Mathematics, 25.01.2021 20:30 awkwardness92

Installation of a certain hardware takes a random amount of time with a standard deviation of 5 minutes. A computer technician installs this hardware on 64 different computers, with the average installation time of 42 minutes. Compute a 95% confidence interval for the mean installation time. Explain your interval in context.

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The 95% confidence interval for the mean installation time is (40.775, 43.225) and this can be determined by using the formula of margin of error.

Given :

Standard deviation is 5 minutes. Sample size is 64.Mean is 42 minutes.95% confidence interval.

The following steps can be used in order to determine the 95% confidence interval for the mean installation time:

Step 1 - The formula of margin of error can be used in order to determine the 95% confidence interval.

$M = z \times \dfrac{\sigma}{\sqrt{n} }$

where z is the z-score, $\sigma$ is the standard deviation, and the sample size is n.

Step 2 - Now, substitute the values of z, $\sigma$ , and n in the above formula.

$M = 1.96 \times \dfrac{5}{\sqrt{64} }$

M = 1.225

Step 3 - So, the 95% confidence interval is given by (M - $\mu$ , M + $\mu$ ) that is (40.775, 43.225).

The 95% confidence interval for the mean installation time is (40.775, 43.225).

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Answer from: kids2006

The 95% confidence interval for the mean installation time is between 40.775 minutes and 43.225 minutes. This means that for all instalations, in different computers, we are 95% sure that the mean time for installation will be in this interval.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of 1-0.025 = 0.975 , so z = 1.96

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.96*\frac{5}{\sqrt{64}} = 1.225$

The lower end of the interval is the sample mean subtracted by M. So it is 42 - 1.225 = 40.775 minutes

The upper end of the interval is the sample mean added to M. So it is 42 + 1.225 = 43.225 minutes

Answer from: Quest

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