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radazharula1
26.02.2020 •
Mathematics
You plan to conduct a survey to find what proportion of the workforce has two or more jobs. You decide on the 98% confidence level and a margin of error of 2%. A pilot survey reveals that 7 of the 50 sampled hold two or more jobs.
1. How many in the workforce should be interviewed to meet your requirement?
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Ответ:
1635 people
Step-by-step explanation:
The proportion of people with two or more jobs in the sample is
7/50 = 0.14
The z-value that gives us a 98% confidence level is 2.33 (this means that 98% of the area under the normal curve N(0;1) is between -2.33 and 2.33)
If we use Simple Random sampling, the size of the sample should be
where z is the z-value which gives us the desired confidence level, p the proportion and e the error. In our case
z = 2.33
p = 0.14
e = 2% = 0.02
So, the sample size should be
rounding up to the nearest integer.
Ответ:
4166 ants
Step-by-step explanation:
To find how many ants were left, multiply 5000 by 5/6, since it decreased by 1/6:
5000(5/6)
= 4166.67
Round this down:
= 4166
So, at the end of April, there were approximately 4166 ants left