no calculators please! need to know how to work it out the long way, thank you!
a) From the given data,
Premimum (Xi) | (Xi-Xbar)^2 | |
1905 | 417316 | |
2725 | 30276 | |
2677 | 15876 | |
2600 | 2401 | |
2962 | 168921 | |
2184 | 134689 | |
2332 | 47961 | |
3112 | 314721 | |
2545 | 36 | |
2525 | 676 | |
2370 | 32761 | |
2545 | 36 | |
2529 | 484 | |
2442 | 11881 | |
2312 | 57121 | |
2981 | 184900 | |
2627 | 5776 | |
2857 | 93636 | |
2675 | 15376 | |
2115 | 190096 | |
Total: | 51020 | 1724940 |
Sample Mean = 51020/20 = 2551
Sample variance (s2) = 1724940 / (20-1) = 90786.31579
SD (s) = 301.3077
a) The point estiamte of the mean annual automobile insurance premium in Michigan is sample mean = $2551
b)
The 95% confidence interval for the mean annual automobile
insurance premium in Michigan is
(xbar - tcrit * s/sqrt(n), xbar + tcrit * s/sqrt(n)) = (2551 -
2.093*301.3077/sqrt(20), 2551 + 2.093*301.3077/sqrt(20)) =
(2409.983667,2692.016333)
c)
c) Given The average annula premium for automobile insurance in the
US is %1503 is not in the above interval. Thus the annula premium
for automobile insurance in the Michigan is not include the nationa
average for the US.
The 95% chance that the annula premium for automobile insurance in the US is lies between 2409.983667 and 2692.016333
no calculators please! need to know how to work it out the long way, thank you!...
no calculators please! need to know how to work it out the long way, thank you! A simple random sample of 50 items from a population with σ-6 resulted in a sample mean of 32. a. Provide a 90% confidence interval for the population mean. b. Provide a 95% confidence interval for the population mean c. Provide a 99% confidence interval for the population mean. 2,