2 1/4 ÷( 3/8 ÷ 1/2 ) help
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The test statistic value lies to the right of the critical value. So we have sufficient evidence to reject the null hypothesis.
<h3>What are null hypotheses and alternative hypotheses?</h3>In null hypotheses, there is no relationship between the two phenomenons under the assumption or it is not associated with the group. And in alternative hypotheses, there is a relationship between the two chosen unknowns.
Students who take a statistics course are given a pre-test on the concepts and skills for the first chapter of a statistics course.
Then they are given a post-test once the professor has concluded lecturing on the material.
Pre-test and post-test scores for 4 students in an elementary statistics class are given below.
Then we have
Then the null hypotheses and alternative hypotheses will be
H₀:
Hₐ:
Then the test statistic will be
Then
The test statistic value is given by
Since this is a right-tailed test, so the critical value is given by
Since the test statistic value lies to the right of the critical value. So we have sufficient evidence to reject the null hypothesis.
Hence, we can conclude that that is test scores have improved.
More about the null hypotheses and alternative hypotheses link is given below.
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Answer:
a) Data given: 42 36 48 51 39 39 42 36 48 33 39 42 45
b) 44.1, 37.8, 50.4, 53.55, 40.95, 44.1, 37.8, 50.4 ,34.65, 40.95, 44.1, 47.25
c) 3.5, 3, 4, 4.25, 3.25, 3.25, 3.5, 3, 4, 2.75, 3.25, 3.5, 3.75
d) As we can see, the average of part b is 1.05 times the average of part a (1.05 * 41.538 = 43.615) and the average of part c is equal to the average obtained in part a divided by 12 (41.538 / 12 = 3.462).
And that happens because we create linear transformations for the parts b and c and the linear transformation affects the mean.
And you have the same interpretation for the deviation, it is affected by the linear transformation as the mean.
Step-by-step explanation:
For this case we can use the following formulas for the mean and standard deviation:
Part a
Data given: 42 36 48 51 39 39 42 36 48 33 39 42 45
And if we calculate the mean we got:
Part b
For this case we know that each value present a 5% of rise so we just need to multiply each value bu 1.05 and we have this new dataset:
44.1, 37.8, 50.4, 53.55, 40.95, 44.1, 37.8, 50.4 ,34.65, 40.95, 44.1, 47.25
And if we calculate the new mean and deviation we got:
Part c
The new dataset would be each value divided by 12 so we have:
3.5, 3, 4, 4.25, 3.25, 3.25, 3.5, 3, 4, 2.75, 3.25, 3.5, 3.75
And the new mean and deviation would be:
Part d
As we can see, the average of part b is 1.05 times the average of part a (1.05 * 41.538 = 43.615) and the average of part c is equal to the average obtained in part a divided by 12 (41.538 / 12 = 3,462).
And that happens because we create linear transformations for the parts b and c and the linear transformation affects the mean.
And you have the same interpretation for the deviation, it is affected by the linear transformation as the mean.