Breaking Strength Test C030C9

1. **State the problem:** We want to test the manufacturer's claim that the average breaking strength of cables is $5000$ Newtons. 2. **Identify the hypotheses:** - Null hypothesis $H_0$: $\mu = 5000$ - Alternative hypothesis $H_a$: $\mu \neq 5000$ 3. **Given data:** - Sample size $n = 100$ - Sample mean $\bar{x} = 4980$ - Population standard deviation (estimated by sample) $s = 100$ - Claimed population mean $\mu_0 = 5000$ 4. **Test statistic formula:** Since the sample size is large, use the $z$-test: $$ z = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} $$ 5. **Calculate the test statistic:** $$ z = \frac{4980 - 5000}{100 / \sqrt{100}} = \frac{-20}{100 / 10} = \frac{-20}{10} = -2 $$ 6. **Determine the significance level:** Commonly $\alpha = 0.05$ for a two-tailed test. 7. **Find critical values:** For $\alpha = 0.05$ two-tailed, critical $z$ values are approximately $\pm 1.96$. 8. **Decision rule:** - If $z < -1.96$ or $z > 1.96$, reject $H_0$. - Otherwise, do not reject $H_0$. 9. **Conclusion:** Since $z = -2$ is less than $-1.96$, we reject the null hypothesis. 10. **Interpretation:** There is sufficient evidence at the $5\%$ significance level to conclude that the average breaking strength differs from $5000$ Newtons.