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laiz [17]
3 years ago
14

An author argued that more basketball players have birthdates in the months immediately following July​ 31, because that was the

age cutoff date for nonschool basketball leagues. Here is a sample of frequency counts of months of birthdates of randomly selected professional basketball players starting with​ January: 390​, 392​, 360​, 318​, 344​, 330​, 322​, 496​, 486​, 486​, 381​, 331 . Using a 0.05 significance​ level, is there sufficient evidence to warrant rejection of the claim that professional basketball players are born in different months with the same​ frequency? Do the sample values appear to support the​ author's claim?

Mathematics
1 answer:
Lady_Fox [76]3 years ago
6 0

Answer:

There is sufficient evidence to warrant rejection of the claim that professional basketball players are born in different months with the same​ frequency.

Step-by-step explanation:

In this case we need to test whether there is sufficient evidence to warrant rejection of the claim that professional basketball players are born in different months with the same​ frequency.

A Chi-square test for goodness of fit will be used in this case.

The hypothesis can be defined as:

<em>H₀</em>: The observed frequencies are same as the expected frequencies.

<em>Hₐ</em>: The observed frequencies are not same as the expected frequencies.

The test statistic is given as follows:

 \chi^{2}=\sum{\frac{(O-E)^{2}}{E}}

The values are computed in the table.

The test statistic value is \chi^{2}=128.12.

The degrees of freedom of the test is:

n - 1 = 12 - 1 = 11

Compute the <em>p</em>-value of the test as follows:

p-value < 0.00001

*Use a Chi-square table.

p-value < 0.00001 < α = 0.05.

So, the null hypothesis will be rejected at any significance level.

Thus, there is sufficient evidence to warrant rejection of the claim that professional basketball players are born in different months with the same​ frequency.

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Step-by-step explanation:

We have the following data set:

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The standard deviation measures how close the set of data is to the mean value of the data set. If data set have high standard deviation than the values are spread out very much. If data set have small standard deviation the data points are very close to the mean.

To find standard deviation we use the following formula

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Create the below table.

Find the sum of numbers in the last column to get.

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