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

the 95% confidence interval would be given by
`-1.882 \leq \mu_(males) -\mu_(fmales) \leq 6.482`

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{((s^2_1)/(n_1)+(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{((s^2_1)/(n_1)+(s^2_2)/(n_2))}`

And replacing we have:

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

Confidence interval

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

`2.3-1.96\sqrt{((24^2)/(227)+(24.3^2)/(293))=-1.882`

`2.3+1.96\sqrt{((24^2)/(227)+(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.