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](https://tex.z-dn.net/?f=%5Coverline%7Bx%7D%20%3D%20%5Cfrac%7B2.2%20%2B%201.5%20%2B%200.5%20%2B%201.3%20%2B%202.4%20%2B%201.5%20%2B%202.7%20%2B%200.3%20%2B%202.0%20%2B%200.3%7D%7B10%7D%20%3D%201.42)
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](https://tex.z-dn.net/?f=s%20%3D%20%5Csqrt%7B%5Cfrac%7B%282.2-1.42%29%5E2%20%2B%20%281.5-1.42%29%5E2%20%2B%20%280.5-1.42%29%5E2%20%2B%20%281.3-1.42%29%5E2%20%2B%20%282.4-1.42%29%5E2%20%2B%20%281.5-1.42%29%5E2%20%2B%20%282.7-1.42%29%5E2%20%2B%20...%7D%7B9%7D%7D%20%3D%200.8772)
Thus, the estimate for the population standard deviation is of 0.8772 million.
Question c:
The sample size is ![n = 10](https://tex.z-dn.net/?f=n%20%3D%2010)
The significance level is ![\alpha = 1 - 0.05 = 0.95](https://tex.z-dn.net/?f=%5Calpha%20%3D%201%20-%200.05%20%3D%200.95)
The estimate, which is the sample standard deviation, is of
.
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](https://tex.z-dn.net/?f=%5Cchi%5E2_%7B%5Cfrac%7B%5Calpha%7D%7B2%7D%2Cn-1%7D%20%3D%20%5Cchi%5E2_%7B0.025%2C9%7D%20%3D%2019.0228)
![\chi^2_{1-\frac{\alpha}{2},n-1} = \chi^2_{0.975,9} = 2.7004](https://tex.z-dn.net/?f=%5Cchi%5E2_%7B1-%5Cfrac%7B%5Calpha%7D%7B2%7D%2Cn-1%7D%20%3D%20%5Cchi%5E2_%7B0.975%2C9%7D%20%3D%202.7004)
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}}](https://tex.z-dn.net/?f=%5Cfrac%7B%28n-1%29s%5E2%7D%7B%5Cchi%5E2_%7B%5Cfrac%7B%5Calpha%7D%7B2%7D%2Cn-1%7D%7D%20%3C%20%5Csigma%5E2%20%3C%20%5Cfrac%7B%28n-1%29s%5E2%7D%7B%5Cchi%5E2_%7B1-%5Cfrac%7B%5Calpha%7D%7B2%7D%2Cn-1%7D%7D)
![\frac{9*0.8772^2}{19.0228} < \sigma^2 < \frac{9*0.8772^2}{2.7004}](https://tex.z-dn.net/?f=%5Cfrac%7B9%2A0.8772%5E2%7D%7B19.0228%7D%20%3C%20%5Csigma%5E2%20%3C%20%5Cfrac%7B9%2A0.8772%5E2%7D%7B2.7004%7D)
![0.3641 < \sigma^2 < 2.5646](https://tex.z-dn.net/?f=0.3641%20%3C%20%5Csigma%5E2%20%3C%202.5646)
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](https://tex.z-dn.net/?f=%5Csqrt%7B0.3641%7D%20%3D%200.6034)
![\sqrt{2.5646} = 1.6014](https://tex.z-dn.net/?f=%5Csqrt%7B2.5646%7D%20%3D%201.6014)
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