Question

a new york times poll asked people in may of 2020 if they favor a nationwide mask mandate. the sample consisted of 750 adults and 55% said "yes". you suspect that the poll was biased and think that it is really 50% of people would favor a mask mandate. if this poll was not biased, what is the probability you would get a sample with 55% or more saying "yes" to a mask mandate?

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Darrell

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45 Answers

We can approach this problem using the normal distribution since the sample size is large enough.

Given:

Sample size,n = 750

Hypothesized population proportion,p = 0.50

Sample proportion,\hat{p} = 0.55

Standard deviation,\sigma=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.50(1-0.50)}{750}}=\sqrt{\frac{0.25}{750}}

To find the probability of getting a sample with 55% or more saying "yes" to a mask mandate, we need to calculate the z-score for this case:

z=\frac{\hat{p} - p}{\sigma}=\frac{0.55-0.50}{\sqrt{\frac{0.25}{750}}}\approx2.7386

Now, we find the probability using a standard normal distribution table:

P(\text{Z}\geq2.7386)\approx0.0031

Therefore, the probability you would get a sample with 55% or more saying "yes" to a mask mandate is approximately\boxed{0.0031} .

Given:

Sample size,

Hypothesized population proportion,

Sample proportion,

Standard deviation,

To find the probability of getting a sample with 55% or more saying "yes" to a mask mandate, we need to calculate the z-score for this case:

Now, we find the probability using a standard normal distribution table:

Therefore, the probability you would get a sample with 55% or more saying "yes" to a mask mandate is approximately

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