Problem Solving. Compute for the hypothesis test values of the given problem. Show the five necessary steps.
A psychiatrist is testing a new anti-anxiety drug, which seems to have the potentially harmful side effect of lowering the heart rate. For a sample of 50 medical students whose pulse was measured after 6 weeks of taking the drug, the mean heart rate was 70 beats per minute (bpm). If the mean heart rate for the population is 72 bpm with a standard deviation of 12, can the psychiatrist conclude that the new drug lowers heart rate significantly? (Set the level of significance to 0.01.)Β
SOLUTIONS:
Step 1: State the hypotheses.
Ho:Β
Ha:
Step 2: The level of significance and the critical region. πΌ = _____, ππππ‘ππππ π£πππ’π = _____.Β
Step 3: Compute for the value of one sample test.
πππππ’π‘ππ π‘ππ π‘ π£πππ’π = _______.Β
Step 4: Decision rule.
Step 5. Conclusion.
The hypotheses to be tested in this scenario are,
"H_0:\\mu=72" "Against" "H_1:\\mu\\lt72"
The sample size, "n=50", sample mean, "\\bar{x}=70" while the population standard deviation, "\\sigma=12".
The level of significance "\\alpha=1\\%=1\/100=0.01" and the critical value is obtained using the standard normal tables. The critical value is the value which leaves an area under the curve of "\\alpha=0.01" to the right and "(1-\\alpha )=1-0.01=0.99" to the left.
For this case, this value is "Z=2.33" and since the alternative hypothesis is left hand one-sided test, we shall negate this value in order to make the required comparisons.
Hence, critical value for this test is "Z=-2.33" .
The test statistic is given as,
"Z_c^*=(\\bar{x}-\\mu)\/(\\sigma\/\\sqrt{n})"
"Z_c^*=(70-72)\/(12\/\\sqrt{50})"
"Z_c^*=-2\/1.6971=-1.18(2\\space decimal\\space places)"
The null hypothesis is rejected if "Z_c^*" is less than the critical value, "Z" . For this case therefore, we fail to reject the null hypothesis since "Z_c^*=-1.18" is greater than "Z=-2.33". Hence, there is no sufficient evidence for the psychiatrist to conclude that the new drug lowers heart rate significantly at "1\\%" level of significance.
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