Question

A guidance counselor at a university is investigating demand for study abroad. The question before her is how engineering and humanities majors compare regarding interest in study abroad during summer. Random samples of 20 engineering and humanities majors each were interviewed. Eight engineering majors and 12 humanities majors expressed interest in study abroad during the summer. The estimated difference in the proportion of engineering and humanities majors, p E − p H , where pE is the proportion of engineering majors interested in study abroad and pH is the proportion for humanities, has sampling distribution:a. with mean μ=0.2 and standard deviation σ=0.1549 shape not known.b. Normal with mean μ= -0.2 and standard deviation σ=0.1549.c. Normal with mean μ= 0.2 and standard deviation σ=0.1549. d. with mean μ=-0.2 and standard deviation σ=0.1549 shape not known

Answer 1

Answer:

$\mu_{p_E -p_H} = 0.4-0.6 =-0.2$

$SE_{p_E -p_H}=\sqrt{\frac{0.4(1-0.4)}{20} +\frac{0.6 (1-0.6)}{20}}=0.1549$

b. Normal with mean μ= -0.2 and standard deviation σ=0.1549

Step-by-step explanation:

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

$p_E$ represent the real population proportion for Engineering

$\hat p_E =\frac{8}{20}=0.4$ represent the estimated proportion for Engineering

$n_E=20$ is the sample size required for Engineering

$p_H$ represent the real population proportion for Humanities  

$\hat p_H =\frac{12}{20}=0.6$ represent the estimated proportion for Humanities

$n_H=20$ is the sample size required for Humanities

We assume that the population proportions follows a normal distribution since we satisfy these conditions:

$np\geq 5$ ,$n(1-p)\geq 5$

$20*0.4= 8\geq 5 , 20(1-0.4)=12\geq 5$

The population proportion have the following distribution  

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$  

And we are interested on the distribution for $p_E-p_H$

For this case we know that the distribution for the differences of proportions is also normal and given by:

$p_E -p_H \sim N(\hat p_E -\hat p_H, \sqrt{\frac{\hat p_E(1-\hat p_E)}{n_E} +\frac{\hat p_H (1-\hat p_H)}{n_H}}$

So then we can find the mean and the deviation like this:

$\mu_{p_E -p_H} = 0.4-0.6 =-0.2$

$SE_{p_E -p_H}=\sqrt{\frac{0.4(1-0.4)}{20} +\frac{0.6 (1-0.6)}{20}}=0.1549$

So then the best option is :

b. Normal with mean μ= -0.2 and standard deviation σ=0.1549