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Question

Assuming the population has an approximate normal distribution if a sample size n=10 has a sample mean x = 38 with a sample standard deviation s=2 find the margin error at a 95% confidence level. Round the answer to two decimal places

Answer 1

Hi Morgan,

We are not given a population standard deviation here, so we have to use a t-confidence interval. Formula for this is:

CI= xbar +/- t*SE

Now, margin of error is only the second part of that:

MoE = t*SE

t*=t-critical value obtained from t table or software

SE=standard error=s/sqrt(n)

Now, to find t*, we first must know degrees of freedom:

df= n-1

n= 10

df=10-1 = 9

Now, go to the t-table--it's available online but Wyzant reviews posts when I link--look up 95% confidence in the column and 9 degrees of freedom in the row:

t* = 2.262

Now, we still need the standard error:

SE=s/sqrt(n)

s=sample standard deviation

n=sample size

s= 2

n= 10

SE= 2/sqrt(10)

SE= 0.63

MoE = (2.262*0.63)

MoE = 1.43

I hope this helps.