Mathematics, 19.12.2019 06:31 Rogeartest4

Abank 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.

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Answer from: mismhan01

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.

Answer from: Quest

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation:

Answer from: Quest

i dont know

step-by-step explanation:

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