# Answer in Statistics and Probability for smilynne #248592

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.)

SOLUTIONS:

Step 1: State the hypotheses.

Ho:

Ha:

Step 2: The level of significance and the critical region. = _____, = _____.

Step 3: Compute for the value of one sample test.

= _______.

Step 4: Decision rule.

Step 5. Conclusion.

The hypotheses to be tested in this scenario are,

“H_0:mu=72” “Against” “H_1:mult72”

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(2space decimalspace 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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