Coin Bias Test 9F532A

1. **Problem Statement:** a) We have observed frequencies of the number of heads in 4 coin tosses repeated 200 times. We want to test if the coins are biased using a $\chi^2$ goodness of fit test. b) We want to test the manufacturer's claim that the average breaking strength of cables is 5000 Newtons based on a sample of 100 cables with mean 4980 and standard deviation 100. --- ### Part a) $\chi^2$ Goodness of Fit Test for Coin Tosses 2. **Formula and Important Rules:** The number of heads in 4 tosses follows a Binomial distribution $B(n=4, p=0.5)$ if the coins are fair. The expected frequency for each number of heads $k$ is: $$E_k = n_{trials} \times P(X=k) = 200 \times \binom{4}{k} (0.5)^k (0.5)^{4-k} = 200 \times \binom{4}{k} (0.5)^4$$ The $\chi^2$ test statistic is: $$\chi^2 = \sum_{k=0}^4 \frac{(O_k - E_k)^2}{E_k}$$ where $O_k$ is the observed frequency. Degrees of freedom $df = \text{number of categories} - 1 = 5 - 1 = 4$. 3. **Calculate Expected Frequencies:** Binomial coefficients: $$\binom{4}{0} = 1, \binom{4}{1} = 4, \binom{4}{2} = 6, \binom{4}{3} = 4, \binom{4}{4} = 1$$ Probability for each $k$: $$P(X=k) = \binom{4}{k} (0.5)^4 = \binom{4}{k} \times 0.0625$$ Expected frequencies: $$E_0 = 200 \times 1 \times 0.0625 = 12.5$$ $$E_1 = 200 \times 4 \times 0.0625 = 50$$ $$E_2 = 200 \times 6 \times 0.0625 = 75$$ $$E_3 = 200 \times 4 \times 0.0625 = 50$$ $$E_4 = 200 \times 1 \times 0.0625 = 12.5$$ 4. **Observed Frequencies:** $$O_0 = 9, O_1 = 42, O_2 = 73, O_3 = 61, O_4 = 15$$ 5. **Calculate $\chi^2$ Statistic:** For each category: $$\frac{(9 - 12.5)^2}{12.5} = \frac{(-3.5)^2}{12.5} = \frac{12.25}{12.5} = 0.98$$ $$\frac{(42 - 50)^2}{50} = \frac{(-8)^2}{50} = \frac{64}{50} = 1.28$$ $$\frac{(73 - 75)^2}{75} = \frac{(-2)^2}{75} = \frac{4}{75} = 0.0533$$ $$\frac{(61 - 50)^2}{50} = \frac{11^2}{50} = \frac{121}{50} = 2.42$$ $$\frac{(15 - 12.5)^2}{12.5} = \frac{2.5^2}{12.5} = \frac{6.25}{12.5} = 0.5$$ Sum: $$\chi^2 = 0.98 + 1.28 + 0.0533 + 2.42 + 0.5 = 5.2333$$ 6. **Decision:** At $\alpha = 0.05$ significance level, the critical value for $df=4$ is approximately 9.488. Since $5.2333 < 9.488$, we **fail to reject** the null hypothesis. **Conclusion:** There is no significant evidence to say the coins are biased. --- ### Part b) Hypothesis Test for Mean Breaking Strength 7. **Problem:** Test the claim that the population mean $\mu = 5000$ Newtons. Sample mean $\bar{x} = 4980$, sample standard deviation $s = 100$, sample size $n=100$. 8. **Hypotheses:** $$H_0: \mu = 5000$$ $$H_a: \mu \neq 5000$$ (two-tailed test) 9. **Test Statistic:** Since $n$ is large, use the $z$-test: $$z = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} = \frac{4980 - 5000}{100/\sqrt{100}} = \frac{-20}{100/10} = \frac{-20}{10} = -2$$ 10. **Critical Value:** At $\alpha = 0.05$ for two-tailed test, critical $z$ values are $\pm 1.96$. 11. **Decision:** Since $-2 < -1.96$, we reject $H_0$. There is sufficient evidence to suggest the mean breaking strength is different from 5000 Newtons. --- **Final answers:** - a) The coins are not biased at 5% significance level. - b) The manufacturer's claim is rejected at 5% significance level; the mean breaking strength differs from 5000 Newtons.