To test the claim that the population mean of garbage weights discarded by households is less than 30 lb, we will perform a one-sample t-test because the population standard deviation is not known, and we will use the sample data provided.
Step 1: Define the null and alternative hypotheses.
- Null hypothesis (\(H_0\)): \(\mu \geq 30\) (the population mean is greater than or equal to 30 lb)
- Alternative hypothesis (\(H_a\)): \(\mu < 30\) (the population mean is less than 30 lb)
Step 2: Identify the sample statistics.
- Sample mean (\(\bar{x}\)) = 27.443 lb
- Sample size (\(n\)) = 62
- Sample standard deviation (\(s\)) = 12.458 lb
- Significance level (\(\alpha\)) = 0.05
Step 3: Calculate the test statistic.
We use the t-test formula for the test statistic: \[ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} \] Where \(\mu_0\) is the hypothesized mean (30 lb).
Plugging in the values: \[ t = \frac{27.443 - 30}{12.458 / \sqrt{62}} = \frac{-2.557}{12.458 / \sqrt{62}} = \frac{-2.557}{1.5835} \approx -1.616 \]
Step 4: Determine the critical t-value (using t-distribution).
For a one-tailed test at a significance level of 0.05 with \(n - 1 = 61\) degrees of freedom, you can look up the critical t-value in a t-table or use a calculator. The critical value for \(t\) with 61 df at the 0.05 level is approximately -1.671.
Step 5: Make the decision.
- If the calculated t-value is less than the critical t-value, we reject the null hypothesis.
In this case:
- Calculated t-value = -1.616
- Critical t-value = -1.671
Since -1.616 is greater than -1.671, we fail to reject the null hypothesis.
Final Conclusion:
B) There is not sufficient evidence to support the claim that the population of households has a mean less than 30 lb.