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Suppose a bowler claims that her bowling score is not equal to 150 points, on average. Several of her teammates do not believe her, so the bowler decides to do a hypothesis test, at a 5% significance level, to persuade them. She bowls 22 games. The mean score of the sample games is 157 points. The bowler knows from experience that the standard deviation for her bowling score is 18 points.

Answer :

Answer:

We conclude that the bowler's score is equal to 150 points.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 150

Sample mean, [tex]\bar{x}[/tex] = 2 157

Sample size, n = 22

Alpha, α = 0.05

Population standard deviation, σ = 18

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 150\text{ points}\\H_A: \mu \neq 150\text{ points}[/tex]

We use Two-tailed z test to perform this hypothesis.

Formula:

[tex]z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]

Putting all the values, we have

[tex]z_{stat} = \displaystyle\frac{157 - 150}{\frac{18}{\sqrt{22}} } = 1.825[/tex]

Now, [tex]z_{critical} \text{ at 0.05 level of significance } = \pm 1.96[/tex]

Since, z-statistic lies within the range of acceptance region that is from -1.96 to 1.96, we fail to reject the null hypothesis and accept the null hypothesis.

Thus, bowler claims that her bowling score is not equal to 150 points is not true and we conclude that the bowler's score is equal to 150 points.

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