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The average speed of a volleyball serve is 57 miles per hour. Natalie practiced a new technique to improve her serving speed. Her coach recorded the speed of 36 random serves during practice and found that her average speed using the new technique was 58.9 miles per hour, with a standard deviation of 2.8 miles per hour.


Part A: State the correct hypotheses if Natalie is trying to prove the new technique is an improvement over the old technique.


Part B: Identify the correct test and check the appropriate conditions.

Answer :

Answer:

a) Null hypothesis: H₀ : μ = 57

Alternative hypothesis  H₁ : μ ≠ 57

b) The calculated value t =4.077 > 1.6896 at 99% of confidence degrees of freedom.

Natalie is trying to prove the new technique is an improvement over the old technique.

Step-by-step explanation:

Step:-(i)

The average speed of a volleyball serve is 57 miles per hour.

μ = 57 miles per hour

Average speed using the new technique was 58.9 miles per hour, with a standard deviation of 2.8 miles per hour.

sample mean x⁻ = 58.9 miles per hour

Standard deviation S = 2.8 miles per hour.

Step:-(ii)

Null hypothesis: H₀ : μ = 57

Alternative hypothesis  H₁ : μ ≠ 57

Degrees of freedom γ=n-1

   The test statistic          

[tex]t= \frac{x^{-} -mean}{\frac{S.D}{\sqrt{n} } }[/tex]

[tex]t= \frac{58.9 -57}{\frac{2.8}{\sqrt{36} } }[/tex]

t = 4.077

Degrees of freedom γ=n-1 = 35-1 =34

tₐ = 2.0322

The calculated value t =4.077 > 1.6896 at 99% of confidence degrees of freedom.

The Natalie is not trying to prove the new technique is an improvement over the old technique.

Conclusion

There is sufficient evidence to support the claim that the average speed of a volleyball serve is 57 miles per hour.

To understand the calculations, check below

Test Hypothesis:

Hypothesis testing is a set of formal procedures used by statisticians to either accept or reject statistical hypotheses.

We have given the claim that

The average speed of a volleyball serve is 57 miles per hour.

Given summary statistics are,

[tex]n=36, \bar{x}=58.9, s=2.8[/tex]

The degree of freedom is,

[tex]d.f=n-1\\=36-1\\=35[/tex]

Part A

The null and alternative hypotheses are,

[tex]H_0:\mu=57\\H_a:\mu\ne 57[/tex]

Part B

The test statistic value is,

[tex]t=\frac{\bar{x}-\mu_0}{\frac{s}{\sqrt{n} } } \\t=\frac{58.9-57}{\frac{2.8}{\sqrt{36} } } \\t=4.08[/tex]

The P-value is,

[tex]P-value=2\times P(t\le-|t_{cal}|)\\=2\times P(t\le-|4.08}|)\\\\=0.0002[/tex]

Assume level of significance = 0.05

Decision

Reject the null hypothesis because P-value (0.002) is less than the level of significance (0.05).

Learn more about the topic Test Hypothesis:

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