Question

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

(b) Two different instruments were used to make a number of replicate measurements. The results were:
Instrument A: 12.06 12.14 12.03 12.09 12.05
Instrument B: 14.62 14.97 14.60 14.51 14.01 14.11
Do these results indicate that either instrument is more precise?

Answer 1

Answer on Question #70464, Math / Statistics and Probability

(a) 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

(b) Two different instruments were used to make a number of replicate measurements. The results were:

Instrument A: 12.06 12.14 12.03 12.09 12.05

Instrument B: 14.62 14.97 14.60 14.51 14.01 14.11

Do these results indicate that either instrument is more precise?

Solution.

(a) We face the challenge of verifying the hypothesis $H_0: a = 20$ with an alternative hypothesis $H_1: a \neq 20$.

Hypothesis $H_0$ is accepted if $Z_{obs} \in \left(-z_{cr(\alpha, n-1)}, z_{cr(\alpha, n-1)}\right)$, where $Z_{obs} = \frac{\overline{x} - 20}{s} \sqrt{n - 1}$, $\overline{x}$ – sample mean, $s$ – the square root of the variance corrected, $n$ – sample size, $z_{cr(\alpha, n-1)}$ – from a table of critical values of Student's distribution.

$$\overline{x} = \frac{1}{n} \sum_{i=1}^{n} x_i = \frac{1}{6} (23 + 19 + 26 + 22 + 18 + 27) = 22,5.$$ $$s^2 = \frac{1}{n - 1} \sum_{i=1}^{n} (x_i - \overline{x})^2 = 13,1; \quad s = 3,62.$$

So,

$$Z_{obs} = \frac{22,5 - 20}{3,62} \sqrt{6 - 1} = 1,54.$$

From the table of critical values of Student distribution at the level of significance $\alpha = 0,05$ we find $z_{cr(0,05,5)} = 2,57$. Since $|Z_{obs}| < z_{cr(0,05,5)}$, the hypothesis $H_0$ is accepted, that is, with the probability of $95\%$ is confirmed the hypothesis that the average dissolution time is 20 seconds.

(b) The instruments are equally accurate if the dispersion of their impressions is the same. Construct hypotheses:

$$H_0: \sigma_A^2 = \sigma_B^2;$$ $$H_1: \sigma_A^2 \neq \sigma_B^2.$$

Find the variances corrected (see the formula in (a)): $s_A^2 = 0,002$, $s_B^2 = 0,126$.

According to Fisher's criterion $F_{obs} = \frac{s_B^2}{s_A^2} = \frac{0,126}{0,002} = 63$, $F_{cr} = F_{(0,05;5;4)} = 6,26$. Since $F_{obs} > F_{cr}$ then the hypothesis $H_0$ reject, which means that the instruments are not exactly precise.

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