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    Hypothesis Testing of Proportions

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    awson and Jick [1976] compare drug prescription in the United States and Scotland.
    (a) In patients with congestive heart failure, two or more drugs were prescribed in 257 of 437 U.S. patients. In Scotland, 39 of 179 patients had two or more drugs prescribed. Test the null hypothesis of equal proportions giving the resulting p- value. Construct a 95% confidence interval for the difference in proportions.

    © BrainMass Inc. brainmass.com December 24, 2021, 10:14 pm ad1c9bdddf
    https://brainmass.com/statistics/hypothesis-testing/hypothesis-testing-proportions-454184

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    Biostatistics
    Awson and Jick [1976] compare drug prescription in the United States and Scotland.
    (a) In patients with congestive heart failure, two or more drugs were prescribed in 257 of 437 U.S. patients. In Scotland, 39 of 179 patients had two or more drugs prescribed. Test the null hypothesis of equal proportions giving the resulting p- value. Construct a 95% confidence interval for the difference in proportions.
    Answer
    The null hypothesis tested is
    H0: There is no significant difference in the proportion of patients had two or more drugs prescribed in the United States and Scotland (P1 = P2)
    The alternative hypothesis is
    H1: There is significant difference in the proportion of patients had two or more drugs prescribed in the United States and Scotland (P1 ≠ P2)
    The Test Statistic used is
    where
    Here p1 = 257/437 = 0.588100686, p2 = 39/179 = 0.217877095, n1 = 437, n2 = 179
    Now = 0.480519481
    Therefore, = 8.350276088
    Rejection criteria: Reject the null hypothesis, if the observed significance (p-value) is less than the significance level 0.05.
    P-value = P ( > 8.350276088) = 0
    Conclusion: Reject the null hypothesis, since the observed significance (p-value) is less than the significance level 0.05. The sample provides enough evidence to conclude that there is significant difference in the proportion of patients had two or more drugs prescribed in the United States and Scotland.
    Details
    Z Test for Differences in Two Proportions

    Data
    Hypothesized Difference 0
    Level of Significance 0.05
    Group 1
    Number of Successes 257
    Sample Size 437
    Group 2
    Number of Successes 39
    Sample Size 179

    Intermediate Calculations
    Group 1 Proportion 0.588100686
    Group 2 Proportion 0.217877095
    Difference in Two Proportions 0.370223592
    Average Proportion 0.480519481
    Z Test Statistic 8.350276088

    Two-Tail Test
    Lower Critical Value -1.959963985
    Upper Critical Value 1.959963985
    p-Value 0
    Reject the null hypothesis

    95% confidence interval is given by,
    , where p1 = 257/437 = 0.588100686, p2 = 39/179 = 0.217877095, n1 = 437, n2 = 179, = 1.959963985
    Therefore, required confidence interval is

    = (0.294154929, 0.446292254)
    Thus with 95% confidence we can claim that the difference in proportions is within (0.294, 0.446).
    Details
    Confidence Interval Estimate
    of the Difference Between Two Proportions

    Data
    Confidence Level 95%

    Intermediate Calculations
    Z Value -1.959963985
    Std. Error of the Diff. between two Proportions 0.038811255
    Interval Half Width 0.076068662

    Confidence Interval
    Interval Lower Limit 0.294154929
    Interval Upper Limit 0.446292254

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 10:14 pm ad1c9bdddf>
    https://brainmass.com/statistics/hypothesis-testing/hypothesis-testing-proportions-454184

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