Based on this, construct a 99% confidence interval for the true population mean watermelon weight.

Question

You measure 21 watermelons' weights, and find they have a mean weight of 67 ounces. Assume the population standard deviation is 10.9 ounces. Based on this, construct a 99% confidence interval for the true population mean watermelon weight.

Give your answers as decimals, to two places

Answer 1

Hi Kaitlyn,

This is one of those "close call" confidence intervals. We do know population standard deviation, but n is relatively small. Not sure if z is appropriate here, so I'll go with a t-confidence interval. Formula for that is:

CI=x-bar +/- t*SE

x-bar=sample mean

t*=t-critical value

SE=standard error

Any time we work with samples and sample means, we have to compute a standard error:

SE=sigma/sqrt(n)

sigma=population standard deviation

n=sample size

Here:

sigma=10.9

n=21

SE=10.9/sqrt(21)

SE=2.38

Returning to our original equation:

CI=x-bar +/-t*SE,

we now have:

x-bar=67

SE=2.38

Now, we need to find t*. To do this, first we must calculate degrees of freedom:

df=n-1

df=21-1

df=20

Now, we can go to the t-table and look for 99% confidence in the column and 20 degrees of freedom in the row. This gives:

t*=2.845

Therefore:

CI=67 +/- (2.38*2.845)

CI=67 +/- 6.77

CI=(60.23, 73.77)

I hope this helps.