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Brrunno [24]
2 years ago
7

Data show that men between the ages of 20 and 29 in a general population have a mean height of 69.3​ inches, with a standard dev

iation of 2.5 inches. a baseball analyst wonders whether the standard deviation of heights of​ major-league baseball players is less than 2.5 inches. the heights​ (in inches) of
Mathematics
1 answer:
dangina [55]2 years ago
4 0

Data show that men between the ages of 20 and 29 in a general population have a mean height of 69.3​ inches, with a standard deviation of 2.5 inches. a baseball analyst wonders whether the standard deviation of heights of​ major-league baseball players is less than 2.5 inches. the heights​ (in inches) of  20 randomly selected players are shown in the table.

72 74 71 72 76

70 77 75 72 72

77 72 75 70 73

74 75 73 74 74

What are the correct hypotheses for this  test?

The null hypothesis is H₀?: ____ 2.5

The alternative hypothesis is H₁?: ____  2.5

Calculate the value of the test statistic.

x² = _____ (Round to three decimal places)

Answer:

Null hypothesis, H₀: σ = 2.5

Alternative hypothesis,  Hₐ: μ<2.5

Test statistic = 12.920

Step-by-step explanation:

Given Data shows that:

men between the ages of 20 and 29 in a general population have a mean height of 69.3​ inches, with a standard deviation of 2.5 inches

We consider a random sample of 20 selected baseball players.

Therefore;

The Null and Alternative hypothesis are as follows:

The Null hypothesis is the standard deviation of the heights of major league baseball players is not less than 2.5 inches.

Null hypothesis, H₀: σ = 2.5

On the other hand: The Alternative hypothesis is the standard deviation of the heights of major league baseball players is less than 2.5 inches.  

Alternative hypothesis,  Hₐ: μ<2.5

The Mean Calculation is:

\bar{x} = \frac{1}{2} \sum x_i

= \frac{1}{20} (72+74+...+74) \\ \\ = \frac{1468}{20} \\ \\ =73.4

The sample standard deviation is:

s = \sqrt{\frac{1}{n-1} \sum (x_1 - \bar{x})^2 }

= \sqrt{\frac{1}{20-1} \sum (72-73.4)^2 + ...+(74-73.4)^2 }  \\ \\ =  \sqrt{4.25}  \\ \\ = 2.06

The test statistics is now determined as :

x^2 = \frac{(n-1)s^2}{\sigma^2} \\ \\ = \frac{(20-1)(2.06)^2}{(2.5)^2}  \\ \\ = \frac{19*4.25}{6.25} \\ \\ = \frac{80.75}{6.25} \\ \\ = 12.920

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