Question

It is believed that using a solid state drive (SSD) in a computer results in faster boot times when compared to a computer with a traditional hard disk (HDD). You sample 10 computers with an HDD and note a sample average of 34.434 seconds with a standard deviation of 5.3833 seconds. A sample of 7 computers with an SSD show an average of 8.946 seconds with a standard deviation of 1.0224 seconds. Construct a 95% confidence interval for the difference between the true average boot times of the two types of hard drives. Assume the difference will represent (HDD-SSD) and that the population standard deviations are statistically the same

1) (21.412, 29.564)
	2) (23.408, 27.568)
	3) (6.785, 44.191)
	4) We only have the sample means, we need to know the population means in order to calculate a confidence interval.
	5) (21.056, 29.92)

Answer 1

Solution:

Given

n1= 10 sample size of HDD computer

n2= 7 sample size of SSD computer

sample mean of HDD computer.

sample mean of SSD computer.

S1= 5.3833 sample standard deviations of HDD computer.

s2 = 1.0224 sample standard deviations of SSD computer.

level of significance

The 95% confidence interval for difference between the true average boot times of the two types of hard drives is

Where

At

from t table

​​​​​​

(25.488 - 4.4314203,. 25.488+4.4314203)

(21.056579, 29.919420)

(21.056, 29.92)

The 95% confidence interval for difference between the true average boot times of the two types of drives is

(21.056, 29.92)

Option 5 is correct