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IRINA_888 [86]
3 years ago
15

Birthdays of hockey players: In Malcolm Gladwell’s book "Outliers" he shares the work of Canadian psychologist Roger Barnsley, w

ho noticed that a disproportionately high percentage of elite ice-hockey players have birthdays between January and March. A group of statistics students would like to test if this is true for the Los Angeles Kings 2010-2015 rosters (22 out of 57). After debating whether this set of hockey players can be viewed as a random sample of hockey players, they decide to run a hypothesis test anyway to practice finding the P‐value. They test the hypotheses LaTeX: H_0H 0: LaTeX: p=0.25p = 0.25 versus LaTeX: H_aH a: LaTeX: p>0.25p > 0.25. They use a significance level of 0.05. Their calculated test statistic is 2.37. Using the applet (at the top of this Checkpoint), what is the P‐value?
a. P‐value = 0.009
b. P‐value = 0.991
c. P‐value = 0.018
d. P‐value = 0.05
Mathematics
1 answer:
alexgriva [62]3 years ago
6 0

Answer:

The P-value is 0.009.

Step-by-step explanation:

We are given that after debating whether this set of hockey players can be viewed as a random sample of hockey players, they decide to run a hypothesis test anyway to practice finding the P‐value.

<em><u>We are given that the following hypothesis below;</u></em>

Null Hypothesis, H_0 : p = 0.25

Alternate Hypothesis, H_A : p > 0.25

Now, the test statistics that was used here for the above hypothesis would be;  

                       T.S.  =  \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

Also, the test statistics given to us is 2.37.

<u></u>

<u>SO, the P-value of the test statistics is given by the following formula;</u>

             P-value = P(Z > 2.37) = 1 - P(Z \leq 2.37)

                           = 1 - 0.99111 = 0.00889 ≈ 0.009

<em>The above probability is calculated by looking at the value of x = 2.37 in the z table which has an area of 0.99111.</em>

Therefore, the required P-value is 0.009.

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Step-by-step explanation:

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