Question

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. 𝛼 = _____, 𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑣𝑎𝑙𝑢𝑒 = _____. 

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

𝑐𝑜𝑚𝑝𝑢𝑡𝑒𝑑 𝑡𝑒𝑠𝑡 𝑣𝑎𝑙𝑢𝑒 = _______. 

Step 4: Decision rule.

Step 5. Conclusion.

Answer 1

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.