Answer:
CI(95%): [205,5;400.8]dolars
Step-by-step explanation:
Hello!
So you need to construct a confidence interval for the average family dental expenses (μ) using the sample given in the problem. To estimate it you need to first choose a statistic. For this small sample, considering that the variable has a normal distribution (I made a quick Shapiro Wilks test, with p-value 0.3234, you can assume normality) the best statistic to use is the Student's t-test.
t= [x(bar)-μ]/S/√n ≈ t₍ₙ₋₁₎
The formula for the confidence interval to estimate the mean is
x(bar)±* (S/√n)
<u>The critical value is from a t-distribution with 11 degrees of freedom </u>
±
<em> >remember since it's two-tailed, to get the right critical value you have to divide α by 2. So in the text, you received a confidence level of 1-α=0.95 so α=0.05 then α/2=0.025 and 1-α/2=0.975</em>
To construct the interval, you need to first calculate the sample mean and the standard derivation.
<u>Sample</u>
115; 370; 250; 593; 540; 225; 117; 425; 318; 182; 275; 228
n= 12
∑xi = 3638 ∑xi² = 1362670
<u>Sample mean</u>
x(bar): (∑xi)/n = 3638/12 = 303.17 dolars
<u>Standard derivation</u>
S²= 1/n-1*[∑xi²- (∑xi)²/n] = 1/11 * [1362670-((3638)²/12)] = 23614.47 dolars²
S= 153.67 dolars
<u>Confidence interval (95%)</u>
303.17± 2.201* (153.67/√12)
[205,5;400.8]dolars
I hope you have a SUPER day!
