Question

Researchers at the Mayo Clinic have studied the effect of sound levels on patient healing and have found a significant association (louder hospital ambient sound level is associated with slower postsurgical healing). Based on the Mayo Clinic's experience, Ardmore Hospital installed a new vinyl flooring that is supposed to reduce the mean sound level (decibels) in the hospital corridors. The sound level is measured at five randomly selected times in the main corridor.

New Flooring	Old Flooring
42	48
41	51
40	44
37	48
44	52
1

(a-1)

Choose the correct null and alternative hypotheses. Assume μ1 refers to the average sound level of the new flooring and μ2 refers to the average sound level of the old flooring.

H₀: μ₁ – μ₂ = 0 vs. H₁: μ₁ – μ₂ ≠ 0

H₀: μ₁ – μ₂ ≤ 0 vs. H₁: μ₁ – μ₂ > 0

H₀: μ₁ – μ₂ ≥ 0 vs. H₁: μ₁ – μ₂ < 0

References

2.

(a-2)

At α = .05, what is the decision rule under the assumption that the variances are equal?

Reject the null hypothesis if t_calc > –1.86 (8 df.)

Reject the null hypothesis if t_calc < –1.86 (8 df.)

3.

(a-3)

Calculate the test statistic. (Round your answer to 2 decimal places. A negative amount should be indicated with a minus sign.)

4.

(a-4)

At α = .05, has the mean been reduced?

(Click to select)RejectDo not reject H0, the mean (Click to select)has not beenhas been reduced.

5.

(b-1)

Now, we are going to test whether or not the two variances are equal, rather than merely assume that they are equal. So, we now need to specify a new set of hypotheses based on the variances themselves. Select the appropriate hypotheses.

H₀: σ₁² / σ₂² = 1vs. H₁: σ₁² / σ₂² ≠ 1.

H₀: σ₁² / σ₂² ≠ 1vs. H₁: σ₁² / σ₂² = 1.

6.

(b-2)

At α = .05, what is the decision rule?

Reject H₀ if F_calc < 9.60 or F_calc > .1042. (d.f.₁ = 4, d.f.₂ = 4.)

Reject H₀ if F_calc > 9.60 or F_calc < .1042. (d.f.₁ = 4, d.f.₂ = 4.)

7.

(b-3)

What is the test statistic? Put the smaller of the two variances in the numerator when you do the calculation. The reason for doing this -- rather than putting the larger of the two variances in the numerator -- may become apparent when you read the explanation that accompanies the last question. (Round the test statistic value to 2 decimal places.)

8.

(b-4)

At α = .05, has the variance changed?

(Click to select)RejectDo not reject H0, the variance (Click to select)has has not  changed.

Answer 1

Solution:

Ardmore Hospital installed a new vinyl flooring that is supposed to reduce the mean sound level (decibels) in the hospital corridors. The sound level is measured at five randomly selected times in the main corridor.

Part 1)

Choose the correct null and alternative hypotheses. Assume μ₁ refers to the average sound level of the new flooring and μ₂ refers to the average sound level of the old flooring.

Since Ardmore Hospital installed a new vinyl flooring that is supposed to reduce the mean sound level (decibels) in the hospital corridors and μ₁ refers to the average sound level of the new flooring and μ₂ refers to the average sound level of the old flooring, thus this is a left tailed test.

Thus correct hypothesis are:

Option c) H₀: μ₁ – μ₂ ≥ 0 vs. H₁: μ₁ – μ₂ < 0

Part 2)

(a-2) At α = .05, what is the decision rule under the assumption that the variances are equal?

df = n1 + n2 - 2 = 5 + 5 - 2 = 8

t = 1.860

Since this is left tailed test, t critical value is negative = -1.86

Thus correct option is:

Reject the null hypothesis if t_calc < –1.86 (8 df.)

Part 3)

(a-3)

Calculate the test statistic.

$t = \frac{\bar{x_{1}}-\bar{x_{2}}}{\sqrt{S_{p}^{2}\times \left ( \frac{1}{n_{1}}+\frac{1}{n_{2}} \right )}}$

where

$S_{p}^{2}=\frac{(n_{1}-1)\times s_{1}^{2}+(n_{2}-1)\times s_{2}^{2}}{n_{1}+n_{2}-2}$

$s_{1}^{2}=\frac{\sum x_{1}^{2}-(\sum x_{1})^{2}/n_{1}}{n_{1}-1}$

$s_{2}^{2}=\frac{\sum x_{2}^{2}-(\sum x_{2})^{2}/n_{2}}{n_{2}-1}$

Thus we need to make following table:

New Flooring x1	Old Flooring x2	x1^2	x2^2
42	48	1764	2304
41	51	1681	2601
40	44	1600	1936
37	48	1369	2304
44	52	1936	2704
$\small \sum x_{1}=204$	$\small \sum x_{2}=243$	$\small \sum x_{1}^{2}=8350$	$\small \sum x_{2}^{2}=11849$

Thus

$s_{1}^{2}=\frac{\sum x_{1}^{2}-(\sum x_{1})^{2}/n_{1}}{n_{1}-1}$

$s_{1}^{2}=\frac{8350 -(204 )^{2}/5}{5-1}=\frac{8350-8323.2 }{4}=\frac{26.8 }{4}=6.7$

$s_{2}^{2}=\frac{\sum x_{2}^{2}-(\sum x_{2})^{2}/n_{2}}{n_{2}-1}$

$s_{2}^{2}=\frac{11849 -(243 )^{2}/5}{5-1}=\frac{11849-11809.8 }{4}=\frac{39.2 }{4}=9.8$

Thus

$S_{p}^{2}=\frac{(n_{1}-1)\times s_{1}^{2}+(n_{2}-1)\times s_{2}^{2}}{n_{1}+n_{2}-2}$

$S_{p}^{2}=\frac{(5-1)\times6.7+(5-1)\times 9.8}{5+5-2}$

$S_{p}^{2}=\frac{4 \times6.7+4 \times 9.8}{8}$

$S_{p}^{2}=\frac{26.8 +39.2 }{8}$

$S_{p}^{2}=\frac{66 }{8}$

$S_{p}^{2}=8.25$

and

$\bar{x}_{1}=\frac{\sum x_{1}}{n_{1}}=\frac{204}{5}=40.8$

$\bar{x}_{2}=\frac{\sum x_{2}}{n_{2}}=\frac{243}{5}=48.6$

Thus t test statistic is:

$t = \frac{\bar{x_{1}}-\bar{x_{2}}}{\sqrt{S_{p}^{2}\times \left ( \frac{1}{n_{1}}+\frac{1}{n_{2}} \right )}}$

$t = \frac{40.8 -48.6 }{\sqrt{8.25 \times \left ( \frac{1}{5}+\frac{1}{5} \right )}}$

$t = \frac{-7.8 }{\sqrt{8.25 \times \left ( \frac{2}{5}\right )}}$

$t = \frac{-7.8 }{\sqrt{3.3 }}$

$t = \frac{-7.8 }{1.8166 }$

Part 4) (a-4) At α = .05, has the mean been reduced?

Since tcal = -4.29 < t critical value = -1.86,

we reject H0. the mean has been reduced.