Confidence interval estimate of population mean HELP Question:

The International Air Transport Association surveys business travelers to develop quality ratings for transatlantic gateway airports. The maximum possible rating is 10. Suppose a simple random sample of 50 business travelers is selected and each traveler is asked to provide a rating for the Miami International Airport. The ratings obtained from the sample of 50 business travelers follow.

Develop a 95% confidence interval estimate of the population mean rating for Miami. Round your answers to two decimal places.

Data: 6
4
6
8
7
7
6
3
3
8
10
4
8
7
8
7
5
9
5
8
4
3
8
5
5
4
4
4
8
4
5
6
2
5
9
9
8
4
8
9
9
5
9
7
8
3
10
8
9
6

My answer:

Sameple: 50
Mean: 6.34
SD: 2.1629

I got -5.73 and +6.95 so    (0.61,13.29)

but it keeps saying that its the wrong answer. Anyone out there have any ideas?
You are doing the arithmetic wrong. For a 95% confidence interval, you need to find the values which are on $\pm 1.96 \times \sigma$ of the mean. I don't know where your -5.73 and +6.95 values are coming from, the confidence interval will be symmetric about the mean on a normal distribution. Try it again and if you are still struggling maybe take a look at this page: http://onlinestatbook.com/2/estimation/mean.html
You are doing the arithmetic wrong. For a 95% confidence interval, you need to find the values which are on $\pm 1.96 \times \sigma$ of the mean. I don't know where your -5.73 and +6.95 values are coming from, the confidence interval will be symmetric about the mean on a normal distribution. Try it again and if you are still struggling maybe take a look at this page: http://onlinestatbook.com/2/estimation/mean.html