Question

A product developer wants to determine if there is a difference in the proportion of people who have a specific brand of smart phone in the U.S. and Poland. A sample of 10,500 people in the U.S. was obtained and 3,000 reported that they owned the type of smart phone in question. In Poland, 5,000 people were surveyed and 1,250 said they owned the smart phone. The developer knows that a z-test for two population proportions is needed to analyze the data but is not sure how to calculate it and he turns to you for help. Based upon the information provided, what is the value of z

Answer 1

`z=\frac{0.286-0.250}{\sqrt{0.274(1-0.274)((1)/(10500)+(1)/(5000))}}=4.696`

Explanation:

1) Data given and notation

`X_(US)=3000` represent the number of people reported that they owned the type of smart phone in question in U.S

`X_(P)=1250` represent the number of people reported that they owned the type of smart phone in question in Poland

`n_(US)=10500` sample of US selected

`n_(P)=5000` sample of Poland selected

`p_(US)=(3000)/(10500)=0.286` represent the proportion of people reported that they owned the type of smart phone in question in U.S

`p_(P)=(1250)/(5000)=0.250` represent the proportion of people reported that they owned the type of smart phone in question in Poland

z would represent the statistic (variable of interest)

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

`\alpha=0.05` significance level given

2) Concepts and formulas to use

We need to conduct a hypothesis in order to check if is there is a difference in the two proportions analyzed, the system of hypothesis would be:

Null hypothesis:
`p_(US) - p_(P)=0`

Alternative hypothesis:
`p_(US) - p_(P) \\eq 0`

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

`z=\frac{p_(US)-p_(P)}{\sqrt{\hat p (1-\hat p)((1)/(n_(US))+(1)/(n_(P)))}}` (1)

Where
`\hat p=(X_(US)+X_(P))/(n_(US)+n_(P))=(3000+1250)/(10500+5000)=0.274`

3) Calculate the statistic

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

`z=\frac{0.286-0.250}{\sqrt{0.274(1-0.274)((1)/(10500)+(1)/(5000))}}=4.696`

4) Statistical decision

Since is a two side test the p value would be:

`p_v =2*P(Z>4.696)= 0.00000265`

Comparing the p value with the significance level assumed
`\alpha=0.05` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that we have significant differences in the two proportions analyzed.