Given the sample mean $=21.725$, sample standard deviation $=5.0738$, and $N=40$ for the low income group, Test the claim that the mean nickel diameter drawn by children in the low income group is greater than $\mathbf{2 1 . 2 1} \mathbf{\mathrm { mm }}$. Test at the 0.05 significance level. a) Identify the correct alternative hypothesis: Give all answers correct to 3 decimal places. b) The test statistic value is: c) Using the Traditional method, the critical value is: d) Based on this, we Reject $H_{0}$ Fail to reject $H_{0}$ e) Which means There is sufficient evidence to warrant rejection of the claim There is not sufficient evidence to support the claim There is not sufficient evidence to warrant rejection of the claim The sample data supports the claim

Step 1 ：We are given the sample mean, sample standard deviation, and the number of samples. We are asked to test the claim that the mean nickel diameter drawn by children in the low income group is greater than 21.21 mm. This is a one-tailed t-test problem because we are testing if the mean is greater than a certain value.

Step 2 ：The null hypothesis, \(H_{0}\), is that the mean nickel diameter is equal to 21.21 mm. The alternative hypothesis, \(H_{1}\), is that the mean nickel diameter is greater than 21.21 mm.

Step 3 ：We can calculate the test statistic using the formula: \(t = \frac{\bar{x} - \mu_{0}}{s / \sqrt{n}}\) where \(\bar{x}\) is the sample mean, \(\mu_{0}\) is the hypothesized population mean, \(s\) is the sample standard deviation, and \(n\) is the number of samples.

Step 4 ：We can then compare the test statistic to the critical value at the 0.05 significance level to decide whether to reject or fail to reject the null hypothesis.

Step 5 ：The test statistic value is approximately 0.642, and the critical value is approximately 1.685. Since the test statistic is less than the critical value, we fail to reject the null hypothesis. This means there is not sufficient evidence to warrant rejection of the claim that the mean nickel diameter drawn by children in the low income group is greater than 21.21 mm.

Step 6 ：Final Answer: a) The correct alternative hypothesis is: \(H_{1}: \mu > 21.21\) b) The test statistic value is: \(\boxed{0.642}\) c) Using the Traditional method, the critical value is: \(\boxed{1.685}\) d) Based on this, we: Fail to reject \(H_{0}\) e) Which means: There is not sufficient evidence to warrant rejection of the claim.