Question

QUESTIONS 1. When operating usually, a manufacturing process produces tablets for which the mean weight of the active ingredient is 5 grams, and the standard deviation is 0.025 gram. For a random sample of 12 tablets the following weights of active ingredient (in grams) were found: 5.01 4.69 5.03 4.98 4.98 4.95 5.00 5.00 5.03 5.01 5.04 4.95 a.Without assuming that the population variance is known, test the null hypothesis that the population mean weight of active ingredient per tablet is 5 grams. Use a two-sided alternative and a 5% significance level. State any assumptions that you make. (20 Points) b.Stating any assumptions that you make, test the null hypothesis that the population standard deviation is 0.025 gram against the alternative hypothesis that the population standard deviation exceeds 0.025 gram. Use a 5% significance level. (15 Points) 2. A random samnle of o02 u

Answer 1

1) $\bar x$ = 4.9725

s = 0.0936

a) H₀: $\mu$ = 5

H₁: $\mu \neq$ 5

The test statistic t = ($\bar x - \mu$)/(s/sqrt(n))

= (4.9725 - 5)/(0.0936/sqrt(12))

= -1.02

At alpha = 0.05, the critical value is t_{0.025, 11} = +/- 2.201

Since the test statistic value is not less than the negative critical value, (-1.02 > -2.201), so we should not reject the null hypothesis.

B) H0: $\sigma$ = 0.025

H1: $\sigma$ > 0.025

The test statistic$\chi^2$ = (n - 1)s²/$\sigma^2$

= 11 * (0.0936)^2/(0.025)^2 = 154.19

P-value = P($\chi^2$ > 154.19)

= 0.0

Since the p-value is less than the significance level (0.0 < 0.05), so we should reject the null hypothesis.

There is sufficient evidence to conclude that the population standard deviation exceeds 0.025 gram.