Answer to Question #183459 in Statistics and Probability for Amoah Henry

Question

. A contractor on the Eastern Corridor road makes a large purchase of industrial cement from Ghacem. Each bag of this industrial cement is supposed to weigh 94 pounds. To see if he is getting his money’s worth, the contractor weighs 10 randomly selected bags and gets the following weights (in pounds):

94.1,93.4, 92.8, 93.4, 95.4, 93.5, 94.0, 93.8, 92.9, 94.2

(a) What are the appropriate null and alternative hypotheses to be tested?

(b) State the assumption that is necessary for performing the test.

(c) Calculate the p−value of the test. Perform the test at a significance level of α = 0.1. 

Answer 1

(a)

Hypotheses:

Null $H_0: µ = 94$

Alternative $H_1: µ < 94$

(b)

Assumptions

1. Sample is randomly drawn

2. Observations (bag weights) follow a normal distribution.

(c)

$n = 10 \\ \mu = 94 \\ \bar{x}= \frac{94.1+93.4+92.8+ 93.4+ 95.4+ 93.5+ 94.0+ 93.8+ 92.9+ 94.2}{10} = 93.75 \\ s = \sqrt{ \frac{\sum (x- \bar{x})^2}{n-1} } \\ s = \sqrt{ \frac{(94.1-93.75)^2+...+(94.2-93.75)^2}{10-1}} = 0.7487$

When n<30 and $\sigma^2$ is unknown, a one sample t-test is used.

$t = \frac{\bar{x}-\mu}{s/\sqrt{n}} \\ t = \frac{93.75-94}{0.7487/\sqrt{10}} \\ = \frac{-0.25}{0.236} \\ = -1.059$

Significance level: 10 %

α=0.1

$t_{α, n-1} = t_{0.1, 9} = 1.383$ (from t-table)

Critical region: One-tailed test. Reject H₀ if t ≤ -1.383

Conclusion: Since t = -1.059 is greater than -1.383, the null hypothesis cannot be rejected.