SOLUTION: A sample of size n=26 has a sample mean x= 14.8 and a sample standard deviation s=2.6
Construct a 95% confidence interval for the population mean μ.
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Question 1103602: A sample of size n=26 has a sample mean x= 14.8 and a sample standard deviation s=2.6
Construct a 95% confidence interval for the population mean μ.
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
The sample size is n = 26, so the degrees of freedom (df) is df = n-1 = 26-1 = 25
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To form the confidence interval, we need to know the t critical value. Why T instead of Z? Because n = 26 is not larger than 30 and because we don't know the population standard deviation sigma.
There are two basic options to find the t critical value: use a calculator or use a table. I prefer a calculator as it is faster and more accurate. Plus, its likely you'll have a calculator handy (on your phone maybe) rather than a bulky table.
I'm going to use this calculator
Type in 25 for the "degrees of freedom" box. Then type 0.05 for the "probability" box. The value 0.05 comes from the fact that alpha = 1-C = 1-0.95 = 0.05, where C is the confidence level (in this case 95%)
Once those values are typed in, hit the "calculate" button, and you'll see two results. The first we'll ignore. The second result shows plus/minus 2.05953856
So the critical t value is approximately t = 2.05953856
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There are many other ways to find this value. If you have a TI83 or TI84, you'll follow these steps
- Hit the key labeled "2ND" up at the top left corner
- Press the VARS key (located just below the arrow keys)
- Scroll down to invT and hit enter
- Type in 0.025 (0.025 is half the value of alpha = 0.05; we cut alpha in half because we equally split alpha among the two tails)
- Type in a comma (the comma key is just above the 7 key) followed by 25 which is the degrees of freedom
- Type in a closing parenthesis. This step is optional, but a good habit to form
After following those steps, you should have this

The result my TI84 shows is roughly -2.059538532
We'll only use the positive version of this number so 2.059538532
Compare this with the other calculator used earlier and we have
2.05953856
2.059538532
The results are nearly identical. The only difference is the 6 at the end doesn't match with the 3 in the second to last slot
Likely due to some kind of rounding error, though I'm not 100% sure.
Since we don't need to be that accurate, lets round the t critical value to 6 decimal places
So we'll go from 2.059538532 to 2.059539
The t critical value we'll use is t = 2.059539
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From the earlier section above, we found the t critical value to be roughly t = 2.059539
The sample mean is xbar = 14.8
The sample standard deviation is s = 2.6
The sample size is n = 26
These four values (t, xbar, s, and n) will be used to compute the lower and upper limits (L and U respectively)
The lower endpoint of the confidence interval L is
L = xbar - t*s/sqrt(n)
L = 14.8 - 2.059539*2.6/sqrt(26)
L = 14.8 - 1.050163
L = 13.749837
L = 13.7
The upper endpoint of the confidence interval U is
U = xbar + t*s/sqrt(n)
U = 14.8 + 2.059539*2.6/sqrt(26)
U = 14.8 + 1.050163
U = 15.850163
U = 15.9
Side Note: the margin of error is roughly 1.050163, which is equal to t*s/sqrt(n)
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Rounded to one decimal place, we got L = 13.7 and U = 15.9
The 95% confidence interval for the population mean mu is (L,U) = (13.7, 15.9)
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