Answer: (1034.29, 1010.95)
Step-by-step explanation:
Sample size (n) = 503
Average sample score (x) = 1019
Sample standard deviation (s) = 95
Even though our population standard deviation is unknown ( we only know of the sample standard deviation), the sample size is greater than 30, hence the critical value for the test will be that of a z test.
The 95% confidence level for population mean is given below as
u = x ± Zα/2 × s/√n
Where
Zα/2 = 1.96 = critical value for a two tailed test at a 5% level of significance ( the confidence level + level of significance = 100, hence a 95% confidence level corresponds to a 5% level of significance).
The upper limit of the interval is given as
u = x + Zα/2 × s/√n
Hence, we have that
u = 1019 + 1.96 × (95/√503)
u = 1019 + 1.96× (4.2358)
u = 1019 + 8.0481
u = 1034.29
The lower limit of the interval is given as
u = x - Zα/2 × s/√n
Hence, we have that
u = 1019 - 1.96 × (95/√503)
u = 1019 - 1.96× (4.2358)
u = 1019 - 8.0481
u = 1010.95
Hence, the 95% confidence level is (1034.29, 1010.95)
