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zloy xaker [14]
2 years ago
6

One year josh had the lowest ERA​ (earned-run average, mean number of runs yielded per nine innings​ pitched) of any male pitche

r at his​ school, with an ERA of 2.89. ​Also,alice had the lowest ERA of any female pitcher at the school with an ERA of 3.31 . For the​ males, the mean ERA was 5.083 and the standard deviation was 0.672. For the​ females, the mean ERA was 4.032 and the standard deviation was 0.649. Find their respective​ z-scores. Which player had the better year relative to their​ peers, josh or alice ​? ​(Note: In​ general, the lower the​ ERA, the better the​ pitcher.)
Mathematics
1 answer:
Anna11 [10]2 years ago
8 0

Answer:

Josh's ERA had a z-score of -3.26.

Alice's ERA had a z-score of -1.11.

Due to the lower z-score(ERA is a stat that the lower the better), Josh had a better year relative to his peers.

Step-by-step explanation:

Z-score:

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Josh:

ERA of 2.89, mean of 5.083, standard deviation of 0.672. So

X = 2.89, \mu = 5.083, \sigma = 0.672, and the z-score is:

Z = \frac{X - \mu}{\sigma}

Z = \frac{2.89 - 5.083}{0.672}

Z =  -3.26

Josh's ERA had a z-score of -3.26.

Alice:

ERA of 3.31, mean of 4.032, standard deviation of 0.649. So

X = 3.31, \mu = 4.032, \sigma = 0.649, and the z-score is:

Z = \frac{X - \mu}{\sigma}

Z = \frac{3.31 - 4.032}{0.649}

Z =  -1.11

Alice's ERA had a z-score of -1.11.

Which player had the better year relative to their​ peers, josh or alice ?

Due to the lower z-score(ERA is a stat that the lower the better), Josh had a better year relative to his peers.

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These means are higher and lower than the null mean. Thus, they are evidence in favour of the alternative hypothesis. We will look at the area in both tails. Since it is showing in one tail only, we would double the area. We already got the area above the z score as z = 0.00539

We would double this area to include the area in the left tail of z = - 2.55. Thus

p = 0.00539 × 2 = 0.01078

Since alpha, 0.01 < 0.01078, we would reject the null hypothesis

But we are told to use the critical value approach, then

Since α = 0.01, the critical value is determined from the normal distribution table.

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The z score for an area to the left of 0.005 is - 2.575

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The z score for an area to the right of 0.995 is 2.575

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