Answer: 90% confidence interval would be (6.022,6.986).
Explanation:
Since we have given that
Number of students = 361
Sample mean = 6.504
Sample standard deviation = 5.584
Standard error of the mean = 0.294
At 90% confidence level,
So, α = 0.01
So, z = 1.64
Margin of error is given by
![z* \text{Standard error}\\\\=1.64* 0.294\\\\=0.48216](https://img.qammunity.org/2020/formulas/mathematics/college/wod11xgf3qxpip415yez48ypqugh46b2fu.png)
So, Lower limit would be
![\bar{x}-0.482\\\\=6.504-0.482\\\\=6.022](https://img.qammunity.org/2020/formulas/mathematics/college/vj75pwu8xt1d3h67sckq93y5k5hjuz5xld.png)
Upper limit would be
![\bar{x}+0.482\\\\=6.504+0.482\\\\=6.986](https://img.qammunity.org/2020/formulas/mathematics/college/rphim6y91v14sh07mek7q5xujqeqfdmgdo.png)
So, 90% confidence interval would be (6.022,6.986).