Confidence Interval

Question

Find the necessary confidence interval for a population mean 𝜇 for the following values. (Round your answers to two decimal places.)

𝛼 = 0.01, n = 36, x = 31, s² = 13

Answer 1

Hi Matthew,

Is your x actually x-bar, a sample mean, or is it just a single observation? If just a single observation, I'm not sure how to compute a confidence interval from that. I suspect, since you were given s, that it is a sample mean. GIven that you don't know population standard deviation, we use a t-confidence interval here. The formula for that is:

CI= x-bar +/- t*SE where:

x-bar=sample mean

t*=t critical value, obtained from t-table or software

SE=standard error, which we also must calculate with formula below:

SE=s/sqrt(n)

Here,

s²=13, so

s=sqrt(13)

n=36

SE=[sqrt(13)]/[sqrt(36]

SE=sqrt(13)/6

SE=0.60

Now, back to our confidence interval formula:

x-bar=31

SE=0.60

Now, we need to compute t*. We can get this from a table, but first we need to know degrees of freedom and confidence level.

The formula:

df=n-1

df=35

Now, for significance level, you were given:

a=0.01

For a two-sided t-interval, which we must assume this is--we weren't told otherwise--we have to divide this by 2 and get 0.005. When we check the t-table--many are available online but Wyzant reviews posts with links so I can't link here--they list the reverse of this 0.995.

On the t-table, look on the left side for degrees of freedom--35 in this case, and along the top for the confidence level--0.995 here. My t-table did not have 35; the closest it has is 30, so we'll use that. You want closest without going over in this case. From table:

df=30, C. Level 0.995;

t*= 2.75

So, plugging back in:

x-bar=31

t*=2.75

SE=0.60

CI=31 +/- (2.75*0.60)

CI=31 +/- 1.65

CI=(29.35, 32.65)

I hope this helps.