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In a study of hormone supplementation to enable oocyte retrieval for assisted reproduction, a team of researchers administered two hormones in different timing strategies to two randomly selected groups of women aged 36 – 40 years. For the Group A treatment strategy, the researchers included both hormones from day 1 . The mean number of oocytes retrieved from the 98 participants in Group A was 9.7 with an 80 % confidence level z ‑interval of ( 8.8 , 10.6 ) .

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Answer:

Note: This question is incomplete and complete question is posted down below with multiple choice questions and from Option 3 is correct

Step-by-step explanation:

Note: This question is incomplete and it is a multiple choice question. With the following multiple choices to select from.

1. The researchers expect that 80% of all similarly constructed intervals will contain the true mean number of oocytes that could be retrieved from the population of women aged 36-40 years.

2. The researchers expect that 80% of all similarly constructed intervals will contain the mean number of oocytes retrieved in the sample of 98 women aged 36-40 years.

3. There is a 80% chance that the the true mean number of oocytes that could be retrieved from the population of women aged 36-40 years is uniquely contained in the reported interval.

4. The researchers expect that 80% of all similarly constructed intervals will contain the range of the number of oocytes that could be retrieved from the population of women aged 36-40 years.

we have to select the appropriate answer from the above options.  

So, in order to choose the correct option, first we need to understand the confidence interval.

Confidence Interval is a range of value for which are pretty sure our true values will lie in that interval. For example, in this question, we have 80% confidence interval and it means there is a 80% confident chance that our true values will lie in that 80% interval. Let's calculate it.

We have following data given from the question.    

Sample = 98

Mean = 9.7

Confidence Interval = 80%

Z = 1.282

And formula for confidence interval is :

Mean ± Z x SD/[tex]\sqrt{n}[/tex]

For this equation, we have all quantities except SD which is standard deviation. Let's calculate it first.

Standard deviation is square root of variance and so we need to calculate variance first.

Mean = Sum/ total number

9.7 = Sum/ 98

Sum = 950.6

For variance we will use this sum.

s² = (950.6 (1 - 9.7)²) ÷ (98-1)

s² = 741. 762 = Variance.

Let's calculate Standard deviation

SD = [tex]\sqrt{variance}[/tex]

SD = [tex]\sqrt{741.762}[/tex]

SD = 27.23 = Standard Deviation.

Now, we have complete data to calculate range of values for the confidence interval of 80%

Mean ± Z x SD/[tex]\sqrt{n}[/tex]

Here, Z is the Z-score and it has predetermined value for different confidence intervals and here in our case for 80% , Value of Z = 1.282

Now, by just plugging in the values in to above formula we will get range of values or confidence interval of 80% in which our true value will likely to lie.

9.7 ± 1.282 x 27.23/[tex]\sqrt{98}[/tex]

9.7 ± 3.52 Or

9.7 + 3.52 = 13.22

9.7 - 3.52 = 6.18

Rangle of Values with confidence interval of 80% = (6.18 to 13.22).

Which means that, our true mean will most likely lie in this interval.

So, the option 3 is correct which meets the correct definition and interpretation of the problem.  

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