Respuesta :
Answer:
1.1264
Step-by-step explanation:
Given : An experiment using 35 guinea pigs is set up to study the weight of the pigs after injecting them with a drug.
To Find: If the population mean is 23.5 grams with a standard deviation of 3.4 grams, what is the margin of error of the sample mean?
Solution:
n = 35
[tex]\sigma = 3.4[/tex]
Confidence interval = 95%. So, z =1.96
Margin of error = [tex]z \times \frac{\sigma}{\sqrt{n}}[/tex]
Margin of error = [tex]1.96\times \frac{3.4}{\sqrt{35}}[/tex]
Margin of error = [tex]1.1264[/tex]
Hence the margin of error of the sample mean is 1.1264.
Answer:
Margin of Error = [tex]\pm 0.1904[/tex]
Step-by-step explanation:
We are given the following information in the question:
Population mean, μ = 23.5 grams
Population standard deviation, σ = 3.4 grams
Sample size, n = 35
Alpha, α = 0.05
Formula:
[tex]\text{Margin of Error} = z\displaystyle\frac{\sigma}{\sqrt{n}}[/tex]
where z is the z-critical at 0.05 level of significance.
Putting the values, we get:
z-critical at 0.05 level of significance = [tex]\pm 1.96[/tex]
Margin of Error =
[tex]\pm 1.96\displaystyle\frac{3.4}{\sqrt{35}} = \pm 1.1264[/tex]
