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Svetach [21]
2 years ago
6

Sarah wants to study the time it takes to complete a lab in a regular biology course compared to the time it takes to complete a

lab in an honors biology course. Which quantities could Sarah use to make her comparisons? Check all that apply.
amount of time needed to complete a cell lab in a regular biology course


amount of time needed to complete a cell lab in an honors biology course


number of students taking a regular biology course


number of students taking an honors biology course


amount of time needed to write a cell lab report in a regular biology course


amount of time needed to write a cell lab report in an honors biology course
Mathematics
2 answers:
Tcecarenko [31]2 years ago
6 0
Just did it...its the first one...second one....fifth one and sixth one...that..is if your on edegenuity..
Olin [163]2 years ago
3 0

Sarah is studying the time it takes to complete a lab in a regular biology course compared to the time it takes to complete a lab in an honors biology course .The question is asking us to select the correct quantities Sarah use to make her comparisons :

Among the given options  Sarah  needs to have:

amount of time needed to complete a cell lab in a regular biology course .

amount of time needed to complete a cell lab in an honors biology course

amount of time needed to write a cell lab report in a regular biology course

amount of time needed to write a cell lab report in an honors biology course

Options 1,2 ,5 ,6 are the right options.




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The subjects of a study by Dugoff et al. (A-5) were 10 obstetrics and gynecology interns at the University of Colorado Health Sc
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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).

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