Step 1 :State the null hypothesis (H0): \(\sigma^2 = 2.89\) cm², and the alternative hypothesis (H1): \(\sigma^2 \neq 2.89\) cm².
Step 2 :Calculate the test statistic: \(\chi^2 = (n - 1) * s^2 / \sigma^2 = (71 - 1) * 2.25 / 2.89 = 54.50\)
Step 3 :Determine the critical values for df = 70 and \(\alpha/2 = 0.025\): \(\chi^2_{lower} = 48.76\) and \(\chi^2_{upper} = 95.02\)
Step 4 :\(\boxed{\text{Final Answer:}}\) Since the test statistic (54.50) falls between the critical values (48.76 and 95.02), we fail to reject the null hypothesis. Therefore, we cannot conclude that the hormone has made a significant difference in the variance of apple diameters at a 5% significance level.