Question

Steve's average golf score at his local course is 93.6 from a random sample of 34 rounds of golf. assume that the population standard deviation for his golf score is 4.2. the margin of error for a 95% confidence interval around this sample mean is ________.

Answer 1

Answer:  The margin of error for a 95% confidence interval around this sample mean is 1.412.

Step-by-step explanation:

We know that the formula to find the margin of error is given by :-

$E=\pm z^*{\dfrac{\sigma}{\sqrt{n}}}$, where $\sigma$ = population standard deviation

n= sample size , z* = critical z-value as per confidence level.

As per given , we have

$\sigma$ =4.2

n= 34

By z-table , for 95% confidence level : z* = 1.96

Then, the margin of error will be :

$E=\pm (1.96)\dfrac{4.2}{\sqrt{34}}\approx\pm1.412$

Hence, the margin of error for a 95% confidence interval around this sample mean is 1.412.

Answer 2

The margin of error is 1.41.

The margin of error is calculated using the formula

$z*\frac{\sigma}{\sqrt{n}}}$
where z is the z-score associated with the confidence level, σ is the standard deviation, and n is the sample size.  

For a confidence level of 95%, we find the z-score by first converting the percent to a decimal and subtracting it from 1:
1 - 0.95 = 0.05

Divide that by 2:
0.05/2 = 0.025
This is the area in each tail.

Subtract this from 1 again, since we want the area between the tails:
1 - 0.025 = 0.975

Looking this up in the z-table (http://www.z-table.com) we see that the z-score is 1.96.

Using our information, we have:
$1.96*(\frac{4.2}{\sqrt{34}}})=1.41$