A study was made to determine whether more Italians than Americans prefer white champagne to pink champagne at weddings. Of the 300 Italians selected at random, 74 preferred white champagne, and of the 400 Americans selected, 60 preferred white champagne. Can we conclude that a higher proportion of Italians than Americans prefer white champagne at weddings? Use a 0.05 level of significance.

Comparing the p value with the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that the proportion of Italians who prefer white champagne at weddings it's significantly higher than the proportion of Americans.

Step-by-step explanation:

1) Data given and notation

[tex]X_{I}=74[/tex] represent the number of Italians that preferred white champagne

[tex]X_{A}=60[/tex] represent the number of Americans that preferred white champagne

[tex]n_{I}=300[/tex] sample of Italians selected

[tex]n_{A}=400[/tex] sample of Americans selected

[tex]p_{I}=\frac{74}{300}=0.247[/tex] represent the proportion of Italians that preferred white champagne

[tex]p_{A}=\frac{60}{400}=0.15[/tex] represent the proportion of Americans that preferred white champagne

z would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the value for the test (variable of interest)

[tex]\alpha=0.05[/tex] significance level given

2) Concepts and formulas to use

We need to conduct a hypothesis in order to check if a higher proportion of Italians than Americans prefer white champagne at weddings, the system of hypothesis would be:

Null hypothesis:[tex]p_{I} - p_{A} \leq 0[/tex]

Alternative hypothesis:[tex]p_{I} - p_{A} > 0[/tex]

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

[tex]z=\frac{p_{I}-p_{A}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{I}}+\frac{1}{n_{A}})}}[/tex] (1)

Where [tex]\hat p=\frac{X_{I}+X_{A}}{n_{I}+n_{A}}=\frac{74+60}{300+400}=0.191[/tex]

3) Calculate the statistic

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

[tex]z=\frac{0.247-0.15}{\sqrt{0.191(1-0.191)(\frac{1}{300}+\frac{1}{400})}}=3.23[/tex]

4) Statistical decision

Since is a right tailed test the p value would be:

[tex]p_v =P(Z>3.23)=0.000619[/tex]

Comparing the p value with the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that the proportion of Italians who prefer white champagne at weddings it's significantly higher than the proportion of Americans.