A candidate for political office wants to determine if there is a difference in his popularity between men and women. To test the claim of this difference, he conducts a survey of voters. The sample contains 250 men and 250 women, of which 44% of the men and 52% of the women favor his candidacy. Do these values indicate a difference in popularity?Use a 0.01 significance level. What are the hypothesis statements?
a) H0:pm=pw
HA:pm b) H0:pm=pw
HA:pm>pw
c) H0:pm=pw
HA:pm≠pw

Using the z-distribution, it is found that since the absolute value of the test statistic is less than the critical value, there values do not indicate a difference in popularity.

At the null hypothesis, it is tested if the proportions are equal, that is, their subtraction is of 0, hence:

At the alternative hypothesis, it is tested if they are different, that is, their subtraction is different of 0, hence:

The proportions and standard errors are:

For the distribution of the differences, the mean and the standard error are given by:

The test statistic is given by:

In which p = 0 is the value tested at the null hypothesis.

Hence:

The critical value, for a two-tailed test, as we are testing if the mean is different of a value, with a significance level of 0.01, is of

Since the absolute value of the test statistic is less than the critical value, there values do not indicate a difference in popularity.

A similar problem, also involving an hypothesis test for a proportion, is given at link

c) H0:pm=pw

HA:pm≠pw

Step-by-step explanation:

We formulate our hypothesis as

H0: pm = pw  " probability of men = probabilityof women" meaning there's no difference in the probabilityof the men and women in favor of his candidacy.

Alternate Hypothesis HA :pm≠pw " probability of men ≠ probabilityof women" meaning there's a difference in the probability of the men and women in favor of his candidacy.

the significance level α= 0.01

The test statistic under H0 is

Z = pm- pw/ √p`q` ( 1/n.m + 1/n.w)

pm= probability of men= 0.44

pw= probability of women = 0.52

p`= n.m pm+ n.w pw/ n.m + n.w

p`= 250 *0.44 + 250 *0.52/ 250 + 250

p`= 110 + 130 /500 = 240 /500 = 0.48

q`= 1- p`= 1-0.48= 0.52

Putting the values

Z= 0.44- 0.52/ √ 0.48 * 0.52

z= 0.08 / √0.2496

z=  0.08/ 0.4995

z= 0.1601

The critical region for α= 0.01 is Z= ± 2.58

Conclusion: Since the calculated z = 0.1601 does not fall in the critical region , so we accept the null hypothesis H0:pm=pw and conclude that the data does not appear to indicate that the tow probabilities are different.

thats not a system of equations, that is a solution