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).
First, add all the numbers together, which is 68. Then divided 100 by 68, which gives you the number 1.47, and that's your answer.
From 0g to 494.7g are rejected as well as those heavier than 505.3grams.
You calculate this by subtracting and adding 5.3 to 500
Then make up a word problem that you can use decimals in.