Answer:
A) An 80% confidence interval for the population mean score of the current group of applicants is (172.8124,202.9876)
B)The confidence level of this interval is 96%
Step-by-step explanation:
Mean =![\mu = 187.9](https://tex.z-dn.net/?f=%5Cmu%20%3D%20187.9)
Standard deviation = s= 32.4
n = 9
A) Find an 80% confidence interval for the population mean score of the current group of applicants.
Degree of freedom = n-1=9-1=8
Significance level = ![\alpha = 0.2](https://tex.z-dn.net/?f=%5Calpha%20%3D%200.2)
So, ![t_(\frac{\alpha}{2},df)=t_(\frac{0.2}{2},8)=t_(0.1,8)=1.397](https://tex.z-dn.net/?f=t_%28%5Cfrac%7B%5Calpha%7D%7B2%7D%2Cdf%29%3Dt_%28%5Cfrac%7B0.2%7D%7B2%7D%2C8%29%3Dt_%280.1%2C8%29%3D1.397)
Formula of confidence interval :
CI : ![\bar{x}-Z_{(\frac{\alpha}{2})} \times \frac{s}{\sqrt{n}} , \bar{x}+Z_{(\frac{\alpha}{2})}\times \frac{s}{\sqrt{n}}](https://tex.z-dn.net/?f=%5Cbar%7Bx%7D-Z_%7B%28%5Cfrac%7B%5Calpha%7D%7B2%7D%29%7D%20%5Ctimes%20%5Cfrac%7Bs%7D%7B%5Csqrt%7Bn%7D%7D%20%2C%20%5Cbar%7Bx%7D%2BZ_%7B%28%5Cfrac%7B%5Calpha%7D%7B2%7D%29%7D%5Ctimes%20%5Cfrac%7Bs%7D%7B%5Csqrt%7Bn%7D%7D)
So, CI: ![187.9-1.397 \times \frac{32.4}{\sqrt{9}} , 187.9+1.397 \times \frac{32.4}{\sqrt{9}}](https://tex.z-dn.net/?f=187.9-1.397%20%5Ctimes%20%5Cfrac%7B32.4%7D%7B%5Csqrt%7B9%7D%7D%20%2C%20187.9%2B1.397%20%5Ctimes%20%5Cfrac%7B32.4%7D%7B%5Csqrt%7B9%7D%7D)
CI:(172.8124,202.9876)
So, an 80% confidence interval for the population mean score of the current group of applicants is (172.8124,202.9876)
B)Based on these sample results, a statistician found for the population mean a confidence interval extending from 165.8 to 210.0 points. Find the confidence level of this interval.
Standard error of sample mean = se =![\frac{s}{\sqrt{n}}= \frac{32.4}{\sqrt{9}} = 10.8](https://tex.z-dn.net/?f=%5Cfrac%7Bs%7D%7B%5Csqrt%7Bn%7D%7D%3D%20%5Cfrac%7B32.4%7D%7B%5Csqrt%7B9%7D%7D%20%3D%2010.8)
Margin of error = 210 - 165.8 = 44.2
Critical test statistic =![\frac{\text{margin of error}}{2 \times \text{standard error}} = \frac{44.2}{2 \times 10.8} = 2.0463](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%7Bmargin%20of%20error%7D%7D%7B2%20%5Ctimes%20%5Ctext%7Bstandard%20error%7D%7D%20%3D%20%5Cfrac%7B44.2%7D%7B2%20%5Ctimes%2010.8%7D%20%3D%202.0463)
Using excel
Confidence interval = ![2 \times [1 - NORMSDIST(2.0463)] = 2 \times (1 - 0.98) = 2 \times 0.02 = 0.04](https://tex.z-dn.net/?f=2%20%5Ctimes%20%5B1%20-%20NORMSDIST%282.0463%29%5D%20%3D%202%20%5Ctimes%20%281%20-%200.98%29%20%3D%202%20%5Ctimes%200.02%20%3D%200.04)
Confidence level= 1 - 0.04 = 0.96
Confidence interval is 96%.
Hence the confidence level of this interval is 96%