Answer to Question #199158 in Statistics and Probability for mandona

Question

1. Nine determinations of the specific heat of iron had a standard deviation of 0.0086. Assuming that these determinations constitute a random sample from a normal population, test the null hypothesis σ = 0.0100 against the alternative hypothesis σ < 0.0100 at the α = .05 significance level. (Hint: What is the relation between standard deviation and variance?) 2. In a random sample, the weights of 24 Black Angus steers of a certain age have a standard deviation of 238 pounds. Assuming that the weights constitute a random sample from a normal population, test the null hypothesis σ = 250 pounds against the two-tailed alternative 휎6= 250 pounds at the α = .01 significance level.

Answer 1

1.

$H_0: \sigma=0.01 \\ H_1: \sigma<0.01 \\ s=0.0086 \\ n=9$

The test statistic for Chi-square test is:

$χ^2 = \frac{(n-1)s^2}{\sigma^2} \\ = \frac{(9-1)(0.0086)^2}{(0.01)^2} \\ = 5.917$

The P-value for the chi-square test at 8 degrees of freedom can be obtained using the excel formula, “=CHI.DIST.RT(5.917,8)”.

The P-value is 0.6565.

Decision rule:

If p-value ≤ α, then reject the null hypothesis. Otherwise, do not reject the null hypothesis.

Conclusion:

Here, p-value (0.6565) is greater than the level of significance (0.05).

Therefore, do not reject the null hypothesis.

There is no sufficient evidence to conclude that the standard deviation has decreased at the 0.05 level of significance.

standard deviation $= \sqrt{variance}$

2.

$n=24 \\ \sigma=250 \\ s=238 \\ α=0.01 \\ H_0: \sigma = 250 \\ H_1: \sigma ≠ 250$

The test statistic:

$Test \;statistic = \frac{(n-1)s^2}{\sigma^2} \\ = \frac{(24-1)(238)^2}{(250)^2} = 20.845 \\ df=n-1=24-1=23$

The P-value using Excel function is

=2*CHISQ.INV.RT(0.005,23)

=0.82

Here, the P-value is greater than the level of significance 0.01.

Thus, the decision is “fail to reject the null hypothesis”.

Therefore, there is no evidence that the population standard deviation is different from 250.