Question

A researcher is interested in religious differences for immigration checkpoint experiences at airports. Polling a sample of 340 Muslims and 190 Christians, he finds that 52% of Muslims and 46% of Christians report having a negative experience. Is there a statistically significant difference between Muslims and Christians who report having a negative experience at airports (α=0.01)?

Answer 1

Answer:

$z=\frac{0.52-0.46}{\sqrt{0.49(1-0.49)(\frac{1}{340}+\frac{1}{190})}}=1.325$    

$p_v =2*P(Z>1.325)= 0.185$  

Comparing the p value with the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can't conclude that we have a significant difference between the two proportions of negative experiences at 1% of significance.

Step-by-step explanation:

Data given and notation  

$n_{M}=340$ sample of Muslims

$n_{C}=190$ sample of Christians

$p_{M}=0.52$ represent the proportion of Muslims with negative experience

$p_{C}=0.46$ represent the proportion of christians with negative experience

z would represent the statistic (variable of interest)  

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

$\alpha=0.01$ significance level given

Concepts and formulas to use  

We need to conduct a hypothesis in order to check if is there is a difference in the two proportions, the system of hypothesis would be:  

Null hypothesis:$p_{M} - p_{C}=0$  

Alternative hypothesis:$p_{M} - p_{C} \neq 0$  

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

$z=\frac{p_{M}-p_{C}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{M}}+\frac{1}{n_{C}})}}$   (1)  

Where $\hat p=\frac{p_{M}+p_{C}}{2}=\frac{0.52+0.46}{2}=0.49$  

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:  

$z=\frac{0.52-0.46}{\sqrt{0.49(1-0.49)(\frac{1}{340}+\frac{1}{190})}}=1.325$    

Statistical decision

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

$p_v =2*P(Z>1.325)= 0.185$  

Comparing the p value with the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can't conclude that we have a significant difference between the two proportions of negative experiences at 1% of significance.