Respuesta :
Answer:
A) sample mean = $1.36 million
B) standard deviation = $0.9189 million
C) confidence interval = ($1.93 million , $0.79 million)
*since the sample size is very small, the confidence interval is not valid.
Step-by-step explanation:
samples:
- $2.7 million
- $2.4 million
- $2.2 million
- $2 million
- $1.5 million
- $1.5 million
- $0.5 million
- $0.5 million
- $0.2 million
- $0.1 million
sample mean = $1.36 million
the standard deviation:
- $2.7 million - $1.36 million = 1.34² = 1.7956
- $2.4 million - $1.36 million = 1.04² = 1.0816
- $2.2 million - $1.36 million = 0.84² = 0.7056
- $2 million - $1.36 million = 0.64² = 0.4096
- $1.5 million - $1.36 million = 0.14² = 0.0196
- $1.5 million - $1.36 million = 0.14² = 0.0196
- $0.5 million - $1.36 million = -0.86² = 0.7396
- $0.5 million - $1.36 million = -0.86² = 0.7396
- $0.2 million - $1.36 million = -1.16² = 1.3456
- $0.1 million - $1.36 million = -1.26² = 1.5876
- total $8.444 million / 10 = $0.8444 million
standard deviation = √0.8444 = 0.9189
95% confidence interval = mean +/- 1.96 standard deviations/√n:
$1.36 million + [(1.96 x $0.9189 million)/√10] = $1.36 million + $0.57 million = $1.93 million
$1.36 million - $0.57 million = $0.79 million
Following are the calculation to the given points:
The average annual pay for head basketball coaches at NCAA Division 1 colleges and universities:
[tex]\to \bar{x}= \Sigma_{i=1}^{10} \ \frac{x_i}{n}\\\\[/tex]
[tex]=\frac{2.2 +0.5+2.4 +2.7 +2.0+1.5 +0.2+1.5 +0.1+0.2}{10}\\\\ = 1.33\\\\[/tex]
Therefore, the mean= 1.33
Calculating the standard deviation:
[tex]\to \sigma= \sqrt{\frac{1}{n} \Sigma (x_i -\bar{x})^2}\\\\[/tex]
[tex]= \sqrt{\frac{1}{10} {(2.2-1.33)^2 +(0.5 - 1.33)^2 + ... +(0.2 - 1.33)^2}}\\\\= 0.9508 \\\\[/tex]
As just a result, our projected standard deviation for annual compensation for head basketball coaches is 0.9508.
Calculating the sample variance:
[tex]\to s^2=\frac{1}{n-1} \Sigma (x_i -\bar{x})^2 \\\\[/tex]
[tex]=\frac{1}{10-1} {(2.2-1.33)^2 +(0.5-1.33)^2 +..........+(0.2-1.33)^2} \\\\=1.005\\\\[/tex]
Calculating the degrees of freedom:
[tex]\to df = n-1 = 10-1 = 9[/tex]
The formula to compute interval estimate of a population variance is as follows:
[tex]\to \frac{(n-1)s^2}{ \chi^2_{\frac{\alpha}{2}} } \leq \sigma^2 \leq \frac{ (n-1) s^2}{ \chi^2_{(1- \frac{\alpha}{2})}}[/tex]
From the chi-square distribution tables the critical value of
[tex]\to \chi^2_{\frac{0.05}{2}}\ \ is\ \ 19.0228[/tex]
From the chi-square distribution tables the critical value of
[tex]\to \chi^2_{(1-(\frac{0.05}{2}))} = 2.7004[/tex]
Thus,
[tex]\to \frac{(10-1)1.005}{19.0228} \leq \sigma^2 \leq \frac{(10 - 1)1.005}{2.7004}\\\\ \to 0.48 \leq \sigma^2 \leq 3.35[/tex]
Therefore, the [tex]95\%[/tex] confidence interval for the population variance is lies between [tex]0.48 \ and \ 3.35[/tex]
Calculating the sample variance:
[tex]\to s^2 =\frac{1}{n-1} \Sigma (x_i-\bar{x})^2\\\\[/tex]
[tex]=\frac{1}{10-1} {(2.2 -1.33)^2 +(0.5-1.33)^2 +...+(0.2 - 1.33)^2}\\\\=1.005 \\\\[/tex]
Calculating the degrees of freedom:
[tex]\to df = n-1 = 10-1 = 9[/tex]
The formula to compute interval estimate of a population variance is as follows:
[tex]\to \sqrt{\frac{(n-1)s^2}{ \chi^2_{\frac{\alpha}{2}}}} \leq \sigma \leq \sqrt{\frac{(n-1) s^2}{ \chi^2_{(1- \frac{\alpha}{2})}}[/tex]
From the chi-square distribution tables the critical value of [tex]\chi^2_{\frac{0.05}{2}} = 19.0228[/tex]
From the chi-square distribution tables the critical value of[tex]\chi^2_{(1-(\frac{0.05}{2}))} = 2.7004[/tex]
Thus,
[tex]\to \sqrt{\frac{(10-1)1.005}{19.0228}} \leq \sigma \leq \sqrt{\frac{(10 - 1)1.005}{2.7004}}\\\\ \to \sqrt{0.48} \leq \sigma \leq \sqrt{3.35}\\\\ \to 0.69 \leq \sigma \leq 1.83\\\\[/tex]
Therefore, the [tex]95\%[/tex] confidence interval for the population variance is lies between [tex]0.69 \ and\ 1.83[/tex]
Learn more:
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