Question

please answer all parts in detail! :)

4. Suppose the weights of parts in one lot has a normal distribution. Six parts were randomly drawn from this lot and weighed. Their weights were 21.6 oz. 22.4 oz. 21.4 oz. 20.2 oz. 22.0 oz 20.8 oz The sample mean x= 21.4 oz and sample standard deviation s = 0.8 oz. (a) (1 point] Calculate the test statistic for testing Ho: u = 22 oz, where u = 22 oz is the weight specification of the part. (b) (1 point] Calculate the test statistic for test Ho: o = 0.5 oz versus H1:0 > 0.5 oz, where o is the standard deviation of the part weight. (c) (3 points) Calculate the p-value for the test Ho:a=0.5 oz versus H1:0 > 0.5 oz. Report your decision when a = 0.01.

Answer 1

a) The test statistic here is computed as:

$t^* = \frac{\bar X - \mu_0}{\frac{s}{\sqrt{n}}} = \frac{21.4 - 22}{\frac{0.8}{\sqrt{6}}} = -1.8371$

Therefore -1.8371 is the test statistic value here.

b) The test statistic for testing for standard deviation here is computed as:

$\chi^2 = \frac{(n - 1)s^2}{\sigma_0^2} = \frac{(6 - 1)0.8^2}{0.5^2} = 12.8$

Therefore 12.8 is the chi square test statistic value here.

c) As this is a one tailed test for n - 1 = 5 degrees of freedom, the p-value here is obtained from chi square distribution tables here as:

$p = P(\chi^2_5 > 12.8) = 0.0253$

As the p-value here is 0.0253 > 0.01 which is the level of significance, therefore the test is not significant here and we cannot reject the null hypothesis here. Therefore we dont have sufficient evidence here that the population standard deviation is more than 0.5 here.