Respuesta :
The value of sample size n in this statement according to a confidence interval is 350.
According to the statement
we have a given that the value of error and population proportion and we have to find the value of sample size.
So, we know that the
confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
Here
p represent the real population proportion of interest
ρ represent the estimated proportion for the sample
n is the sample size required (variable of interest)
z represent the critical value for the margin of error
So, The population proportion have the following distribution
P≈n ( p, ([p (1-p)]/n)^1/2)
Here In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by α = 1 - 0.95 = 0.05
and α/2 = 0.025 And the critical value would be given by:
z = 1.96
And on this case we have that ME = ±0.05 and we are interested in order to find the value of n, if we solve n from equation (a) we got the value
n = 0.35(1-0.35) / (0.05/1.96)^2
so, the value becomes
n = 350.
So, The value of sample size n in this statement is 350.
Learn more about sample size here https://brainly.com/question/17203075
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Question:
In a survey, the planning value for the population proportion is p* = 0.35. How large a sample should be taken to provide a 95% confidence interval with a margin of error of 0.05?
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