Question

Which of these statements are true of inference for one or two proportions?

Answer 1

Answer:

b. II only

Step-by-step explanation:

Hello!

I. The populations from which the samples are drawn must be at least 10 times greater than the samples.

Incorrect, when sampling the size of the population only matters if it is a finite population (For example if you are studying an endangered species whose number is known, you have to take that into account when calculating the sample size) for infinite populations there are no restrictions about the sample size or its proportion in relation with it. As long as the sample is representative of the population to allow you to infer about it.

II. How you compute the standard error depends on whether you are constructing a confidence interval or performing a significance test.

Correct, the standard deviation for the sample proportion is $\sqrt{\frac{p(1-p)}{n} }$ and for two proportions $\sqrt{\frac{p_1(1-p_1}{n_1}+\frac{p_2(1-p_2)}{n_2} }$, as you see the standard deviation contains the population proportion.

When doing an hypothesis test for one population, you state the possible number of the proportion in the hypothesis and therefore you can use it to calculate the standard deviation. $\sqrt{\frac{p(1-p)}{n} }$

When making an hypothesis test for the difference between two population proportions the standard deviation is computed with a pooled sample proportion. $\sqrt{p'(1-p'*(\frac{1}{n_1} + \frac{1}{n_2})}$

When estimating the population proportion/s, since there is no information about the population value/s (If it was the estimation would be pointless) the standard deviation is also estimated and you calculate it using the information of the sample proportion:

One population: $\sqrt{\frac{^p(1-^p)}{n} }$

Two populations: $\sqrt{\frac{^p_1}{n_1} + \frac{^p_2}{n_2} }$

III. When you are doing inference for proportions, you use z-procedures because of the sampling distribution of pˆis normal.

Incorrect, the distribution of the sampling proportions is binomial, you can approximate it to normal by applying the Central Limit Theorem.

I hope it helps!