1.80
C 0.3 28 30 0.13
M 0.1 6 10 1.60
P 0.4 52 40 3.60
Total 1 100 100 7.13
The test statics = chi square = = 7.13
statistics = 7.13
P-value = 0.067767248
P -value >
Therefore, we do not reject the . There is sufficient evidence to conclude that the data follows the given distribution. So there is sufficient evidence to conclude the given claim is true.