for each problem, find:\n1. null, alternate hypothesis, type of test & level of significance\n2. check the…

for each problem, find:\n1. null, alternate hypothesis, type of test & level of significance\n2. check the conditions.\n3. compute the sample test statistic, draw a picture and find the p-value.\n4. state the conclusion about the null hypothesis.\n5. interpret the conclusion.\n1) in a clinical study of an allergy drug, 108 of the 202 subjects reported experiencing significant relief from their symptoms. at the 0.01 significance level, test the claim that more than half of all those using the drug experience relief.

for each problem, find:\n1. null, alternate hypothesis, type of test & level of significance\n2. check the conditions.\n3. compute the sample test statistic, draw a picture and find the p-value.\n4. state the conclusion about the null hypothesis.\n5. interpret the conclusion.\n1) in a clinical study of an allergy drug, 108 of the 202 subjects reported experiencing significant relief from their symptoms. at the 0.01 significance level, test the claim that more than half of all those using the drug experience relief.

Answer

Explanation:

Step1: State hypotheses, test type, α

Null Hypothesis: $H_0: p = 0.5$ Alternate Hypothesis: $H_a: p > 0.5$ Type of test: Right-tailed z-test for population proportion Level of significance: $\alpha = 0.01$

Step2: Check conditions

  1. Random sample (implied by clinical study)
  2. $np_0 = 202 \times 0.5 = 101 \geq 10$
  3. $n(1-p_0) = 202 \times 0.5 = 101 \geq 10$
  4. Population size > 10n (reasonable, as allergy drug users are far more than 2020) All conditions are satisfied.

Step3: Calculate test statistic

Sample proportion: $\hat{p} = \frac{108}{202} \approx 0.5347$ Test statistic: $$ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} = \frac{0.5347 - 0.5}{\sqrt{\frac{0.5 \times 0.5}{202}}} \approx \frac{0.0347}{0.0351} \approx 0.99 $$ For right-tailed test, P-value = $P(Z > 0.99) = 1 - P(Z \leq 0.99)$ From z-table, $P(Z \leq 0.99) = 0.8389$, so P-value = $1 - 0.8389 = 0.1611$

Step4: Make conclusion

Compare P-value to $\alpha$: $0.1611 > 0.01$ Fail to reject $H_0$.

Step5: Interpret conclusion

At the 0.01 significance level, there is not sufficient evidence to support the claim that more than half of all those using the drug experience relief.

Answer:

  1. Null Hypothesis: $H_0: p = 0.5$; Alternate Hypothesis: $H_a: p > 0.5$; Right-tailed z-test for proportion; $\alpha = 0.01$
  2. All conditions for the z-test are satisfied
  3. Test statistic $z \approx 0.99$; P-value = 0.1611
  4. Fail to reject the Null Hypothesis
  5. There is insufficient evidence at the 0.01 level to support the claim that over half of drug users experience symptom relief.