Installation of a certain hardware takes random time with a standard deviation o
ID: 3133213 • Letter: I
Question
Installation of a certain hardware takes random time with a standard deviation of 5 minutes.
(a) A computer technician installs this hardware on 64 different computers with an average installation
time of 42 minutes. Compute a 95% confidence interval for the mean installation time.
(b) Suppose that the population mean installation time is 40 minutes. A technician
installs the hardware on your PC. What is the probability that the installation time will be within the
interval computed in (a)?
(c) A manager questions the assumption of the above. Her pilot sample of 40 installation times has a sample standard deviation of s = 6.2 min, and she says that it is significantly different from the assumed value of standard deviation = 5 min. Do you agree with the manager? Conduct a suitable test of a standard deviation.
Explanation / Answer
a)
Note that
Margin of Error E = z(alpha/2) * s / sqrt(n)
Lower Bound = X - z(alpha/2) * s / sqrt(n)
Upper Bound = X + z(alpha/2) * s / sqrt(n)
where
alpha/2 = (1 - confidence level)/2 = 0.025
X = sample mean = 42
z(alpha/2) = critical z for the confidence interval = 1.959963985
s = sample standard deviation = 5
n = sample size = 64
Thus,
Margin of Error E = 1.22497749
Lower bound = 40.77502251
Upper bound = 43.22497749
Thus, the confidence interval is
( 40.77502251 , 43.22497749 ) [ANSWER]
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b)
We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as
x1 = lower bound = 40.77502251
x2 = upper bound = 43.22497749
u = mean = 40
n = sample size = 64
s = standard deviation = 5
Thus, the two z scores are
z1 = lower z score = (x1 - u) * sqrt(n) / s = 1.240036016
z2 = upper z score = (x2 - u) * sqrt(n) / s = 5.159963984
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.892518963
P(z < z2) = 0.999999877
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.107480913 [ANSWER]
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c)
Let us test at 0.05 level.
Formulating the null and alternative hypotheses,
Ho: sigma = 5
Ha: sigma =/ 5
As we can see, this is a two tailed test.
Thus, getting the critical chi^2, as alpha = 0.05 ,
alpha/2 = 0.025
df = N - 1 = 39
chi^2 (crit) = 23.65432456 and 58.12005973
Getting the test statistic, as
s = sample standard deviation = 6.2
sigmao = hypothesized standard deviation = 5
n = sample size = 40
Thus, chi^2 = (N - 1)(s/sigmao)^2 = 59.9664
As chi^2 is not between the two critical values, we REJECT THE NULL HYPOTHESIS.
Hence, there is significant evidence that the true population standard deviation is not 5 min. [CONCLUSION]