Answer:
[tex]z=\frac{0.7-0.07}{\sqrt{0.385(1-0.385)(\frac{1}{215}+\frac{1}{215})}}=13.42[/tex]
[tex]p_v =P(Z>13.42)\approx 0[/tex]
So the p value is a very low value and using any significance level given [tex]\alpha=0.01[/tex] we se that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that vinyl gloves have a greater virus leak rate than latex gloves at 1% of significance.
Step-by-step explanation:
Data given and notation
[tex]n_{V}=215[/tex] sample of vinyl gloves selected
[tex]n_{L}=215[/tex] sample of latex gloves selected
[tex]p_{V}=0.7[/tex] represent the proportion of vynil gloves with leaked viruses
[tex]p_{L}=0.07[/tex] represent the proportion of latex gloves with leaked viruses
z would represent the statistic (variable of interest)
[tex]p_v[/tex] 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 vinyl gloves have a greater virus leak rate than latex gloves , the system of hypothesis would be:
Null hypothesis:[tex]p_{V} \leq p_{L}[/tex]
Alternative hypothesis:[tex]p_{V} > \mu_{L}[/tex]
We need to apply a z test to compare proportions, and the statistic is given by:
[tex]z=\frac{p_{V}-p_{L}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{V}}+\frac{1}{n_{L}})}}[/tex] (1)
Where [tex]\hat p=\frac{X_{V}+X_{L}}{n_{V}+n_{L}}=\frac{0.7+0.07}{2}=0.385[/tex]
Calculate the statistic
Replacing in formula (1) the values obtained we got this:
[tex]z=\frac{0.7-0.07}{\sqrt{0.385(1-0.385)(\frac{1}{215}+\frac{1}{215})}}=13.42[/tex]
Statistical decision
Since is a one right tailed test the p value would be:
[tex]p_v =P(Z>13.42)\approx 0[/tex]
So the p value is a very low value and using any significance level given [tex]\alpha=0.01[/tex] we se that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that vinyl gloves have a greater virus leak rate than latex gloves at 1% of significance.
