Question

According to a poll of adults about 41% work during their summer vacation. Suppose that this claim about the population proportion is true. Now if we take a sample of 75 adults, and find the sample proportion [^(p)] of adults who work during summer vacation.

What is the expected value of sample proportion [^(p)]?

Tries 0/5

What is the standard deviation of sample proportion [^(p)]?

Tries 0/5

The shape of the sampling distribution of sample proportion [^(p)] will be roughly like a
normal distribution
uniform distribution
Poisson distribution
student's t distribution
binomial distribution

Tries 0/3

P(0.30 ≤ [^(p)] ≤ 0.47) =

Tries 0/5

Is the sample large enough to compute the above probability?
Yes, because np ≥ 10 and n(1−p) ≥ 10.
No, the sample size is not sufficiently large. We actually had to assume normally distributed population.
Yes, because n ≥ 30
Yes, because np ≥ 10.
Yes, because the number of successes and failures are both larger than 10.

Tries 0/3

Find c such that P([^(p)] ≥ c) = 0.7

Answer 1

Since the sample size is greater than 30, The shape of the sampling distribution of sample proportion is a roughly normal distribution.

From Z-table, Lookup for Z-value corresponding to area 0.7 to the right of the normal curve.