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)

If we compare the p value and using any significance level for example always 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  

sample 1 male selected

sample 2 female selected

represent the proportion of male selected

represent the proportion of female selected

z would represent the statistic (variable of interest)  

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:  

Alternative hypothesis:  

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

  (1)

Where

Calculate the statistic

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

Statistical decision

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

If we compare the p value and using any significance level for example always 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.

blue marble is 15 times out of 50 = 15/50 reduces to 3/10

landing on tails is 24 out of 50 = 24/50 reduces to 12/25

to find the probability of both multiply together:

3/10 x 12/25 = 3 x 12 / 10 x 25 = 36/250

if you multiplied the numerator and denominator by 10 you get 360/2500.

the answer is 360/2500