A study compared tooth health and periodontal damage in a group of 40 young adult males with a tongue piercing and a control group of 50 young adult males without tongue piercing. Here are the summary statistics:GroupSample sizeMean of enamel cracksStDev of enamel cracksTongue piercing404.13.7No Tongue piercing501.31.6 State the null and alternative hypotheses: Give the test statistic: Find the p-value: Decision and Conclusion:

Given information, Sample size of young adult males with a tongue piercing n1=40 Sample size of young adult males without a tongue piercing n2=50 Sample mean of young adult males with a tongue piercing X¯1=4.1 Sample mean of young adult males without a tongue piercing X¯2=1.3 Sample standard deviation of young adult males with a tongue piercing s1=3.7 Sample standard deviation of young adult males without a tongue piercing s2=1.6Defining the hypotheses, H0:μ1=μ2H1:μ1≠μ2 This corresponds to a two-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used. The degrees of freedom will be calculated as follows: df=n1+n2-2=40+50-2=88 The critical value tc, at 0.05 significance level is ±1.987. The test statistic will be computed as follows: t=X¯1-X¯1sp1n1+1n2 where, sp=n1-1s12+n2-1s22n1+n2-2=40-1*3.72+50-1*1.6240+50-2=2.737 Solving for the test statistic, t=X¯1-X¯1sp1n1+1n2=4.1-1.32.737140+150=4.822 Since the p-value is less than 0.00001 (that is, p<0.00001), therefore, the p-value can be approximated to 0. Since the test statistic (4.822) is greater than the critical value (1.987), and the p-value (0) is less than the significance level (0.05), therefore, there is sufficient evidence to reject the null hypothesis. Hence, it can be concluded that the tooth health and periodontal damage in a group of 40 young adult males with a tongue piercing is significantly different from the tooth health and periodontal damage of a control group of 50 young adult males without tongue piercing.