Answer:
The 95 percent confidence interval for the mean of the population from which the study subjects may be presumed to have been drawn is (19.1269, 32.6730).
Step-by-step explanation:
Intern No. of Breast
Number Exams Performed X²
1 30 900
2 40 1600
3 8 64
4 20 400
5 26 676
6 35 1225
7 35 1225
8 20 400
9 25 625
<u>10 20 400 </u>
<u> </u><u> ∑ 259 ∑ 7515</u>
Mean= X`= ∑x/n= 259/10= 25.9
Variance = s²= 1/n-1[∑X²- (∑x)²/n]
= 1/0[7515- (259)²/10]= 1/9[7515- 6708.1]
= 806.9/9=89.655= 89.66
Standard Deviation= √89.655= 9.4687
Hence
The value of t with significance level alpha= 0.05 and 9 degrees of freedom is t(0.025,9)= 2.262
The 95 % Confidence interval is given by
x`±t(∝,n-1) s/√n
So Putting the values
25.9± 2.262( 9.4687/√10)
= 25.9 ±2.262 (2.9943)
= 25.9 ± 6.7730
= 25.9 +6.7730=32.6730
25.9 -6.7730= 19.1269
= 19.1269, 32.6730
The 95 percent confidence interval for the mean of the population from which the study subjects may be presumed to have been drawn is (19.1269, 32.6730).
Answer:
Step-by-step explanation:
Undefined.
The answer is alternate interior angles theorem and substitution property of equality.
