Question

The concentration of active ingredient in a liquid laundry detergent is thought to be affected by the type of catalyst used in the process. The standard deviation of active concentration is known to be 3 grams per liter regardless of the catalyst type. Ten observations on concentration are taken with each catalyst, and the data follow:
Catalyst 1: 57.9, 66.2, 65.4, 65.4, 65.2, 62.6, 67.6, 63.7, 67.2, 71.0
Catalyst 2: 66.4, 71.7, 70.3, 69.3, 64.8, 69.6, 68.6, 69.4, 65.3, 68.8.

a. Find a 95% confidence interval on the difference in mean active concentrations for the two catalysts. Find the P-value.
b. Is there any evidence to indicate that the mean active concentrations depend on the choice of catalyst? Base your answer on the results of part (a).
c. Suppose that the true mean difference in active concentration is 5 grams per liter. What is the power of the test to detect this difference if α = 0.05?
d. If this difference of 5 grams per liter is really important, do you consider the sample sizes used by the experimenter to be adequate? Does the assumption of normality seem reasonable for both samples?

Answer 1

Answer:

a. The 95% C.I. is approximately -5.83 < μ₁ - μ₂ < -0.57

b. Yes

c. The power of the test is 0.03836

d. The sample size can be considered adequate

Step-by-step explanation:

a. The given data in the study

Catalyst 1: 57.9, 66.2, 65.4, 65.4, 65.2, 62.6, 67.6, 63.7, 67.2, 71.0

Catalyst 2: 66.4 , 71.7, 70.3, 69.3, 64.8, 69.6, 68.6, 69.4, 65.3, 68.8

Using the Average, Standard Deviation function from Microsoft Excel, we have;

The mean for Catalyst 1, $\overline x_1$ = 65.22

The standard deviation for catalyst 1, σ₁ = 3.444416

The mean for Catalyst 2, $\overline x_2$ = 68.42

The standard deviation for catalyst 2, σ₂ = 2.22401

The 95% confidence interval on the difference in mean is given as follows;

$\left (\bar{x}_1-\bar{x}_{2} \right ) \pm z_{c}\sqrt{\dfrac{\sigma _{1}^{2}}{n_{1}} + \dfrac{\sigma_{2}^{2}}{n_{2}}}$

The critical-z for a 95% confidence interval = 1.96

Therefore, we have;

$\left (65.22-68.42 \right ) \pm 1.96 \times \sqrt{\dfrac{3^{2}}{10} + \dfrac{3^{2}}{10}}$

Therefore, we have;

The 95% C.I. is approximately -5.83 < μ₁ - μ₂ < -0.57

The test statistics is given as follows;

$z=\dfrac{(\bar{x}_{1}-\bar{x}_{2})}{\sqrt{\dfrac{\sigma_{1}^{2} }{n_{1}}-\dfrac{\sigma _{2}^{2}}{n_{2}}}}$

Therefore we have;

$z=\dfrac{(65.22-68.42)}{\sqrt{\dfrac{3^{2} }{10}+\dfrac{3^{2}}{10}}} = -2.385$

The p-value = P(Z<-2.39 or Z > 2.39) = 2P(Z<-2.39) = 2 × 0.00842 = 0.01684

b. From the confidence interval which range from approximately -5.83 to -0.57 and does not include 0, therefore, there is a difference in mean active concentration which depends on the choice of catalyst

c. The power of the test

 The sample mean difference is given as follows;

$\left (\bar{x}_1-\bar{x}_{2} \right ) = \pm z_{c}\sqrt{\dfrac{\sigma _{1}^{2}}{n_{1}} + \dfrac{\sigma_{2}^{2}}{n_{2}}}$

Therefore, we have;

$\left (\bar{x}_1-\bar{x}_{2} \right ) = \pm 1.96 \times \sqrt{\dfrac{3^{2}}{10} + \dfrac{3^{2}}{10}} = \pm 2.6296$

The z-value is given as follows;

$z=\dfrac{(\bar{x}_{1}-\bar{x}_{2})-(\mu_{1}-\mu _{2} )}{\sqrt{\dfrac{\sigma_{1}^{2} }{n_{1}}-\dfrac{\sigma _{2}^{2}}{n_{2}}}}$

μ₁ - μ₂ = 5

When $\bar{x}_1-\bar{x}_{2} = 2.6296$

$z=\dfrac{2.6296-5}{\sqrt{\dfrac{3^{2} }{10}-\dfrac{3^{2}}{10}}} \approx -1.7667918$

When $\bar{x}_1-\bar{x}_{2} = -2.6296$

$z=\dfrac{-2.6296-5}{\sqrt{\dfrac{3^{2} }{10}-\dfrac{3^{2}}{10}}} \approx -5.686768$

The power of the test is given by P = P(Z<-5.69) + P(Z>-1.77) = 0 + 0.03836 = 0.03836

The power of the test = 0.03836

d. The sample size is statistically adequate because the confidence interval of -5.83 < μ₁ - μ₂ < -0.57 has a value of -5 as a possible population difference in mean

The assumption of normality seems adequate because the confidence interval obtained by using the sample standard deviation is given as follows;

(-5.74, -0.66) which also contains -5 which is the difference in the population mean