Question

> Question 9 3 pts In a randomized, double-blind, placebo-controlled trial of children, echinacea was tested as a treatment for upper respiratory infections in children. "Days of fever" was one criterion used to measure effects. Among 337 children treated with echinacea, the mean number of days with fever was 0.81, with a standard deviation of 1.50 days. Among 370 children given a placebo, the mean number of days with fever was 0.64 with a standard deviation of 1.16 days. Assuming that both samples are from a normally distributed population with unknown but unequal standard deviations. Use a 0.07 significance level to test the claim that he echinacea affects the number of days with fever. Make your decision on the degrees of freedom by choosing the smaller sample's size, n. A. Test the claim with the p-value of the test statistic. i. What is the value of the test statistic to three decimal places? ii. What is the test statistic's p-value? B. Construct the confidence interval appropriate for the hypothesis test in part (A). iii. Lower Limit: iv. Upper Limit: C Conduct the same test of the claim but now consider the population standard deviations are EQUAL Does the conclusion to Reject or Fail to Reject the Null hypothesis change? HTML Editor BIVA-A-I EX31 x *, ! EE

Answer 1

X1 = 0.81 X2 = 0.64 Also, the provided sample standard deviations are: $i = 1.5 S2 = 1.16 and the sample sizes are ni = 337 and na = 370. .(1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: Mi = 12 Ha: Mi 7 M2 This corresponds to a two-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.07, and the degrees of freedom are df = 631.203. In fact, the degrees of freedom are computed as follows, assuming that the population variances are unequal: = 1.815, for Hence, it is found that the critical value for this two-tailed test is to a = 0.07 and df = 631.203. The rejection region for this two-tailed test is R={t:\t\ > 1.815}. (3) Test Statistics Since it is assumed that the population variances are unequal, the t-statistic is computed as follows: X1 - X2 t= V ni n2 0.81 0.64 1.674 1.52 V 337 + 1.162 370 (4) Decision about the null hypothesis Since it is observed that |t| = 1.674 <tc = 1.815, it is then concluded that the null hypothesis is not rejected. = 0.0946 > Using the P-value approach The p-value is p 0.0946, and since p 0.07, it is concluded that the null hypothesis is not rejected. (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean Mi is different than um, at the 0.07 significance level. Confidence Interval The 93% confidence interval is -0.014 < Mi – M2 < 0.354.

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