Question

Monica works for a company that creates wooden furniture and claims that only 2.5% of their furniture is defective. She wants to test this claim to see if the actual percentage of defective furniture differs from the company's claim, using a significance level (α) of 0.05. Monica takes a sample of 780 pieces of furniture and observes that 40 of them are defective. Assume a normal sampling distribution.

(a) Formulate the null and alternative hypotheses for Monica's hypothesis test.
(b) Calculate the test statistic for Monica's hypothesis test, rounding to two decimal places if needed.
(c) Using a statistical table, determine the p-value for Monica's hypothesis test, rounding the answer to four decimal places if needed.
(d) Based on the p-value, should Monica reject the null hypothesis? Explain.
(e) What conclusion can Monica draw from this data regarding the company's claim about the percentage of defective furniture?

Answer 1

Monica can reject the null hypothesis and conclude that there is enough evidence to suggest that the actual percentage of defective furniture differs from the company's claim of 2.5%.

Step-by-step explanation:

1. The null hypothesis (H0) is that the actual percentage of defective furniture is 2.5%. The alternative hypothesis (Ha) is that the actual percentage of defective furniture differs from 2.5%.
2. To calculate the test statistic, we use the formula: Test statistic = (observed proportion - hypothesized proportion) / standard deviation of the proportion. In this case, the observed proportion is 40/780 = 0.0513, the hypothesized proportion is 0.025, and the standard deviation of the proportion is sqrt((0.025 * (1-0.025)) / 780) = 0.00718. Plugging in these values, we get: Test statistic = (0.0513 - 0.025) / 0.00718 = 3.649.
3. Using a statistical table or calculator, we find that the p-value for a test statistic of 3.649 with degrees of freedom equal to 779 (assuming a normal distribution) is approximately 0.0003.
4. Based on the p-value of 0.0003, which is less than the significance level of 0.05, Monica should reject the null hypothesis.
5. From this data, Monica can conclude that there is enough evidence to suggest that the actual percentage of defective furniture differs from the company's claim of 2.5%.