Question

10 COLUOPULU PL . 5. In a survey, 8236 out of 9858 doctors recommend aspirin for for their patients with headaches. Let p denote the percentage of doctors who would recommend aspirin for their patients with headache. (a) Construct a 95% confidence interval for p based on the sample information (b) It has been claimed that 9 out of 10 doctors recommend aspirin for their patients with headaches. Is this claim accurate? Perform a statistical test at the significance level a = 0.05.

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Answer 1

EXCEL (Mega Stat) output: Confidence interval - proportion 95% confidence level 0.835463583 proportion 9858 n 1.960 z 0.007 half-width 0.843 upper confidence limit 0.828 lower confidence limit The 95% confidence interval for p is 0.828 < p <0.843. (6) Null hypothesis: Ho: p=0.9 That is, 9 out of 10 doctors recommend aspirin for their patients with headaches. Alternative hypothesis:

H:p0.9 The claim is not accurate. Rejection rule: If p-value <a(=0.05), then reject the null hypothesis H. EXCEL (Mega Stat) output: Hypothesis test for proportion vs hypothesized value Observed Hypothesized 0.8355 0.9 p (as decimal) 8236/9858 8872/9858 p (as fraction) 8236 8872.2 X 9858 9858 n 0.003 std. error -21.36 z 0.00E+00 p-value (two-tailed) Test Statistic: z =-21.36. P-value = 0.

Conclusion: Use a significance level, a = 0.05. Here, p-value is 0, which is less than the value of a -0.05. That is, p-value (0) Sal=0.05). Therefore, by the rejection rule, reject the null hypothesis H. Interpretation: There is insufficient evidence at level of significance a = 0.05. The claim is not accurate.