Question

A bank is trying to determine which model of safe to install. The bank manager believes that each model is equally resistant to safe crackers, but sets up a test to be sure. He hires nine safe experts to break into each of the models, timing each endeavor. The results (in seconds) are given next, paired by expert. Let D be the difference: Time to break Safe 1 minus Time to break Safe 2.
Safe 1: 103, 90, 64, 120, 104, 92, 145, 106, 76
Safe 2: 101, 94, 58, 112, 103, 90, 140, 110, 74

Refer to Exhibit 10.10. Should we conclude that the average time it takes experts to crack the safes does not differ by model at the 5% significance level?

A) Yes, the 95% confidence interval for ?D contains 0.
B) No, the 95% confidence interval for ?D does not contain 0.
C) No, the 95% confidence interval for ?D contains 0, but the absolute value of the upper bound is larger than the absolute value of the lower bound.
D) Not enough information to determine.

Answer 1

Answer:

So on this case the 95% confidence interval would be given by (-1.152;5.152).

$-1.152 < \mu_{safe1}-\mu_{safe2}$  

Should we conclude that the average time it takes experts to crack the safes does not differ by model at the 5% significance level?

A) Yes, the 95% confidence interval for ?D contains 0.

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".  

Solution

Let put some notation  

x=value for Safe 1 , y = value for Safe 2

x: 103, 90, 64, 120, 104, 92, 145, 106, 76

y: 101, 94, 58, 112, 103, 90, 140, 110, 74

The first step is calculate the difference $d_i=x_i-y_i$ and we obtain this:

d: 2, -4, 6, 8, 1, 2, 5, -4, 2

The second step is calculate the mean difference  

$\bar d= \frac{\sum_{i=1}^n d_i}{n}= \frac{18}{9}=2$

The third step would be calculate the standard deviation for the differences, and we got:

$s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =4.093$

The next step is calculate the degrees of freedom given by:

$df=n-1=9-1=8$

Now we need to calculate the critical value on the t distribution with 8 degrees of freedom. The value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$, so we need a quantile that accumulates on each tail of the t distribution 0.025 of the area.

We can use the following excel code to find it:"=T.INV(0.025;8)" or "=T.INV(1-0.025;8)". And we got $t_{\alpha/2}=\pm 2.31$

The confidence interval for the mean is given by the following formula:  

$\bar d \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)  

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

$2-2.31\frac{4.093}{\sqrt{9}}=-1.152$  

$2+2.31\frac{4.093}{\sqrt{9}}=5.152$  

So on this case the 95% confidence interval would be given by (-1.152;5.152).

$-1.152 < \mu_{safe1}-\mu_{safe2}$  

Should we conclude that the average time it takes experts to crack the safes does not differ by model at the 5% significance level?

A) Yes, the 95% confidence interval for D contains 0.