Question

Each person in random samples of 227 male and 293 female working adults living in a certain town in Canada was asked how long, in minutes, his or her typical daily commute was. (Use a statistical computer package to calculate the P-value. Use μmales − μfemales. Round your test statistic to two decimal places, your df down to the nearest whole number, and your P-value to three decimal places.)

Answer 1

Answer:

the 95% confidence interval would be given by $-1.882 \leq \mu_{males} -\mu_{fmales} \leq 6.482$  

Step-by-step explanation:

Previous concepts  

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

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".  

$\bar X_1 =30.2$ represent the sample mean 1  (males)

$\bar X_2 =27.9$ represent the sample mean 2  (females)

n1=227 represent the sample 1 size (males)  

n2=293 represent the sample 2 size (females)

$s_1 =24$ sample standard deviation for sample 1  (males)

$s_2 =24.3$ sample standard deviation for sample 2 (females)

$\sigma =3$ represent the population standard deviation

$\mu_1 -\mu_2$ parameter of interest.

Confidence interval

The confidence interval for the difference of means is given by the following formula:  

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{(\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2})}$ (1)  

The point of estimate for $\mu_1 -\mu_2$ is just given by:

$\bar X_1 -\bar X_2 =30.2-27.9=2.3$

Since the sample size is large enough we can assume that th t distirbution is approximately equal to the normal distribution in order to find the quantile.

Let's assume a Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$, and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $z_{\alpha/2}=1.96$  

The standard error is given by the following formula:

$SE=\sqrt{(\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2})}$

And replacing we have:

$SE=\sqrt{(\frac{24^2}{227}+\frac{24.3^2}{293})}=2.134$

Confidence interval

Now we have everything in order to replace into formula (1):  

$2.3-1.96\sqrt{(\frac{24^2}{227}+\frac{24.3^2}{293})=-1.882$  

$2.3+1.96\sqrt{(\frac{24^2}{227}+\frac{24.3^2}{293})=6.482$  

So on this case the 95% confidence interval would be given by $-1.882 \leq \mu_{males} -\mu_{females} \leq 6.482$  

R code

> barmale=30.2

> barfemale=27.9

> diff=barmale-barfemale

> smale=24

> sfemale=24.3

> nmale=227

> nfemale=293

> SE=sqrt((smale^2)/nmale +(sfemale^2)/nfemale)

> ME=qnorm(1-0.025)*SE

> lower=diff-ME;lower

[1] -1.882018

> upper=diff+ME;upper

[1] 6.482018

And we got the same results.