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Please help me!!!

Suppose the proportion p of a school’s students who oppose a change to the school’s dress code is 73%. Nicole surveys a random sample of 56 students to find the percent of students who oppose the change. What are the values of p that she is likely to obtain?

User Jayshao
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Since Nicole is taking a sample of 56 students, she may obtain a sample proportion that is different from the true population proportion of 73%. The range of sample proportions that Nicole is likely to obtain can be calculated using confidence intervals.

Assuming a normal distribution for the sample proportion, the 95% confidence interval can be calculated as:

p ± 1.96 * sqrt(p*(1-p)/n)

where:

p is the population proportion (given as 0.73)
n is the sample size (given as 56)
1.96 is the z-score for a 95% confidence level (which is a standard value used in statistics)
Substituting the given values, we get:

0.73 ± 1.96 * sqrt(0.73*(1-0.73)/56)
= 0.73 ± 0.112

Therefore, Nicole is likely to obtain sample proportions in the range of 0.618 to 0.842 (or equivalently, 61.8% to 84.2%). This means that if Nicole were to take many random samples of 56 students from the school, 95% of the sample proportions she obtains would fall within this range.
User Mark P Neyer
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