Find the 98% confidence interval for a sample of size 416 with a mean of 54.3 and a standard deviation of 13.4.

Question

Assume that a sample is used to estimate a population mean. Find the 98% confidence interval for a sample of size 416 with a mean of 54.3 and a standard deviation of 13.4. Enter your answer as a tri-linear inequality accurate to one decimal place (because the sample statistics are reported accurate to one decimal place).

Answer should be obtained without any preliminary rounding. However, the critical value may be rounded to 3 decimal places.

Answer 1

Hi Kaitlyn,

I think your sample size is large enough that we can use a z confidence interval, but if you need t, let me know. The problem with t is that most tables don't give an accurate value for our degrees of freedom, which would be n-1, 459-1=458 in this case. They tend to go 100, then 1000, which is not ideal. So, we'll go with z.

Formula for this type of confidence interval is:

CI=x-bar +/- z*SE

Let's break this down.

x-bar=sample mean

z*=z-critical value, available at z table

SE=standard error, which we must account for when dealing with samples

Standard error has its own formula:

SE=sigma/sqrt(n), so let's compute that now.

sigma=13.4

n=459

SE=13.4/sqrt(459)

SE=0.625

Now, returning to our original formula:

CI=x-bar +/- z* SE

x-bar=54.3

SE=0.625

To find z*, look at interior of z-table for 0.99, since we have a two-sided interval, we need to divide (1-.98)/2 where 0.98 is our confidence level. Closest on my z-table is 0.9901 where:

z=2.33

So:

x-bar=54.3

z*=2.33

SE=0.625

CI=(54.3+/-(2.33*0.625))

CI=54.3 +/- 1.5

Written as a trilinear inequality, this would be:

(54.3-1.5)<x<(54.3+1.5)

Good luck. I hope this helps.