Question

Assume that you plan to use a significance level of alpha α equals =0.05 to test the claim that p1 equals = p2. Use the given sample sizes and numbers of successes to find the pooled estimate p. Round your answer to the nearest thousandth. n1 equals =​100, n2 equals =100 x1 equals =​42, x2 equals =45.

Answer 1

Answer:

$z=\frac{0.42-0.45}{\sqrt{\frac{0.42(1-0.42)}{100}+\frac{0.45(1-0.45)}{100}}}=-0.428$    

$p_v =2*P(Z$  

And we can use the following R code to find it: "2*pnorm(-0.428)"

The p value is a very high value and using any significance given $\alpha=0.05$ always $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the two proportions are not significantly different.  

Step-by-step explanation:

1) Data given and notation  

$X_1 =42$ people with some characteristic desired on sample 1

$X_2 =45$ people with some characteristic desired on sample 2

$n_{1}=100$ sample selected 1

$n_{2}=100$ sample selected 2

$p_{1}=\frac{42}{100}=0.42$ represent the proportion estimated for 1

$p_{2}=\frac{45}{100}=0.45$ represent the proportion estimated for 2

$\alpha=0.05$ represent the significance level

z would represent the statistic (variable of interest)  

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

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to check if the proportion for the two samples are different , the system of hypothesis would be:  

Null hypothesis:$p_{1} = p_{2}$  

Alternative hypothesis:$p_{1} \neq p_{2}$  

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

$z=\frac{p_{1}-p_{2}}{\sqrt{\frac{p_1 (1-p_1)}{n_{1}}+\frac{p_2 (1-p_2)}{n_{2}}}}$   (1)  

3) Calculate the statistic  

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

$z=\frac{0.42-0.45}{\sqrt{\frac{0.42(1-0.42)}{100}+\frac{0.45(1-0.45)}{100}}}=-0.428$    

4) Statistical decision

We can calculate the p value for this test.    

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

$p_v =2*P(Z$  

And we can use the following R code to find it: "2*pnorm(-0.428)"

The p value is a very high value and using any significance given $\alpha=0.05$ always $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the two proportions are not significantly different.