Question

document or pdf with your answer. Your solution to each problem must have the following: • The null and alternative hypotheses with the appropriate symbols. • The significance level of the test. If it is not given use a significance level of 0.05. • The P-value for the test. • State whether you reject or fail to reject the null hypothesis. In other words, compare the P-value to the significance level to decide whether to reject or fail to reject the null hypothesis. • A complete sentence that states your conclusion to the problem.

Answer 1

1)

Z-test for two population proportion

Hypothesis

The null and alternative hypothesis are,

$H_0:p_1=p_2$

$H_A:p_1\neq p_2$

This is a two-tailed test.

Where,

$p_1=\frac{203}{750}=0.2707$

$p_2=\frac{270}{750}=0.36$

Test statistic

The z-statistic is,

$z=\frac{p_1-p_2}{\sqrt{\overline{p}(1-\overline{p})\left ( \frac{1}{n_1}+\frac{1}{n_2} \right )}}$

Where,

$\overline{p}=\frac{x_1+x_2}{n_1+n_2}=\frac{203+270}{750+750}=0.3153$

$z=\frac{0.3707-0.36}{\sqrt{0.3153(1-0.3153)\left ( \frac{1}{750}+\frac{1}{750} \right )}}=-3.723$

P-value

The P-value for the z-statistic = -3.723 is obtained from the z distribution table.

$\text{P-value}=0.0002$

Decision

Since the p-value = 0.0002 is less than 0.05 at a 5% significance level, the null hypothesis is rejected.

Conclusion

There is sufficient evidence to conclude that the proportion of men who are overweight is different from the proportion of women who are overweight.

3)

T-test for two population mean

Hypothesis

The null and alternative hypothesis are,

$H_0:\mu_1=\mu_2$

$H_a:\mu_1<\mu_2$

This is a left-tailed test.

The significance level = 0.05

Test statistic

The t statistic is obtained using the formula,

$t=\frac{\overline{X}_1-\overline{X}_2}{\sqrt{\left ( \frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2} \right )\left ( \frac{1}{n_1}+\frac{1}{n_2} \right )}}$

$t=\frac{41-54}{\sqrt{\left ( \frac{(513-1)2.6^2+(513-1)2.9^2}{513+513-2} \right )\left ( \frac{1}{513}+\frac{1}{513} \right )}}$

$t=-75.598$

P-value

The P-value for the t-statistic = -75.598 is obtained from the t distribution table for the degree of freedom = n1+n2-2= 1024

$\text{P-value}=0.0000$

Decision

Since the p-value = 0.0000 is less than 0.05 at a 5% significance level, the null hypothesis is rejected.

Conclusion

There is sufficient evidence to conclude that the Democrats believe fewer tax dollars are wasted compared to Republicans.