Consider the number of blocked intrusions below:
Before firewall change: 56, 47, 49, 37, 38, 60, 50, 43, 43, 59, 50, 56, 54, 58
After firewall change: 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 setting (assume equal variance)
(b) Can we claim significant reduction in the rate of intrusion attempts? The number of intrusion each day has approximately normal distribution. Compute P-values and state your conclusions under the assumption of equal variances and without it. Does this assumption make a difference?

Question

contestada

Respuesta :

contexto1025 contexto1025 · Answer 1 · 2022-11-29T06:35:49+01:00

a)The 95% confidence interval are

0-44,44-88

b))Yes, we can claim significant reduction in the rate of intrusion attempts

We have firewall change of 14 people.

We need to compute total sum before firewall change which can be done by

(56+47+49+37+38+60+50+43+59+50+56+54+58)/14

=657/14

=46.92

Now, similarly, after firewall change total sum is

(53+21+32+49+45+38+44+33+32+43+53+46+36+48+39+35+37+36+39+45)/20

=756/20

=37.8

Now the difference between average number of intrusion attempts before and after firewall change is

(46.92-37.8)

=9.12

Now, we need to make confidence interval to only 95% of this difference=

(95×9.12)/100

=44.196=44(approx)

It means our interval range varies from 0-44,44-88 with a difference of 44.

b)Yes, we can easily claim that rate of intrusion attempts decreases.

Before firewall change, we can see average number of intrusion attempt is =46.92

After firewall change, we can see average number of intrusion attempt is

37.8

So, we can see rate of intrusion decreases on increasing the number of blocked intrusions.

For computing z-value, we just need to take square root of the difference of average number of intrusions attempst

=>Z-value= √9.12=3.01

Using the z-table, we get that for the above z-value, corresponding p-value is 0.067

Hence, the interval range is 0-44,44-88 and b)Yes we can claim, p-value is 0.067.No, this assumption does not make a difference.

To know more about confidence interval, visit here:

https://brainly.com/question/24131141

#SPJ4