Answer :
Answer:
1)
[tex]H_0: p_1-p_2=0\\\\H_a: p_1-p_2>0[/tex]
2) z=3.164
3) Critic value z₀=1.282.
4) P=0.00078
5) The correct answer is: "Since the p-value is less than the given value of alpha, there is sufficient evidence to reject H_0"
Step-by-step explanation:
5.1) Being:
p₁: proportion of men who keep track of the deadlines in their head
p₂: proportion of women who keep track of the deadlines in their head
If we want to test if p₁ is larger than p₂, the null hypothesis and the alternative hypothesis should be:
[tex]H_0: p_1-p_2=0\\\\H_a: p_1-p_2>0[/tex]
In this way, if we reject the null hypothesis, it can be claimed that p₁ is larger than p₂.
5.2) Compute the test statistic for the test.
First, we have to estimate a proportion as if the null hypothesis is true. This means the average of proportion of the samples taken from men and women, weighted by the sample size.
[tex]\bar p=\frac{n_1p_1+n_2p_2}{n_1+n_2}=\frac{500*0.54+500*0.44}{500+500}= 0.49[/tex]
Then, we used this average to estimate the standard error
[tex]s=\sqrt{\frac{p(1-p)}{n_1}+{\frac{p(1-p)}{n_2}}}=\sqrt{\frac{0.49(1-0.49)}{500}+{\frac{0.49(1-0.49)}{500}}}=\sqrt{0.0004998+0.0004998}\\\\s= 0.0316[/tex]
Lastly, we calculate the statistic z
[tex]z=\frac{p_1-p_2}{s}=\frac{0.54-0.44}{0.0316}=\frac{0.10}{0.0316}=3.164[/tex]
5.3) Give the rejection region for the test, using α = 0.10
For a one-tailed test with α = 0.10, the z value to limit the rejection region is z=1.282.
For every statistic larger than 1.282, the null hypothesis should be rejected.
5.4) Find the p-value for the test.
The p-value for a z=3.164 is P=0.00078 (corresponding to the area ot the standard normal distribution for a z larger than 3.164).
5.5) Choose the correct answer below.
The correct answer is: "Since the p-value is less than the given value of alpha, there is sufficient evidence to reject H_0"
The difference between the proportions is big enough to be statistically significant and enough evidence to reject the null hypothesis.