Question

Q(3)A sample of dissolving times (in seconds) of a drug in water generated the following data:

42.3, 43.4, 44.4, 45.1, 46.6, 45.2, 46.8, 47.6

Assume that the dissolving time is normally distributed and the population standard deviation is 1.5 seconds.

(a) Determine a 95% confidence interval for the mean dissolving time.

(b) At the 0.1 level of significance, test the hypothesis that the mean dissolving time under the conditions experienced is equal to 46 seconds using the probability-

value approach.

(c) At the 0.01 level of significance, test the hypothesis that the mean dissolving time

under the conditions experienced is greater than 45 seconds using the classical approach.

Answer 1

= 45.175

s = 1.7957

a) At 95% confidence interval the critical value is z_0.025 = 1.96

The 95% confidence interval forpopulation mean is

+/- z_0.025 *

= 45.175 +/- 1.96 * 1.5/

= 45.175 +/- 1.039

= 44.136, 46.214

b) H₀: = 46

   H₁: 46

The test statistic z = ()/()

                              = (45.175 - 46)/(1.5/)

                              = -1.56

P-value = 2 * P(Z < -1.56)

             = 2 * 0.0594

             = 0.1188

Since the P-value is greater than the significance level (0.1188 > 0.1), we should not reject H₀.

So there is sufficient evidence to conclude that the mean dissolving time under the conditions experienced is equal to 46 seconds .

c) H₀: = 46

   H₁: 45

The test statistic z = ()/()

                              = (45.175 - 45)/(1.5/)

                              = 0.33

At 0.01 significance level, the critical value is z_0.99 = 2.33

Since the test statistic value is not   greater than the critical value (0.33 < 2.33), we should not reject the null hypothesis.

So there is not sufficient evidence to conclude that the mean dissolving time under the conditions experienced is greater than 45 seconds