Question

Two machines are used for filling plastic bottles with a net volume of 16.0 ounces. The filling processes can be assumed to be normal, with standard deviations of ????1 = 0.015 and ????2 = 0.018. The quality engineering department suspects that both machines fill to the same net volume, whether or not this volume is 16.0 ounces. An experiment is performed by taking a random sample 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) State the hypotheses that should be tested in this experiment. (b) Test these hypotheses using ???? = 0.05. What are your conclusions? (c) Find the P-value for this test. (d) Find a 95 percent confidence interval on the difference in mean fill volume for the two machines.

Answer 1

Answer:

a) Null hypothesis:$\mu_{1}-\mu_{2}=0$  

Alternative hypothesis:$\mu_{1} - \mu_{2}\neq 0$  

b) $z=\frac{(16.015-16.005)-0}{\sqrt{\frac{0.015^2}{10}+\frac{0.018^2}{10}}}=1.350$

c) Since is a bilateral test the p value would be:  

$p_v =2*P(z>1.350)=0.177$  

Comparing the p value with the significance level $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and the difference between the two groups is NOT significantly different.  

d) $0.01-1.96\sqrt{\frac{0.015^2}{10}+\frac{0.018^2}{10}}=-0.0045$  

$0.01+1.96\sqrt{\frac{0.015^2}{10}+\frac{0.018^2}{10}}=0.025$  

So on this case the 95% confidence interval would be given by $-0.0045 \leq \mu_1 -\mu_2 \leq 0.025[/tex] Step-by-step explanation:For this case we have the following info:Machine 1: 16.03 16.01 16.04 15.96 16.05 15.98 16.05 16.02 16.02 15.99Machine 2: 16.02 16.03 15.97 16.04 15.96 16.02 16.01 16.01 15.99 16.00We can calculate the sample mean with the following formula:[tex] \bar X = \frac{\sum_{i=1}^n X_i}{n}$

$\bar X_{1}=16.015$ represent the mean for sample 1  

$\bar X_{2}=16.005$ represent the mean for sample 2  

$\sigma_{1}=0.015$ represent the population standard deviation for 1

$\sigma_{2}=0.018$ represent the sample standard deviation for 2  

$n_{1}=10$ sample size for the group 2  

$n_{2}=10$ sample size for the group 2  

Significance level provided

z would represent the statistic (variable of interest)  

Concepts and formulas to use

Part a

We need to conduct a hypothesis in order to check if the difference in the population means, the system of hypothesis would be:  

Null hypothesis:$\mu_{1}-\mu_{2}=0$  

Alternative hypothesis:$\mu_{1} - \mu_{2}\neq 0$  

We have the population standard deviation's, so for this case is better apply a z test to compare means, and the statistic is given by:  

$z=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}$ (1)

Part b  

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.  

With the info given we can replace in formula (1) like this:  

$z=\frac{(16.015-16.005)-0}{\sqrt{\frac{0.015^2}{10}+\frac{0.018^2}{10}}}=1.350$

Part c  

P value

Since is a bilateral test the p value would be:  

$p_v =2*P(z>1.350)=0.177$  

Comparing the p value with the significance level $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and the difference between the two groups is NOT significantly different.  

Part d

The confidence interval for the difference of means is given by the following formula:  

$(\bar X_1 -\bar X_2) \pm z_{\alpha/2}\sqrt{\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2}}$ (1)  

The point of estimate for $\mu_1 -\mu_2$ is just given by:  

$\bar X_1 -\bar X_2 =16.015-16.005=0.01$  

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $z_{\alpha/2}=1.96$  

Confidence interval  

Now we have everything in order to replace into formula (1):  

$0.01-1.96\sqrt{\frac{0.015^2}{10}+\frac{0.018^2}{10}}=-0.0045$  

$0.01+1.96\sqrt{\frac{0.015^2}{10}+\frac{0.018^2}{10}}=0.025$  

So on this case the 95% confidence interval would be given by $-0.0045 \leq \mu_1 -\mu_2 \leq 0.025$