Chapter 25, Problem 25E

(a)

To determine

To create a $90\%$ confidence interval for the mean difference in cost.

Answer to Problem 25E

We are $90\%$ confident that the mean difference in cost is between $-4409.0722\text{ and }-2759.3489$ .

Explanation of Solution

The table of the universities and colleges fees are given that they charged from the resident and the non-resident students. It is also given that,

  $n=19$

And we will find the difference between the two samples as:

Resident	Non-resident	Difference
4200	8800	-4600
1900	3600	-1700
3400	8600	-5200
3200	7000	-3800
3400	12700	-9300
2600	5700	-3100
3300	5900	-2600
2900	3400	-500
2200	4600	-2400
3400	7300	-3900
3200	6000	-2800
1600	8300	-6700
3300	7700	-4400
6100	10700	-4600
1600	5400	-3800
1700	4400	-2700
2000	4800	-2800
800	1000	-200
2800	5800	-3000

Now, the sample mean and the standard deviation of the difference are as follows:

  $\begin{array}{l}\overline{d}=\frac{-4600-1700-\mathrm{....}-200-3000}{19}\\ =-3584.211\\ s_d=\sqrt{\frac{{(-4600-(-3584.211))}^2+\mathrm{....}+{(-3000-(-3584.211))}^2}{19-1}}\\ =2073.4467\end{array}$

Now the degree of freedom is then as:

  $\begin{array}{l}df=n-1\\ =19-1\\ =18\end{array}$

Now let us find out the t -value by looking in the row starting with degree of freedom and in table T of appendix F is as:

  $t=1.734$

Now the confidence interval will be as:

  $\begin{array}{l}\overline{d}-t_{\alpha /2}\times \frac{s_d}{\sqrt{n}}\\ =-3584.211-1.734\times \frac{2073.4467}{\sqrt{19}}\\ =-4409.0722\\ \overline{d}+t_{\alpha /2}\times \frac{s_d}{\sqrt{n}}\\ =-3584.211+1.734\times \frac{2073.4467}{\sqrt{19}}\\ =-2759.3489\end{array}$

Thus, we conclude that we are $90\%$ confident that the mean difference in cost is between $-4409.0722\text{ and }-2759.3489$ .

(b)

To determine

To interpret the interval in your context.

Explanation of Solution

The table of the universities and colleges fees are given that they charged from the resident and the non-resident students. It is also given that,

  $n=19$

And we will find the difference between the two samples as:

Resident	Non-resident	Difference
4200	8800	-4600
1900	3600	-1700
3400	8600	-5200
3200	7000	-3800
3400	12700	-9300
2600	5700	-3100
3300	5900	-2600
2900	3400	-500
2200	4600	-2400
3400	7300	-3900
3200	6000	-2800
1600	8300	-6700
3300	7700	-4400
6100	10700	-4600
1600	5400	-3800
1700	4400	-2700
2000	4800	-2800
800	1000	-200
2800	5800	-3000

Then the t -value by looking in the row starting with degree of freedom and in table T of appendix F is as:

  $t=1.734$

Now the confidence interval will be as:

  $\begin{array}{l}\overline{d}-t_{\alpha /2}\times \frac{s_d}{\sqrt{n}}\\ =-3584.211-1.734\times \frac{2073.4467}{\sqrt{19}}\\ =-4409.0722\\ \overline{d}+t_{\alpha /2}\times \frac{s_d}{\sqrt{n}}\\ =-3584.211+1.734\times \frac{2073.4467}{\sqrt{19}}\\ =-2759.3489\end{array}$

Thus, we conclude that we are $90\%$ confident that the mean difference in cost is between $-4409.0722\text{ and }-2759.3489$ .

(c)

To determine

To explain what does your confidence interval indicates about this assertion.

Explanation of Solution

The table of the universities and colleges fees are given that they charged from the resident and the non-resident students. It is also given that,

  $n=19$

And we will find the difference between the two samples as:

Resident	Non-resident	Difference
4200	8800	-4600
1900	3600	-1700
3400	8600	-5200
3200	7000	-3800
3400	12700	-9300
2600	5700	-3100
3300	5900	-2600
2900	3400	-500
2200	4600	-2400
3400	7300	-3900
3200	6000	-2800
1600	8300	-6700
3300	7700	-4400
6100	10700	-4600
1600	5400	-3800
1700	4400	-2700
2000	4800	-2800
800	1000	-200
2800	5800	-3000

Then the t -value by looking in the row starting with degree of freedom and in table T of appendix F is as:

  $t=1.734$

Now the confidence interval will be as:

  $\begin{array}{l}\overline{d}-t_{\alpha /2}\times \frac{s_d}{\sqrt{n}}\\ =-3584.211-1.734\times \frac{2073.4467}{\sqrt{19}}\\ =-4409.0722\\ \overline{d}+t_{\alpha /2}\times \frac{s_d}{\sqrt{n}}\\ =-3584.211+1.734\times \frac{2073.4467}{\sqrt{19}}\\ =-2759.3489\end{array}$

Thus, we conclude from the interval that we are $90\%$ confident that the mean difference in cost is between $-4409.0722\text{ and }-2759.3489$ and therefore the institution claiming that that they charge $\$3500$ less than out-of state residents is true as this cost lies between the confidence interval.