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Danielyanez
31.12.2019 •
Mathematics
Arandom sample of 125 students is chosen from a population of 4,000 students. if the mean iq in the sample is 100 with a standard deviation of 8, what is the 98% confidence interval for the students' mean iq score?
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Ответ:
98.33 - 101.67
Step-by-step explanation:
The given Sample mean n= 100
Standard deviation![\sigma=8](/tpl/images/0437/7586/86912.png)
Mean IQ in the sample![\bar{x}=125](/tpl/images/0437/7586/f86b5.png)
Standard error of mean =![\frac{\sigma}{\sqrt{\bar{x}}}](/tpl/images/0437/7586/244a2.png)
Standard error of mean =![\frac{8}{\sqrt{125}}](/tpl/images/0437/7586/7e420.png)
Thus, standard error =![\frac{8}{11.180}](/tpl/images/0437/7586/3783d.png)
Standard error of mean (SE)= 0.7155
We know that z - score for 98% confidence interval is 2.33 .
Now, 98% confidence interval will be
n-z(SE) and n+z(SE)
Hence, the 98% confidence interval for the students' mean IQ score will be 98.33 - 101.67
Ответ: