Mannn what the heck...it’s so confusing sorry I couldn’t read.its messing
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).
If you do not mind me asking, what did Seth write? Us helpers cannot answer it if we do not have the full question. I apologize if this seems rude.
