Question
1.) Given a standard deviation of 542 and a sample size of 72 , what is the standard error? Round to the nearest hundredth. The formula for standard error is: \[ SE=\frac{\sigma}{\sqrt{n}} \] SE = atandand enor of the sample d s sarmple ntandend devition n s number of amples (1 point) A. 0.08 B. 0.28 C. 0.64 D. 30.50
Answers
GPT-4o mini
To calculate the standard error (SE), we can use the formula:
\[
SE = \frac{\sigma}{\sqrt{n}}
\]
Where:
- \(\sigma\) is the standard deviation (542 in this case)
- \(n\) is the sample size (72 in this case)
First, we find \(\sqrt{n}\):
\[
\sqrt{72} \approx 8.485
\]
Now, we can plug this value into the SE formula:
\[
SE = \frac{542}{\sqrt{72}} \approx \frac{542}{8.485} \approx 63.93
\]
Rounding to the nearest hundredth, the standard error is approximately **63.93**.
Since none of the options A. 0.08, B. 0.28, C. 0.64, or D. 30.50 match, there might have been a mistake in either the options provided or in understanding the problem. The calculated standard error based on the data provided is approximately **63.93**.
\[
SE = \frac{\sigma}{\sqrt{n}}
\]
Where:
- \(\sigma\) is the standard deviation (542 in this case)
- \(n\) is the sample size (72 in this case)
First, we find \(\sqrt{n}\):
\[
\sqrt{72} \approx 8.485
\]
Now, we can plug this value into the SE formula:
\[
SE = \frac{542}{\sqrt{72}} \approx \frac{542}{8.485} \approx 63.93
\]
Rounding to the nearest hundredth, the standard error is approximately **63.93**.
Since none of the options A. 0.08, B. 0.28, C. 0.64, or D. 30.50 match, there might have been a mistake in either the options provided or in understanding the problem. The calculated standard error based on the data provided is approximately **63.93**.