Question

A survey of 501 randomly chosen US adults found that 40% of the 225 men and 50% of the 276 women made an online purchase during the past month. Do these data provide statistical evidence at the α = 0.05 level that women are more likely than men to make online purchases? Be sure to state the parameter, check conditions, perform calculations, and make conclusion(s). (10 points)

Answer 1

Answer:

$z=\frac{0.5-0.4}{\sqrt{0.45(1-0.45)(\frac{1}{225}+\frac{1}{276})}}=2.24$  

$p_v =P(Z>2.24)=0.013$  

If we compare the p value and using any significance level for example $\alpha=0.05$ always $p_v<\alpha$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that at 5% of significance the proportion for women made is higher than the proportion for male.

Step-by-step explanation:

Data given and notation  

$n_{1}=225$ sample 1 male selected

$n_{2}=276$ sample 2 female selected

$p_{1}=0.40$ represent the proportion of male selected

$p_{2}=0.5$ represent the proportion of female selected

z would represent the statistic (variable of interest)  

$p_v$ represent the value for the test (variable of interest)

Concepts and formulas to use  

We need to conduct a hypothesis in order to check if the proportion for female is greater than males , the system of hypothesis would be:  

Null hypothesis:$p_{2} \leq p_{1}$  

Alternative hypothesis:$p_{2} > p_{1}$  

We need to apply a z test to compare proportions, and the statistic is given by:  

$z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}$   (1)

Where $\hat p=\frac{\hat p_1 +\hat p_2}{2}=\frac{0.4+0.5}{2}=0.45$

Calculate the statistic

Replacing in formula (1) the values obtained we got this:  

$z=\frac{0.5-0.4}{\sqrt{0.45(1-0.45)(\frac{1}{225}+\frac{1}{276})}}=2.24$  

Statistical decision

Since is a one sided test the p value would be:  

$p_v =P(Z>2.24)=0.013$  

If we compare the p value and using any significance level for example $\alpha=0.05$ always $p_v<\alpha$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that at 5% of significance the proportion for women made is higher than the proportion for male.

Answer 2

Answer:

Step-by-step explanation:

Hello!

You need to test at 5% that the proportion of women that purchased online past month is greater than the proportion of men that purchased online past month.

There are two study variables:

X₁: Number of men that purchased online past month, out of 225

n₁= 225

p'₁= 0.40

X₂: Number of women that purchased online past month, out of 276

n₂= 276

p'₂= 0.50

Since both variables met the Binomial criteria:

1. The number of observations of the trial is fixed.

2. The experiment has only two possible outcomes, "success" and "failure"

3. Each observation in the trial is independent, this means that none of the trials will affect the probability of the next trial.

4. The probability of success in the same from one trial to another.

We can say that X₁~ Bi(n₁;ρ₁) and X₂~ Bi(n₂;ρ₂)

Since n₁≥30; n₁*ρ₁≥5; n₁*(1-ρ₁)≥5 and n₂≥30; n₂*ρ₂≥5; n₂(1-ρ₂)≥5 it is valid to use the Central Limit Theorem to approximate the distribution of both sample proportions p'₁ and p'₂ to normal and use the approximation of the standard normal with pooled sample proportion to analyze if the claim is true or not.

The statistic hypotheses are:

H₀: ρ₁ ≥ ρ₂

H₁: ρ₁ < ρ₂

α: 0.05

This test is one-tailed left, the critical value is:

$Z_{\alpha }= Z_{0.05}= -1.648$

You will reject the null hypothesis to values of $Z_{H_0}$ ≤ -1.648

$Z_{H_0}= \frac{(p'_1-p'_2)-(p_1-p_2)}{\sqrt{p'(1-p')*(\frac{1}{n_1} +\frac{1}{n_2} )} }$

$p'= \frac{(p'_1+p'_2)}{2} = 0.45$

$Z_{H_0}= \frac{(0.40-0.50)-0}{\sqrt{0.45*0.55(\frac{1}{225} +\frac{1}{276} )} } = -2.24$

Since the obtained value is less than the critical value the decision is to reject the null hypothesis.

At a level of 5%, you can say that the population proportion of women that purchased online past month is greater than the population proportion of men that purchased online past month.

I hope it helps!