Enter your answer as an open-interval (i.e., parentheses) accurate to one decimal place (because the sample statistics are reported accurate to one decimal place).

Question

Assume that a sample is used to estimate a population mean. Find the 90% confidence interval for a sample of size 59 with a mean of 48.3 and a standard deviation of 17.9. Enter your answer as an open-interval (i.e., parentheses) accurate to one decimal place (because the sample statistics are reported accurate to one decimal place).

90% C.I. = 

Answer 1

Hi Kaitlyn,

I think you can use a z-confidence interval for this, since you have n>30, but your instructor may prefer a t-confidence interval. Let me know if that is the case. Anyhow, proceeding for z, the confidence interval formula is:

CI=x-bar +/- z*SE

where

x-bar=sample mean

z*=z-critical value

SE=standard error

SE=s/sqrt n

So, we need to compute standard error first:

s=17.9

n=59

SE=17.9/sqrt(59)

SE=2.33

Now, we need z*. You can get it from a z-table, but I recommend memorizing it since 90% confidence intervals are relatively common. It's z*=1.645. If asked for 95%, it's z*=1.96. Anyhow, returning to our original formula:

CI=x-bar +/- z*SE

We now know:

x-bar=48.3

z*=1.645

SE=2.33

Thus:

CI=48.3 +(1.645*2.33)

CI=48.3 +/- 3.8

CI= (44.5, 52.1)

I hope this helps.