# Answer in Statistics and Probability for blossomqt #248886

Two machines are used for filling plastic bottles with a net volume of 16.0 ounces. The fill volume can be assumed normal, with standard deviation ! = 0.020 and ” = 0.025 ounces. A member of the quality engineering staff suspects that both machines fill to the same mean net volume, whether or not this volume is 16.0 ounces. A random sample of 10 bottles is taken from the output of each machine. Use = 0.10

Machine 1 16.67 16.29 16.87 16.51 15.92 16.54

16.29 15.88

15.43 16.99

16.19 15.49

16.49 16.09

Machine 2

15.46 16.04 15.56 16.78 16.39 16.91

QUESTION

Two machines are used for filling plastic bottles with a net volume of 16.0 ounces. The fill volume can be assumed normal, with standard deviation “sigma” 1 = 0.020 and “sigma” 2 = 0.025 ounces. A member of the quality engineering staff suspects that both machine fill to the same mean net volume, whether or not this volume is 16.0 ounces. A random sample of 10 bottles is taken from the output of each machine.

Machine 1 Machine 2

16.03 16.01 16.02 16.03

16.04 15.96 15.97 16.04

16.05 15.98 15.96 16.02

16.05 16.02 16.01 16.01

16.02 15.99 15.99 16.00

(a) Do you think the engineer is correct? Use “alpha” = 0.05.

(b) What is the P-value for this test?

(c) What is the power of the test in part (a) for a true difference in mean of 0.04?

SOLUTION

(a) Do you think the engineer is correct? Use “alpha” = 0.05

“bar{x_1}={16.03+16.01+16.04+15.96+16.05over 10}+”“+{15.98+16.05+16.02+16.02+15.99over 10}=16.015”“bar{x_2}={16.02+16.03+15.97+16.04+15.96over 10}+”“+{16.02+16.01+16.01+15.99+16.00over 10}=16.005”

Given that “sigma_1=0.020, sigma_2=0.025.”

a.The null and alternative hypothesis

“H_0:mu_1-mu_2=0”

“H_1:mu_1-mu_2not=0”

This corresponds to a two-tailed test, for which a z-test for two population means, with known population standard deviations will be used.

The test statistic is

“z={bar{x_1}-bar{x_2} over sqrt{{sigma_1^2 over n_1}+{sigma^2 over n_2}}}={16.015-16.005 over sqrt{{(0.020)^2 over10}+{(0.025)^2 over 10}}}approx0.9877”

b. Calculate p-value

“p=P(Z<-0.9877 or Z>0.9877)=”“=2P(Z<-0.9877)approx2(0.16165)=0.3233”

Since “p=0.3233>0.05,” it is concluded that the null hypothesis is not rejected.

There is not sufficient evidence to support the claim that the population means are not equal.

Therefore, the engineer is correct that both machine fill to the same mean net volume, whether or not this volume is 16.0 ounces.

c. The power is the probability of rejecting the null hypothesis when the alternative hypothesis is true.

“z_{alpha/2}=pm1.96”“bar{x_1}-bar{x_2}=(mu_1-mu_2)-z_{alpha/2}cdotsqrt{{sigma_1^2 over n_1}+{sigma^2 over n_2}}=”“=0-1.96cdotsqrt{{(0.020)^2 over10}+{(0.025)^2 over 10}}=-0.0198”“bar{x_1}-bar{x_2}=(mu_1-mu_2)+z_{alpha/2}cdotsqrt{{sigma_1^2 over n_1}+{sigma^2 over n_2}}=”“=0+1.96cdotsqrt{{(0.020)^2 over10}+{(0.025)^2 over 10}}=0.0198”“z={(bar{x_1}-bar{x_2})-(mu_1-mu_2) over sqrt{{sigma_1^2 over n_1}+{sigma^2 over n_2}}}={-0.0198-0.04over sqrt{{(0.020)^2 over10}+{(0.025)^2 over 10}}}approx-5.9066”“z={(bar{x_1}-bar{x_2})-(mu_1-mu_2) over sqrt{{sigma_1^2 over n_1}+{sigma^2 over n_2}}}={0.0198-0.04over sqrt{{(0.020)^2 over10}+{(0.025)^2 over 10}}}approx-1.9952”

Determine the probability of rejecting of the null hypothesis

“P(Z<-5.9066 or Z>-1.9952)=”“=P(Z<-5.9066)+P(Z>-1.9952)=”“=P(Z<-5.9066)+P(Z<1.9952)approx”“approx0.0000000017462+0.976989approx0.9770”

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