Question

21. In a certain chemical process, it is very important that a particular solution that is to be used as a reactant have a pH of exactly 8.2. A method for determining pH that is available for solutions of this type is known to that are normally distributed with a mean equal to the actual pH. Suppose 10 ylelded the following pH values: 8.18,8.21, 8.17,8.22, 8.16, 8.16, 8.17, 8.19, give 8.15, 8.18 Fill in the blanks. Use ? = 0.05. Indicate the parameter of interest: State the null Hypothesis: State the alternative hypothesis: u a. Define the appropriate statistic: Define the rejection criteria (using critical values) Calculate the numerical value of the statistic (and its components) defined previously Conclusion: (reject or not the Null hypothesis) Conclusion in the context of the problem: b. Find the P-value (the na possible range of yalues glven your statistical tables). c. Calculate a two-tails CI for the true pH. Conclude. to.os, 174-.00323, 9.17 LO dronlets or aerosols) is generated when metal-removing fluids are used in machining operations

Answer 1

a) Parameter of Interest : Population mean of pH value of solutions

Null Hypothesis

Alternative Hypothesis $H1: \mu \neq 8.2$

The test statsitic is

$t = \frac{\overline{x} -\mu }{s/\sqrt{n}} \sim t_{n-1}$

Degrees of freedom = n-1 = 10-1 = 9

Significance level $\alpha = 0.05$

Rejection Criteria

Reject H0. if t > 2.262 ot t < -2.262

Sample mean

81.79 10 8.1798 8.18 x= 7n

Sample Standard deviation

$s= \sqrt{\frac{1}{n-1}\sum \left (x -\overline{x} \right )^2}=\sqrt{ \frac{0.0045}{9}}=0.0224$

The value of test statistic is

$t = \frac{8.18-8.2}{0.0224/\sqrt{10}}= -2.828$

Conclusion : Since |t| calculated is greater than critical value . Reject H0.

At 5% significance level, the pH value of particular solution is significantly differ from 8.2

b) The P-Value is 0.019787

c) The 95% CI

$\overline{x} \pm t\times \frac{s}{\sqrt{n}} = 8.18\pm 2.262\times \frac{0.0224}{\sqrt{10}}$

$= 8.18\pm 0.0160$

Answer: The 95% CI is = (8.164, 8.196)