Question

Using Minitab Test the hypothesis that the average dissolution time is 20 seconds, using the random sample of dissolution times below.
Dissolution time (secs): 23 19 26 22 18 27

Answer 1

Answer on Question #41393– Math – Statistics and Probability

The sample mean is $\bar{x} = \frac{23 + 19 + 26 + 22 + 18 + 27}{6} = \frac{135}{8} = 22.5$ (or via an Excel function $= \text{AVERAGE}(23; 19; 26; 22; 18; 27)$ )

The sample standard deviation is

$$\begin{array}{l} s = \sqrt {\frac {\sum_ {i = 1} ^ {n} (x _ {i} - \bar {x}) ^ {2}}{n - 1}} = \\ = \sqrt {\frac {(2 3 - 2 2 . 5) ^ {2} + (1 9 - 2 2 . 5) ^ {2} + (2 6 - 2 2 . 5) ^ {2} + (2 2 - 2 2 . 5) ^ {2} + (1 8 - 2 2 . 5) ^ {2} + (2 7 - 2 2 . 5) ^ {2}}{5}} = 3. 6 2 \text {(o r v i a a n E x c e l f u n c t i o n} \\ = \text {S T D E V} (2 3; 1 9; 2 6; 2 2; 1 8; 2 7)). \\ \end{array}$$

The formulation of the null and alternative hypotheses should be

$$H _ {0} \colon \mu = 2 0 \text { versus } H _ {1} \colon \mu \neq 2 0.$$

The $t$ test statistic is

$$T = \frac {\bar {x} - \mu_ {0}}{\frac {x}{\sqrt {n}}} = \frac {2 2 . 5 - 2 0}{\frac {3 . 6 2}{\sqrt {8}}} = 1. 9 5, \text { degrees of freedom d . f .} = n - 1 = 6 - 1 = 5.$$

We test at the level of significance $\alpha = 0.05$ . Since $H_{1}$ is two-tailed, we set the rejection region

$$R \colon | T | \geq t _ {0. 0 2 5}.$$

From the $t$ table we find that $t_{0.025}$ with d.f. = 5 is 2.571. Because the observed value $t = 1.95$ is smaller than 2.571, the null hypothesis is not rejected at $\alpha = 0.05$ .

Conclusion: there is strong evidence that the average dissolution time is 20 seconds (with $\alpha = 0.05$ ).

Minitab

Data: C1

Stat>Basic Statistics>1-Sample t.

Type C1 in Samples in.

Following Test mean, type 20, the value of the mean under the null hypothesis. Click Options and type 95 in Confidence level.

In the Alternative cell select greater than, the direction of the alternative hypothesis. Click OK. Click OK.

If the sample size, mean, and standard deviation are available, instead of the second step, type these values in the corresponding cells.

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