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Answer:
There is no convincing evidence at α = 0.10 level that the average vertical jump of students at this school differs from 15 inches.
Step-by-step explanation:
We have to make a hypothesis test to prove the claim that the average vertical jump of students differs from 15 inches.
The null and alternative hypothesis are:
[tex]H_0: \mu=15\\\\H_a: \mu\neq15[/tex]
The significance level is 0.10.
The sample mean is 17 and the sample standard deviation is 5.37.
The degrees of freedom are df=(20-1)=19.
The t-statistic is:
[tex]t_{19}=\frac{M-\mu}{s_M/\sqrt{n}} =\frac{17-15}{5.37/\sqrt{20}}=\frac{2}{5.37/4.47} =2/1.20=1.67[/tex]
The two-sided P-value for t=1.67 is P=0.11132.
This P-value is bigger than the significance level, so the effect is not significant. The null hypothesis can not be rejected.
There is no convincing evidence at α = 0.10 level that the average vertical jump of students at this school differs from 15 inches.
There is no convincing evidence at α = 0.10 level that the average vertical jump of students at this school differs from 15 inches.
What are null hypotheses and alternative hypotheses?
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.
Student researchers Haley, Jeff, and Nathan saw an article on the Internet claiming that the average vertical jump for teens was 15 inches.
They wondered if the average vertical jump of students at their school differed from 15 inches, so they obtained a list of student names and selected a random sample of 20 students.
We have to make a hypothesis test to prove the claim the vertical jump of the students differs from 15 inches.
The null hypotheses and alternative hypotheses will be
H₀: μ = 15
Hₐ: μ ≠ 15
The significance level is 0.10.
The sample mean is 17 and the sample deviation is 5.37.
The degrees of freedom are df = 20 - 1 = 19
The t-statistic will be
[tex]\rm t_{19} = \dfrac{M - \mu}{\frac{ \sigma }{\sqrt{n}}} = \dfrac{17- 15}{\frac{ 5.37}{\sqrt{12}}}\\\\\\t_{19} = \dfrac{2}{1.20} = 1.67[/tex]
The two-sided P-value for t = 1.67 is P = 0.11132.
This P-value is bigger than the significance level, so the effect is not significant. The null hypothesis can not be rejected.
There is no convincing evidence at α = 0.10 level that the average vertical jump of students at this school differs from 15 inches.
More about the null hypotheses and alternative hypotheses link is given below.
https://brainly.com/question/9504281
