Hypothesis Testing sample essay
Task 1: Solve the following problems:
A student of the author surveyed her friends and found that among 20 males, 4 smoke and among 30 female friends, 6 smoke. Give two reasons why these results should not be used for a hypothesis test of the claim that the proportions of male smokers and female smokers are equal. First, the sample is not random, it is a convenience sample and therefore cannot be trusted. Second, np҄ = x = 4 < 5, therefore the normal distribution cannot be used to approximate the binomial distribution. Given a simple random sample of men and a simple random sample of women, we want to use a 0.05 significance level to test the claim that the percentage of men who smoke is equal to the percentage of women who smoke. One approach is to use the P-value method of hypothesis testing; a second approach is to use the traditional method of hypothesis testing; and a third approach is to base the conclusion on the 95% confidence interval estimate of p1—p2. Will all three approaches always result in the same conclusion? Explain. As long as the test is two tailed (so that a confidence interval can be used) and since α = .05 and percent confidence = 1 – .05 = .95, then yes, the results will NEARLY ALWAYS be the same between traditional approach and confidence interval approach. Also it will NEARLY ALWAYS be the same between the p-value approach and the confidence interval approach. The reason that I say “almost always” is that the standard error formula for p҄1-p҄2 is slightly different for the confidence interval approach than for the hypothesis testing approach. The result will ALWAYS be the same between the traditional and p-value approaches whether the test is two tailed or not. Criteria for rejecting Ho: Traditional approach: α = .05, reject Ho if z < -1.96 or z > 1.96 P-value approach: Reject Ho if p-value < .05 (which only occurs if z < -1.96 or z > 1.96). Confidence interval approach: Reject Ho if 0 is not within the confidence interval.
Task 2: Solve the following problems:
The mean tar content of a simple random sample of 25 unfiltered king-size cigarettes is 21.1 mg, with a standard deviation of 3.2 mg. The mean tar content of a simple random sample of 25 filtered 100 mm cigarettes is 13.2 mg with a standard deviation of 3.7 mg. Assume that the two samples are independent, simple, random samples, selected from normally distributed populations. Do not assume that the population standard deviations are equal. o Use a 0.05 significance level to test the claim that unfiltered king-size cigarettes have mean tar content greater than that of filtered 100 mm cigarettes.
Ho: µ1 ≤ µ2
Ha: µ1 > µ2
α = .05
df = nsmaller – 1 = 25-1 = 24.
Reject Ho if t > 1.711
n1 = 25, x̅1 = 21.1, s1 = 3.2
n2 = 25, x̅2 = 13.2, s2 = 3.7
t = (x̅1-x̅2)/√[s12/n1+s22/n2] = (21.1-13.2)/√[3.22/25+3.72/25] = 8.075. Since 8.075 > 1.711, then reject Ho.
There is sufficient evidence to support the claim that the mean tar content for king size cigarettes is greater than that of filtered 100 mm cigarettes. o What does the result suggest about the effectiveness of cigarette filters? It appears that cigarette filters are effective in reducing tar. Listed below are systolic blood pressure measurements (mm Hg) taken from the right and left arms of the same woman. Use a 0.05 significance level to test for a difference between the measurements from the two arms. What do you conclude? Assume that the paired sample data are simple random samples and that the differences have a distribution that is approximately normal.
Right Arm Left Arm
Ho: µd = 0
Ha: µd ≠ 0
α = .05
Reject Ho if p-value < .05 t-Test: Paired Two Sample for Means Right Arm Left Arm Mean 91 163.2 Variance 129.5 297.7 Observations 5 5 Pearson Correlation 0.867087288 Hypothesized Mean Difference 0 df4 t Stat -17.33854504 P(T<=t) one-tail 3.24714E-05 t Critical one-tail 2.131846782 P(T<=t) two-tail 6.49427E-05 t Critical two-tail 2.776445105 t = -17.338 P-value = .0000 Since .000 < .05, then reject Ho. There is sufficient evidence to indicate a significant difference in mean systolic blood pressure between the two arms.
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