Answer:
The 95% confidence interval for the population mean rating is (5.73, 6.95).
Step-by-step explanation:
We start by calculating the mean and standard deviation of the sample:
![M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M=\dfrac{1}{50}(6+4+6+. . .+6)\\\\\\M=\dfrac{317}{50}\\\\\\M=6.34\\\\\\s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{49}((6-6.34)^2+(4-6.34)^2+(6-6.34)^2+. . . +(6-6.34)^2)}\\\\\\s=\sqrt{\dfrac{229.22}{49}}\\\\\\s=\sqrt{4.68}=2.16\\\\\\](https://tex.z-dn.net/?f=M%3D%5Cdfrac%7B1%7D%7Bn%7D%5Csum_%7Bi%3D1%7D%5En%5C%2Cx_i%5C%5C%5C%5C%5C%5CM%3D%5Cdfrac%7B1%7D%7B50%7D%286%2B4%2B6%2B.%20.%20.%2B6%29%5C%5C%5C%5C%5C%5CM%3D%5Cdfrac%7B317%7D%7B50%7D%5C%5C%5C%5C%5C%5CM%3D6.34%5C%5C%5C%5C%5C%5Cs%3D%5Csqrt%7B%5Cdfrac%7B1%7D%7Bn-1%7D%5Csum_%7Bi%3D1%7D%5En%5C%2C%28x_i-M%29%5E2%7D%5C%5C%5C%5C%5C%5Cs%3D%5Csqrt%7B%5Cdfrac%7B1%7D%7B49%7D%28%286-6.34%29%5E2%2B%284-6.34%29%5E2%2B%286-6.34%29%5E2%2B.%20.%20.%20%2B%286-6.34%29%5E2%29%7D%5C%5C%5C%5C%5C%5Cs%3D%5Csqrt%7B%5Cdfrac%7B229.22%7D%7B49%7D%7D%5C%5C%5C%5C%5C%5Cs%3D%5Csqrt%7B4.68%7D%3D2.16%5C%5C%5C%5C%5C%5C)
We have to calculate a 95% confidence interval for the mean.
The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.
The sample mean is M=6.34.
The sample size is N=50.
When σ is not known, s divided by the square root of N is used as an estimate of σM:
The degrees of freedom for this sample size are:
![df=n-1=50-1=49](https://tex.z-dn.net/?f=df%3Dn-1%3D50-1%3D49)
The t-value for a 95% confidence interval and 49 degrees of freedom is t=2.01.
The margin of error (MOE) can be calculated as:
![MOE=t\cdot s_M=2.01 \cdot 0.305=0.61](https://tex.z-dn.net/?f=MOE%3Dt%5Ccdot%20s_M%3D2.01%20%5Ccdot%200.305%3D0.61)
Then, the lower and upper bounds of the confidence interval are:
The 95% confidence interval for the mean is (5.73, 6.95).