A random sample of six observations from a Normal population with unknown mean μ has sample mean 38.40 and sample standard deviation 3.50.
A 95% prediction interval for a future observation X is
38.40 ± 2.45×3.50×√(1+1/5)
38.40 ± 2.57×(1+3.50/√6)
38.40 ± 1.96×3.50×√(1+1/6)
38.40 ± 1.96×3.50
38.40 ± 2.45×(1+ 3.50)
38.40 ± 2.57×3.50×√(1+1/6)
I tried to attempt this question twice and still got it wrong and now im just extremely confused. At first when i read that the population mean was unknown I thought the answer would be 38.40 ± 1.96×3.50×√(1+1/6) since if the mean is unknown but the standard deviation was known then the prediction interval would be x-bar±1.96×σ√(1+1/n).
But then I thought that since they give you the sample mean it would indicate the both the mean and standard deviation was known so i thought that the answer would be 38.40 ± 1.96×3.50, but once again it was wrong.
Can someone please help me! thanks :)
This is a t test so you need to look up the t value for df 5 at the .05 probability ----2 tail. And then use that number with the appropriate formula with $\sqrt 6$