1. **State the problem:** A medical researcher claims that a new vaccine will decrease the number of colds in adults. We have data on the number of colds before and after vaccination for 14 adults. We want to test if the vaccine significantly decreases the number of colds. 2. **Identify the claim and hypotheses:** - Claim: The vaccine decreases the number of colds. - Null hypothesis $H_0$: The mean difference in colds before and after vaccination is zero or more, i.e., $\mu_d \geq 0$. - Alternative hypothesis $H_a$: The mean difference in colds before and after vaccination is less than zero, i.e., $\mu_d < 0$. 3. **Level of significance:** Typically, $\alpha = 0.05$ unless otherwise specified. 4. **Sample size $n$:** There are 14 adults, so $n = 14$. 5. **Calculate differences $d = \text{before} - \text{after}$ for each adult:** 1: $3 - 2 = 1$ 2: $4 - 1 = 3$ 3: $2 - 0 = 2$ 4: $1 - 1 = 0$ 5: $3 - 1 = 2$ 6: $6 - 3 = 3$ 7: $4 - 3 = 1$ 8: $5 - 2 = 3$ 9: $2 - 2 = 0$ 10: $0 - 2 = -2$ 11: $2 - 3 = -1$ 12: $5 - 4 = 1$ 13: $3 - 3 = 0$ 14: $3 - 2 = 1$ 6. **Calculate sample mean difference $\bar{d}$:** $$\bar{d} = \frac{1 + 3 + 2 + 0 + 2 + 3 + 1 + 3 + 0 - 2 - 1 + 1 + 0 + 1}{14} = \frac{13}{14} \approx 0.9286$$ 7. **Calculate sample standard deviation $s_d$ of differences:** First, calculate each $(d_i - \bar{d})^2$: $(1 - 0.9286)^2 = 0.0051$ $(3 - 0.9286)^2 = 4.2867$ $(2 - 0.9286)^2 = 1.1479$ $(0 - 0.9286)^2 = 0.8622$ $(2 - 0.9286)^2 = 1.1479$ $(3 - 0.9286)^2 = 4.2867$ $(1 - 0.9286)^2 = 0.0051$ $(3 - 0.9286)^2 = 4.2867$ $(0 - 0.9286)^2 = 0.8622$ $(-2 - 0.9286)^2 = 8.5929$ $(-1 - 0.9286)^2 = 3.7184$ $(1 - 0.9286)^2 = 0.0051$ $(0 - 0.9286)^2 = 0.8622$ $(1 - 0.9286)^2 = 0.0051$ Sum of squares $= 30.9712$ Sample variance $s_d^2 = \frac{30.9712}{14 - 1} = \frac{30.9712}{13} \approx 2.3824$ Sample standard deviation $s_d = \sqrt{2.3824} \approx 1.543$ 8. **Calculate test statistic $t$:** $$t = \frac{\bar{d} - 0}{s_d / \sqrt{n}} = \frac{0.9286}{1.543 / \sqrt{14}} = \frac{0.9286}{1.543 / 3.7417} = \frac{0.9286}{0.4124} \approx 2.252$$ 9. **Determine critical value:** For a left-tailed test with $\alpha = 0.05$ and $df = n - 1 = 13$, the critical value $t_{crit} \approx -1.771$. 10. **Decision:** Since $t = 2.252$ is greater than $-1.771$, it is NOT in the rejection region. 11. **Interpretation:** We fail to reject the null hypothesis. There is not enough evidence at the 0.05 significance level to support the claim that the vaccine decreases the number of colds. **Final answer:** The data do not provide sufficient evidence to support the researcher's claim that the vaccine decreases colds.