I'll go with A and B
Hope this helps
Data:
15 16 14 15 19 17
n=6 points
sum is 96
mean is 96/6 = 16
Now we look at the absolute deviations, each of which is the absolute value of a score minus the mean, basically the distance of the score to the mean .
Scores 15 16 14 15 19 17
AbsDev 1 0 2 1 3 1
The sum of the absolute deviations is 8 and there are six of them so the
Mean Absolute Deviation = 8/6 = 4/3
Answer: 2. 8/6
7^50
= 1.798 × 10^42
i am a mathematics teacher. if anything to ask please pm me
The result of the respective questions are:
- This chi-square test only takes into consideration one variable.
- The type of chi-square test this is is a Goodness of Fit
- df= 3
- NO
<h3>How many variables are involved in the chi-square test?</h3>
a)
This chi-square test only takes into consideration one variable.
b)
The type of chi-square test this is, is a Goodness of Fit
To test the hypothesis, we must determine whether the actual data conform to the assumed distribution.
The "Goodness-of-Fit" test is a statistical hypothesis test that determines how well the data that was seen resembles the data that was predicted.
c)
Parameter
n = 4
Therefore
Degrees of freedom
df= n - 1
df= 4 - 1
df= 3
d)
In conclusion
Parameters

df = 3
Hence
Critical value = 7.814728
Test statistic = 6.6
Test statistic < Critical value, .
NO, the result of this test is not statistically significant.
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