Question

Oxygen bubble velocity in a purification process. Refer to the Chemical Engineering Research and Design (March 2013) study of a method of purifying nuclear fuel waste, Exercise 6.35 (p. 349). Recall that the process involves oxidation in molten salt and tends to produce oxygen bubbles with a rising velocity. To monitor the process, the researchers collected data on bubble velocity (measured in meters per second) for a random sample of 25 photographic bubble images. These data (simulated) are reproduced in the accompanying table. When oxygen is inserted into the molten salt at a rate (called the sparging rate) of \(3.33 \times {10^{ - 6}}\) , the researchers discovered that the true mean bubble rising velocity \(\mu = .338\)

a. Conduct a test of hypothesis to determine if the true mean bubble rising velocity for the population from which the sample is selected is\(\mu = .338\)Use\(\alpha = .10\).

0.275 0.261 0.209 0.266 0.265 0.312 0.285 0.317 0.229 0.251 0.256 0.339 0.213 0.178 0.217 0.307 0.264 0.319 0.298 0.169 0.342 0.270 0.262 0.228 0.22

Short answer

a) The test statistic is -9.60.

Step-by-step solution

Constructing the hypothesis

The null and alternative hypothesis are given by

\(\begin{aligned}{H_0}:\mu = .338\\{H_a}:\mu \ne .338\end{aligned}\)

Computing the mean and standard deviation

a) The mean and standard deviation is calculated as

\(\begin{aligned}\bar x &= \frac{\begin{aligned}{l}0.275 + 0.261 + 0.209 + 0.266 + 0.265 + 0.312 + 0.285 + 0.317 + 0.229\\ + 0.251 + 0.256 + 0.399 + 0.213 + 0.178 + 0.217 + 0.307 + 0.264 + 0.319\\ + 0.298 + 0.169 + 0.342 + 0.270 + 0.262 + 0.228 + 0.22\end{aligned}}{{25}}\\ &= \frac{{6.016}}{{25}}\\ &= 0.24\end{aligned}\)

\(\begin{aligned}sd &= \sqrt {\frac{\begin{aligned}{l}{\left( {0.035} \right)^2} + {\left( {0.021} \right)^2} + {\left( { - 0.031} \right)^2} + {\left( {0.026} \right)^2} + {\left( {0.025} \right)^2} + {\left( {0.072} \right)^2} + \\{\left( {0.045} \right)^2} + {\left( {0.077} \right)^2} + {\left( { - 0.011} \right)^2} + {\left( {0.011} \right)^2} + {\left( {0.016} \right)^2} + {\left( {0.099} \right)^2} + \\{\left( { - 0.027} \right)^2} + {\left( { - 0.062} \right)^2} + {\left( { - 0.023} \right)^2} + {\left( {0.067} \right)^2} + {\left( {0.024} \right)^2} + {\left( {0.079} \right)^2} + \\{\left( {0.058} \right)^2} + {\left( { - 0.071} \right)^2} + {\left( {0.102} \right)^2} + {\left( {0.03} \right)^2} + {\left( {0.022} \right)^2} + {\left( { - 0.012} \right)^2} + {\left( {0.02} \right)^2}\end{aligned}}{{24}}} \\ &= \sqrt {\frac{\begin{aligned}{l}0.001225 + 0.000441 + 0.000961 + 0.000676 + 0.000625 + 0.005184 + \\0.002025 + 0.005929 + 0.000121 + 0.000121 + 0.000256 + 0.009801 + \\0.000729 + 0.003844 + 0.000529 + 0.004489 + 0.000576 + 0.006241 + \\0.003364 + 0.005041 + 0.010404 + 0.0009 + 0.000484 + 0.000144 + 0.0004\end{aligned}}{{24}}} \\ &= \sqrt {\frac{{0.06451}}{{24}}} \\ &= 0.051\end{aligned}\)

Therefore, the mean and standard deviation are 0.24 and 0.051.

Test statistic

The test statistic is calculated as

\(\begin{aligned}t &= \frac{{\bar x - \mu }}{{\frac{s}{{\sqrt n }}}}\\ &= \frac{{0.24 - .338}}{{\frac{{0.051}}{{\sqrt {25} }}}}\\ &= \frac{{ - 0.098}}{{0.0102}}\\ &= - 9.60\end{aligned}\)

Therefore, the test statistic is -9.60.

Degrees of freedom are

\(\begin{aligned}df &= n - 1\\ &= 25 - 1\\ &= 24\end{aligned}\)

For \(\alpha = .10\,and\,df = 5\)

The tabulated value is -1.47.

The calculated value is less than the tabulated value.

Therefore, we reject the null hypothesis.