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Grace [21]
2 years ago
9

In 2018, Mike Krzyewski and John Calipari topped the list of highest paid college basketball coaches (Sports Illustrated website

). The following sample shows the head basketball coach's salary for a sample of 10 schools playing NCAA Division I basketball. Salary data are in millions of dollars.
University Coach's Salary University Coach's Salary
North Carolina State 2.2 Miami (FL) 1.5
Iona 0.5 Creighton 1.3
Texas A&M 2.4 Texas Tech 1.5
Oregon 2.7 South Dakota State 0.3
Iowa State 2.0 New Mexico State 0.3
a. Use the sample mean for the 10 schools to estimate the population mean annual salary for head basketball coaches at colleges and universities playing NCAA Division I basketball.
b. Use the data to estimate the population standard deviation for the annual salary for head basketball coaches.
c. What is the 95% confidence interval for the population variance?
d. What is the 95% confidence interval for the population standard deviation?
Mathematics
1 answer:
expeople1 [14]2 years ago
4 0

From the data given, we estimate the population mean and population standard deviation. Then, we use this estimate to find a 95% confidence interval for the population variance and the population standard deviation.

Sample:

Salaries in millions of dollars: 2.2, 1.5, 0.5, 1.3, 2.4, 1.5, 2.7, 0.3, 2.0, 0.3

Question a:

The mean is the sum of all values divided by the number of values. So

\overline{x} = \frac{2.2 + 1.5 + 0.5 + 1.3 + 2.4 + 1.5 + 2.7 + 0.3 + 2.0 + 0.3}{10} = 1.42

The sample mean salary is of 1.42 million.

Question b:

The standard deviation is the square root of the difference squared between each value and the mean, divided by one less than the number of values.

So

s = \sqrt{\frac{(2.2-1.42)^2 + (1.5-1.42)^2 + (0.5-1.42)^2 + (1.3-1.42)^2 + (2.4-1.42)^2 + (1.5-1.42)^2 + (2.7-1.42)^2 + ...}{9}} = 0.8772

Thus, the estimate for the population standard deviation is of 0.8772 million.

Question c:

The sample size is n = 10

The significance level is \alpha = 1 - 0.05 = 0.95

The estimate, which is the sample standard deviation, is of s = 0.8772.

Now, we have to find the critical values for the Pearson distribution. They are:

\chi^2_{\frac{\alpha}{2},n-1} = \chi^2_{0.025,9} = 19.0228

\chi^2_{1-\frac{\alpha}{2},n-1} = \chi^2_{0.975,9} = 2.7004

The confidence interval for the population variance is:

\frac{(n-1)s^2}{\chi^2_{\frac{\alpha}{2},n-1}} < \sigma^2 < \frac{(n-1)s^2}{\chi^2_{1-\frac{\alpha}{2},n-1}}

\frac{9*0.8772^2}{19.0228} < \sigma^2 < \frac{9*0.8772^2}{2.7004}

0.3641 < \sigma^2 < 2.5646

Thus, the 95% confidence interval for the population variance is (0.3641, 2.5646)

Question d:

Standard deviation is the square root of variance, so:

\sqrt{0.3641} = 0.6034

\sqrt{2.5646} = 1.6014

The 95% confidence interval for the population standard deviation is (0.6034, 1.6014).

For more on confidence intervals for population mean/standard deviation, you can check brainly.com/question/13807706

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a) 99.7% of the widget weights lie between 45 and 75 ounces

b) 97.2% of the widget weights lie between 50 and 75 ounces

c) 84% of the widget weights lie above 55

Step-by-step explanation:

The Empirical Rule states that:

50% of the values of a measure is above the mean, and the other 50% is below the mean.

99.7% of the values of a measure lie between 3 standard deviations of the mean.

95% of the values of a measure lie between 2 standard deviations of the mean.

68% of the values of a measure lie between 1 standard deviations of the mean.

In this problem, we have that: The widget weights have a mean of 60 ounces and a standard deviation of 5 ounces.

(a) 99.7% of the widget weights lie between

3 standard deviations of the mean, so:

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(b) What percentage of the widget weights lie between 50 and 75 ounces?

We have to find the percentage that are below 75 and subtract by the percentage that are below 50. So

75 is 3 standard deviations above the mean. So 99.7% of the measures are below 75.

50 is 2 standard deviations below the mean. So only 5% of the measures that are below the mean are below 50.

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(c) What percentage of the widget weights lie above 55?

55 is one standard deviation below the mean.

50% of the widget weights are above the mean.

Of the 50% that is below, 68% lie between one standard deviation(So from 55 to 60)

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50% + 68%(50%) = 84%

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