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Savatey [412]
3 years ago
8

Kara Danvers (Supergirl) has always relied on her strength to win fights. But what happens when she meets an alien just as stron

g? Her sister is training her to be a more technical fighter so that Supergirl can meet any challenge. The data below record the significant strikes during randomly selected training sessions 6 months apart. Is Kara showing improvement in her fighting?
Mathematics
1 answer:
attashe74 [19]3 years ago
7 0

Answer:

The answer is below

Step-by-step explanation:

The corresponding data are missing, which are the following:

Strikes (pre):

29

32

44

34

19

Strikes (post):

51

45

68

92

64

We have to say the difference between the post-pre values of the strike. The d will be the average of the differences between the post and pre values. If Kara is to show improvement, her post-workout attacks should be more than the pre-workout values. Let m be the population mean of the difference:

H0: m = 0 the mean difference in Strikes between post and pre is zero.

H0: m>0, the mean difference in strikes between post and pre is more than zero.

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26. Students who take a statistics course are given a pre-test on the concepts and skills for the first chapter of a statistics
hjlf

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

\mu _d = \mu _{post} - \mu _{pre}

Then the null hypotheses and alternative hypotheses will be

H₀: \mu _d = 0

Hₐ: \mu _d > 0

Then the test statistic will be

\rm \overline{x} _d = \dfrac{\Sigma x_d}{n} = \dfrac{15+12+10+1}{4}\\\\\overline{x} _d = 9.5

Then

\rm S_d = 6.02

The test statistic value is given by

\rm t = \dfrac{\overline{x} _d }{\dfrac{S_d}{\sqrtn}} \\\\t = \dfrac{9.5}{\dfrac{6.02}{\sqrt4}}\\\\t = 3.16

Since this is a right-tailed test, so the critical value is given by

\rm t_{n-1}(\alpha ) = t_3 (0.05) = 2.353

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 \mu _d > 0 that is test scores have improved.

More about the null hypotheses and alternative hypotheses link is given below.

brainly.com/question/9504281

#SPJ1

7 0
2 years ago
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