Mathematics, 06.05.2020 05:36 kaffolter25

5. The numbers of blocked intrusion attempts on each day during the first two weeks of the month were 56, 47, 49, 37, 38, 60, 50, 43, 43, 59, 50, 56, 54, 58 After the change of firewall settings, the numbers of blocked intrusions during the next 20 days were 53, 21, 32, 49, 45, 38, 44, 33, 32, 43, 53, 46, 36, 48, 39, 35, 37, 36, 39, 45. (a) Construct a 95% confidence interval for the difference between the average number of intrusion attempts per day before and after the change of firewall settings. (b) Can we claim a significant reduction in the rate of intrusion attempts

Answers: 3

Show answers

Answers

Answer from: rscott2649

a) The 95% confidence interval for the true mean reduction of intrusion attemps per day is (-2.898, 11.698).

b) As negative values are also included in the confidence interval, we can not claim that there is a significant reduction in the rate of intrusion attempts.

Step-by-step explanation:

The mean and standard deviation for the blocked intrusion attempts on each day during the first two weeks of the month are:

$M_1=\dfrac{1}{14}\sum_{i=1}^{14}(56+47+49+37+38+60+50+43+43+59+50+56+54+58)\\\\\\ M_1=\dfrac{700}{14}=50$

$s_1=\sqrt{\dfrac{1}{(n-1)}\sum_{i=1}^{14}(x_i-M)^2}\\\\\\s_1=\sqrt{\dfrac{1}{13}\cdot [(56-(50))^2+(47-(50))^2+(49-(50))^2+...+(54-(50))^2+(58-(50))^2]}\\\\\\$

$s_1=\sqrt{\dfrac{1}{13}\cdot [(36)+(9)+...+(64)]}\\\\\\s_1=\sqrt{\dfrac{754}{13}}=\sqrt{58}\\\\\\s_1=7.616$

The mean and standard deviation for the blocked intrusion attempts on each day during the next 15 days of the month are:

$M_2=\dfrac{1}{15}\sum_{i=1}^{15}(53+21+32+49+45+38+44+33+32+43+53+46+36+48+39+35+37+36+39+45)\\\\\\ M_2=\dfrac{684}{15}=45.6$

$s_2=\sqrt{\dfrac{1}{(n-1)}\sum_{i=1}^{15}(x_i-M)^2}\\\\\\s_2=\sqrt{\dfrac{1}{14}\cdot [(53-(45.6))^2+(21-(45.6))^2+...+(45-(45.6))^2]}\\\\\\$

$s_2=\sqrt{\dfrac{1}{14}\cdot [(54.76)+(605.16)+(184.96)+...+(0.36)]}\\\\\\s_2=\sqrt{\dfrac{1786.4}{14}}=\sqrt{127.6}\\\\\\s_2=11.296$

a) We have to calculate a 95% confidence interval for the difference between means.

The sample 1 (first two weeks), of size n1=14 has a mean of 50 and a standard deviation of 7.616.

The sample 2 (last two weeks), of size n2=15 has a mean of 45.6 and a standard deviation of 11.296.

The difference between sample means is Md=4.4.

M_d=M_1-M_2=50-45.6=4.4

The estimated standard error of the difference between means is computed using the formula:

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{7.616^2}{14}+\dfrac{11.296^2}{15}}\\\\\\s_{M_d}=\sqrt{4.143+8.507}=\sqrt{12.65}=3.557$

The critical t-value for a 95% confidence interval is t=2.052.

The margin of error (MOE) can be calculated as:

$MOE=t\cdot s_{M_d}=2.052 \cdot 3.557=7.298$

Then, the lower and upper bounds of the confidence interval are:

$LL=M_d-t \cdot s_{M_d} = 4.4-7.298=-2.898\\\\UL=M_d+t \cdot s_{M_d} = 4.4+7.298=11.698$

The 95% confidence interval for the true mean reduction of intrusion attemps per day is (-2.898, 11.698).

b) We could claim a significant reduction in the rate of intrusion if the confidence interval bounds were both positive but, in this case, the confidence interval for the true mean reduction also includes negative values, so we can not be sure that there is significant improvement.

Answer from: Quest

there is no graph so i cant sorry

step-by-step explanation:

Answer from: Quest

since we know that the two functions are equal, f (x) = g

(x), therefore when we equate the two equations the result should be similar to

3x = x – 4:

a. f(x) = 3x, g(x) = x – 4

equating the two functions:

3x = x – 4 (correct)

so we now got our answer (a)