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Sagot :
The value of the test statistic is 0.711, for the test at the 5% level of significance that the population variances differ in a random sample of 20 patients.
Since the information given to us is from a random sample of 20 patients who visited a clinic at medical center 1, a researcher found that the variance of the waiting time (in minutes) was 128.0. in a random sample of 15 patients in the clinic of medical center 2, the researcher found the variance to be 178.8, and tested at the 5% level of significance that the population variances differ. The test statistic for this problem is calculated using the F-test. The formula for the F-test is:
F-test = (variance of sample 1) / (variance of sample 2). Therefore, the test statistic for this problem is F-test = 128.0 / 178.8 = 0.711.
To determine the p-value, we need to look up the F-distribution with (df1 = 20-1 = 19, df2 = 15-1 = 14) at a significance level of 5
%. The p-value is 0.347. Therefore, at a 5% level of significance, the test statistic is 0.711 and the p-value is 0.347. We cannot reject the null hypothesis that the population variances are equal.
To know more about test statistic refer to the link brainly.com/question/14128303
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