Answer:
a)
H₀: ρ₁ ≤ ρ₂
H₁: ρ₁ > ρ₂
b)
Z= [tex]\frac{('p_{1} - 'p_{2}) - (p_{1} - p_{2})}{\sqrt{'p(1-'p)[\frac{1}{n_1} + \frac{1}{n_2} ]} }[/tex] ≈ N(0;1)
c)
p-value < .00001
d)
Reject the null hypothesis.
The population proportion vinyl gloves that leak viruses is greater than the population proportion of latex gloves that leak viruses.
Step-by-step explanation:
Hello!
The objective is to test if the population proportion of vinyl gloves that leak viruses (population 1) is greater then the population proportion of latex gloves that leak viruses (population 2).
Sample 1 (Vinyl)
n₁= 225
Sample proportion 'ρ₁= 0.60
Sample 2 (Latex)
n₂= 225
Sample proportion 'ρ₂= 0.09
Pooled proportion: 'ρ= (x₁+x₂)/(n₁+n₂)= ( 'ρ₁ + 'ρ₂)/2= (0.60 + 0.09)/2= 0.345 ≅0.35
The hypothesis is:
H₀: ρ₁ ≤ ρ₂
H₁: ρ₁ > ρ₂
α: 0.05
The statistic to use is a pooled Z
Z= [tex]\frac{('p_{1} - 'p_{2}) - (p_{1} - p_{2})}{\sqrt{'p(1-'p)[\frac{1}{n_1} + \frac{1}{n_2} ]} }[/tex] ≈ N(0;1)
The critical region is one-tailed to the right (positive), the critical value is:
[tex]Z_{1-\alpha } = Z_{0.95} = 1.64[/tex]
If Z ≥ 1.64, you will reject the null hypothesis.
If Z < 1.64, you will not reject the null hypothesis.
Z= [tex]\frac{(0.6 - 0.09) - (0}{\sqrt{0.35(1-0.35)[\frac{1}{225} + \frac{1}{225} ]} }[/tex] = 11.34
Since the calculated Z- value is greater than the critical value, the decision is to reject the null hypothesis. The population proportion vinyl gloves that leak viruses is greater than the population proportion of latex gloves that leak viruses.
P-value approach:
p-value < .00001
If you compare with the level of signification, you reach the same decision.
I hope it helps!
Z-value is greater than the critical value, therefore reject the null hypothesis. Theerefore, population proportion vinyl gloves that leak viruses is greater than the population proportion of latex gloves that leak viruses.
Step-by-step explanation:
Given :
Sample 1 (Vinyl) ---- [tex]n_1=225[/tex]
Sample proportion ---- [tex]'p_1 = 0.60[/tex]
Sample 2 (Latex) ---- [tex]n_1 =225[/tex]
Sample proportion ---- [tex]'p_2=0.09[/tex]
Calculation :
Pooled proportion -
[tex]'p = \dfrac{x_1+x_2}{n_1+n_2}=\dfrac{'p_1\;+\;'p_2}{2}[/tex]
[tex]'p = \dfrac{0.60+0.09}{2}=0.345\cong0.35[/tex]
The hypothesis is:
[tex]H_0: \;p_1\leq p_2[/tex]
[tex]H_1:\;p_1>p_2[/tex]
[tex]\alpha =0.05[/tex]
The statistic to use is a pooled Z
[tex]Z=\dfrac{('p_1\;-\;'p_2)-(p_1-p_2)}{\sqrt{'p(1\;-\;'p)[\dfrac{1}{n_1}+\dfrac{1}{n_2}]} }[/tex]
[tex]\rm Z\approx N(0,1)[/tex]
[tex]Z_1_-_\alpha =Z_0_._9_5=1.64[/tex]
If [tex]Z \geq 1.64[/tex], reject the null hypothesis.
If [tex]Z<1.64[/tex], you will not reject the null hypothesis.
[tex]Z=\dfrac{0.6-0.09}{\sqrt{0.35(1-0.35)[\dfrac{1}{225}+\dfrac{1}{225}]}} }=11.34[/tex]
Z-value is greater than the critical value, therefore reject the null hypothesis. Therefore, population proportion vinyl gloves that leak viruses is greater than the population proportion of latex gloves that leak viruses.
For more information, refer the link given below
https://brainly.com/question/14790912?referrer=searchResults
