Question

Preliminary data analyses indicate that you can reasonably consider the assumptions for using pooled procedures satisfied. Independent random samples of released prisoners in the fraud and firearms offense categories yielded the following information on time served in months. Obtain a 98% confidence interval for the difference between the mean times served by prisoners in the fraud and firearms offense categories Fraud Firearms 8 9 16 2 21.1 122 33 99 20.1 222 127 41 179 16.5 5.7 54 155 157 11.9 133 12.6 212 (Note: x-9.14, s, = 4.39, X, - 17.61, and s, = 3.54.) The 98% confidence interval is from to (Round to three decimal places as needed.)

Answer 1

Here we have given that

X1i: Time served by prisoners in the fraud

X2i: Time served by prisoners in the firearms

n1= number of prisoners in the fraud group = 10

$\bar x1$ =sample mean fraud group =9.14                         

S1=sample standard deviation of fraud group =4.39

n2= number of prisoners in the firearms group = 10                    

$\bar x2$=sample mean firearms group =17.61

S2=sample standard deviation of firearms group =4.39

Here, The below-mentioned assumptions are satisfied to calculate the confidence interval :

• The sample of released prisoners in fraud and firearms groups are independent random sample.
• we consider that the assumption of polled t-procedures satisfied. i.e. the two population variances are equal.
• Population standard deviation $\sigma 1$ and $\sigma 2$ are not known i.e. here we use the t-test with pooled standard deviation approach.

Now we want to find the 98% confidence interval for the difference between the mean times served by prisoners in the fraud and firearms offence categories ($\mu 1- \mu 2$):

The formula is as follows,

$(\bar x1 - \bar x2 ) - t_{critical} *Sp \sqrt{ \frac{1}{n1}+\frac{1}{n2}} \leq \mu 1 -\mu 2 \leq (\bar x1 - \bar x2 ) + t_{critical} *Sp \sqrt{\frac{1}{n1}+\frac{1}{n2}}$

Where,

Sp=Pooled Sample standard deviation =$\sqrt \frac{(n1-1)S1^2+(n2-1)S2^2}{n1+n2-2}$

Now we find the S-pooled

$Sp =\sqrt \frac{(n1-1)S1^2+(n2-1)S2^2}{n1+n2-2}$

     =$\sqrt \frac{(10-1)4.39^2+(10-1)3.54^2}{10+10-2}$

= 3.9877

Now, we can find the critical value

c=confidence level = 0.98

$\alpha$= level of significance = 1- c= 1=0.98 = 0.02

Degrees of freedom = n1+n2-2 = 10+10-2 =18

t-critical = 2.552 Using t-table see two tailed value corresponding to D.F =18.

We get the 95% confidence interval is

$(\bar x1 - \bar x2 ) - t_{critical} *Sp \sqrt{ \frac{1}{n1}+\frac{1}{n2}} \leq \mu 1 -\mu 2 \leq (\bar x1 - \bar x2 ) + t_{critical} *Sp \sqrt{\frac{1}{n1}+\frac{1}{n2}}$

$(9.14 - 17.61 ) - 2.552*3.9877* \sqrt{ \frac{1}{10}+\frac{1}{10}} \leq \mu 1 -\mu 2 \leq (9.14 - 17.61 ) + 2.552*3.9877* \sqrt{ \frac{1}{10}+\frac{1}{10}}$

$-13.021 \leq \mu 1 -\mu 2 \leq -3.919$

The 98% confidence interval is from -13.021 to -3.919

Interpretation:

This confidence interval shows we are 98% confident that the difference between the mean times served by prisoners in the fraud and firearms offence categories ($\mu 1- \mu 2$) will falls within that interval.