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:

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:

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:

Then, the lower and upper bounds of the confidence interval are:
The 95% confidence interval for the mean is (5.73, 6.95).