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bigcm3497
19.06.2020 •
Mathematics
In a random survey of 226 college students, 20% reported being "only" children (no siblings). The standard deviation is 2%. Determine the confidence interval that is likely to contain the percent of all students nationwide who are only children.
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Ответ:
The confidence interval that is likely to contain the percent of all students nationwide who are only children is (0.1505,0.2554)
Step-by-step explanation:
Total number of students = n = 226
We are given that 20% reported being "only" children (no siblings).
Number of students being "only" children (no siblings)![= 20\% \times 226=\frac{20}{100} \times 226=45.6\sim46](/tpl/images/0690/4108/5c5c1.png)
So, x = 46
Z at 95% is 1.96
Formula for confidence interval for sample proportion :
Substitute the values in the formula :
Hence the confidence interval that is likely to contain the percent of all students nationwide who are only children is (0.1505,0.2554)
Ответ:
Step-by-step explanation:
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