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ive patients entering a hospital with a high fever were selected at random and were administered a new drug which is designed to reduce fever. The body temperature (°F) of each patient was measured before the new drug was administered and 2 hours after the administration. The data are given in the following table.Patient 1 2 3 4 5Before 103 99 102 101 103After 100 99 98 100 100 Assume the differences between before and after measurements follow a normal distribution. Let μ1 denote the mean body temperature of high fever patients before taking the new drug and let μ2 denote the mean body temperature of high fever patients after taking the new drug. Also let μD = μ1 - μ2. Suppose we want to test whether the new drug is effective in reducing fever at the 5% level of significance. What is the observed test statistic for this test?Question 2Select one:a.2.678b.-2.678c.-2.994d.2.994

Question

ive patients entering a hospital with a high fever were selected at random and were administered a new drug which is designed to reduce fever. The body temperature (°F) of each patient was measured before the new drug was administered and 2 hours after the administration. The data are given in the following table.Patient 1 2 3 4 5Before 103 99 102 101 103After 100 99 98 100 100 Assume the differences between before and after measurements follow a normal distribution. Let μ1 denote the mean body temperature of high fever patients before taking the new drug and let μ2 denote the mean body temperature of high fever patients after taking the new drug. Also let μD = μ1 - μ2. Suppose we want to test whether the new drug is effective in reducing fever at the 5% level of significance. What is the observed test statistic for this test?Question 2Select one:a.2.678b.-2.678c.-2.994d.2.994

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Solution

To answer this question, we first need to calculate the differences in temperatures before and after the drug was administered for each patient. Then, we calculate the mean and standard deviation of these differences.

Here are the differences for each patient:

Patient 1: 103 - 100 = 3 Patient 2: 99 - 99 = 0 Patient 3: 102 - 98 = 4 Patient 4: 101 - 100 = 1 Patient 5: 103 - 100 = 3

The mean (μD) of these differences is (3 + 0 + 4 + 1 + 3) / 5 = 2.2

The standard deviation is calculated as follows:

sqrt[((3-2.2)^2 + (0-2.2)^2 + (4-2.2)^2 + (1-2.2)^2 + (3-2.2)^2) / (5-1)] = sqrt[(0.64 + 4.84 + 3.24 + 1.44 + 0.64) / 4] = sqrt[10.8 / 4] = sqrt[2.7] = 1.643

The observed test statistic (t) is calculated as (mean difference - hypothesized difference) / (standard deviation / sqrt(n)). In this case, the hypothesized difference is 0 (because we're testing whether the drug has an effect), and n is 5 (the number of patients).

So, t = (2.2 - 0) / (1.643 / sqrt(5)) = 2.2 / (1.643 / 2.236) = 2.2 / 0.734 = 2.994

So, the observed test statistic for this test is 2.994. Therefore, the correct answer is d. 2.994.

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