Answer :
Answer:
[tex]z=\frac{0.094-0.04}{\sqrt{0.0642(1-0.0642)(\frac{1}{180}+\frac{1}{225})}}=2.203[/tex]
[tex]p_v =P(Z>2.203)= 0.0138[/tex]
Comparing the p value with the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so then the claim from the retailer makes sense at 1% of significance.
Step-by-step explanation:
Use a significance level of α=0.01 for the test.
Data given and notation
[tex]X_{1}=17[/tex] represent the number of defectives from the retailer
[tex]X_{2}=9[/tex] represent the number of defectives from the competitor
[tex]n_{1}=180[/tex] sample 1 selected
[tex]n_{2}=225[/tex] sample 2 selected
[tex]p_{1}=\frac{17}{180}=0.094[/tex] represent the proportion estimated for defectives from the retailer
[tex]p_{2}=\frac{9}{225}=0.04[/tex] represent the proportion estimated for defectives from the competitor
[tex]\hat p[/tex] represent the pooled estimate of p
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the value for the test (variable of interest)
[tex]\alpha=0.01[/tex] significance level given
Concepts and formulas to use
We need to conduct a hypothesis in order to check if the percentage of defective cellular phones found among his products, ( p1), will be no higher than the percentage of defectives found in a competitor's line, ( p2), the system of hypothesis would be:
Null hypothesis:[tex]p_{1} \leq p_{2}[/tex]
Alternative hypothesis:[tex]p_{1} > p_{2}[/tex]
We need to apply a z test to compare proportions, and the statistic is given by:
[tex]z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}[/tex] (1)
Where [tex]\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{17+9}{180+225}=0.0642[/tex]
z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.
Calculate the statistic
Replacing in formula (1) the values obtained we got this:
[tex]z=\frac{0.094-0.04}{\sqrt{0.0642(1-0.0642)(\frac{1}{180}+\frac{1}{225})}}=2.203[/tex]
Statistical decision
Since is a right sided test the p value would be:
[tex]p_v =P(Z>2.203)= 0.0138[/tex]
Comparing the p value with the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so then the claim from the retailer makes sense at 1% of significance.