Question

Let the mean of the population be 38 instances of from 6" - 9" hatchings per nest, and let the standard deviation of the mean be 4. What sample mean would have a confidence level of 95% or a 2.5% margin of error? 32 34 36 38

Answer 1

Answer with explanation:

Mean of the population having 6"-9" Hatching =38

Standard Deviation =4

Confidence interval of 95% means that 95% of data lie within two standard deviations.

That is data values lie between 38-2×4 to 38 +2×4.

⇒Which is 38-8 to 38+8.

⇒30 to 46.

it is also given that margin of error is 2.5%.

So, the data value will lie between

$30-30*\frac{2.5}{100} {\text{to}} 46 +46*\frac{2.5}{100}\\\\ 30-0.75 {\text{to}}46+\frac{115}{100}\\\\ 29.25 {\text{to}}47.15$

Mean of beginning value to extreme value of 95% confidence level

   $=\frac{29.25+47.15}{2}\\\\=38.20$

So, Sample mean of this observation having 2.5% margin of error = 38

Option D:→ 38

   

Answer 2

I saw 34 I'm trying that now