The following data refers to airborne bacteria count (number of colonies/ft^3) b
ID: 3135081 • Letter: T
Question
The following data refers to airborne bacteria count (number of colonies/ft^3) both for m = 8 carpeted hospital rooms and for n = 8 uncarpeted rooms. Make any necessary assumptions and answer the questions below. Is the evidence sufficient to support the variance for carpeted rooms being greater than 7 colonies/ft^3 at level 0.05? Give a 90% confidence interval for the standard deviation of carpeted rooms Is the evidence sufficient to support the variance for uncarpeted rooms being greater than 9 colonies/ft^3 at level 0.05? What is the P-value for the test in part "c."? Is the evidence sufficient to support the variance for uncarpeted rooms is greater than the variance for carpeted rooms at level 0.05? What is the P-value for the test in part "a."? Give a 90% confidence interval for the variance of bacteria count in carpeted rooms over the variance of bacteria count in ucarpeted rooms.Explanation / Answer
a)
The sample standard deviation is
s = 2.6774188
Formulating the null and alternative hypotheses,
Ho: sigma <= 2.645751311
Ha: sigma > 2.645751311
As we can see, this is a right tailed test.
Thus, getting the critical chi^2, as alpha = 0.05 ,
alpha = 0.05
df = N - 1 = 7
chi^2 (crit) = 14.06714045
Getting the test statistic, as
s = sample standard deviation = 2.6774188
sigmao = hypothesized standard deviation = 2.645751311
n = sample size = 8
Thus, chi^2 = (N - 1)(s/sigmao)^2 = 7.168571431
Hence, as chi^2 < 14.067, we fail to reject Ho.
There is no significant evidence that the variance for carpeted rooms is greater than 7 colonies/ft^3.
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b)
As
df = n - 1 = 7
alpha = (1 - confidence level)/2 = 0.05
Then the critical values for chi^2 are
chi^2(alpha/2) = 14.06714045
chi^2(alpha/2) = 2.167349909
Thus, as
lower bound = (n - 1) s^2 / chi^2(alpha/2) = 3.567178432
upper bound = (n - 1) s^2 / chi^2(1 - alpha/2) = 23.15269897
Thus, the confidence interval for the variance is
( 3.567178432 , 23.15269897 )
Also, for the standard deviation, getting the square root of the bounds,
( 1.888697549 , 4.811725155 ) [ANSWER]
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c)
The sample standard deviation is
s = 3.210001113
Formulating the null and alternative hypotheses,
Ho: sigma <= 3
Ha: sigma > 3
As we can see, this is a right tailed test.
Thus, getting the critical chi^2, as alpha = 0.05 ,
alpha = 0.05
df = N - 1 = 7
chi^2 (crit) = 14.06714045
Getting the test statistic, as
s = sample standard deviation = 3.210001113
sigmao = hypothesized standard deviation = 3
n = sample size = 8
Thus, chi^2 = (N - 1)(s/sigmao)^2 = 8.014305558
Hence, as chi^2 < 14.067, we fail to reject Ho.
There is no significant evidence that the variance for carpeted rooms is greater than 7 colonies/ft^3.
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d)
Using technology for the right tailed area of chi^2 = 8.014305558 at df = 7,
P = 0.331334143 [ANSWER]
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