Question

The viscosity of a liquid detergent is supposed to average 800 centistokes at 25° C. A random sample of 16 batches of detergent is collected, and the average viscosity is 812. Suppose we know that the standard deviation of viscosity is σ = 25 centistokes: (a) State the hypotheses that should be tested, (b) Test these hypotheses using a = 0.05. What are your conclusions?, (c) What is the P-value for the test?, (d) Find a 95 percent confidence interval on the mean.   

Answer 1

To, test whether the average viscocity is 800 centistokes at the 5% level of significance or not using the z-test at the level of significance is a = 0.05. We need to perform two-tail hypothesis test. Step1: The null and alternative hypotheses are: #9:41= 800 (The population mean is equal to 800) versus H: 4+800(The population mean is not equal to 800) Step 2: The level of significance is a =0.05. Step3: The decision rule: The significance level is 0.05. The symbol + sign in the alternative hypothesis indicates that the test is two-tailed with two rejection regions, one in each tail of the normal distribution curve of because the total area of both rejection regions is 0.05 (the significance level), the area of the rejection region in left tail is 0.05; that is, To find the z values for these critical point, we look for 0.05 area in the normal distribution table. From Normal table, the z values of the right-tail critical point, is approximately 1.9599. The table value of the z- test statistic is zonas = +1.9599. From Normal table, the z values of the left-tail critical point, is approximately - 1.9599. The table value of the z-test statistic is 20.025 = -1.9599. Step4: We reject the null hypothesis H, ifz > 1.9599. Otherwise, we do not reject the null hypothesis HA Sample statistics: n=16, X=812, and o= 25 centistokes. --------------- ------------------------ (b) Step 5: The formula for calculating the z-test statistic is 2-X-4 812-800 25 lal - 6.25 12 = 1.92 zs (1.92 -----------

(c) The calculation of the P-value: Step6: The P-value approach: 2P(2>2 ) = 2(1-P(z<zo)) =2(1-P(Z <1.92)) =2(1-0.972571) =2(0.027429) => 0.054858 P-value = 0.055) > a = 0.05). We do not reject the null hypothesis H. - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Step7: Conclusion: Since, the p-value = 0.055> a= 0.05 We do not reject the null hypothesis Ho. Since, the test statistic is z = 1.92 <20.025 = 11.9599 We do not reject the null hypothesis H, There is insufficient statistical evidence to conclude that the population mean is not equal to 800 at the 5% level of significance.

The sample size is n=16. The sample mean is X = 812. The sample standard deviation is o = 25. The formula for calculating the standard deviation is SE PO = 6.25 SE, - 16.25 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - The 95% confidence interval for the population mean of the viscocity: The level of significance is a =0.05. The critical values of the normal distribution is Zax2 = 200512 = 2002 20025 = 1.9599 ----- --- The formula for calculating the 95% confidence interval for the population mean is X-Zanz Toys usX+Zane =812–1.9599(SE,) s u5812+1.9599 (SE,) =812-1.9599(6.25) <45812+1.9599(6.25) =812– 12.24938 Su5812+12.24938 = 799.7506 SU 824.2434 X-Zw2 mai sus&+2v2 = 1799.7506 5 4 5824.2494 --------------- -- Interpretation: We can be 95% confident that the population mean of the average viscosity is between 799.7506 and 824.2494