Questions: The director of research and development is testing a new medicine. She wants to know if there is evidence at the 0.05 level that the medicine relieves pain in less than 384 seconds. For a sample of 71 patients, the mean time in which the medicine relieved pain was 382 seconds. Assume the standard deviation is known to be 23. State the null and alternative hypotheses for the above scenario.

Solution Steps

We define the null and alternative hypotheses as follows: \[ H_0: \mu \geq 384 \quad \text{(The mean time to relieve pain is 384 seconds or more)} \] \[ H_a: \mu < 384 \quad \text{(The mean time to relieve pain is less than 384 seconds)} \]

The standard error \(SE\) is calculated using the formula: \[ SE = \frac{\sigma}{\sqrt{n}} = \frac{23}{\sqrt{71}} \approx 2.7296 \]

The Z-test statistic is calculated using the formula: \[ Z = \frac{\bar{x} - \mu_0}{SE} = \frac{382 - 384}{2.7296} \approx -0.7327 \]

For a left-tailed test, the P-value is given by: \[ P = T(z) \approx 0.2319 \]

At a significance level of \(\alpha = 0.05\), we compare the P-value with \(\alpha\): \[ 0.2319 > 0.05 \] Since the P-value is greater than the significance level, we fail to reject the null hypothesis.

Final Answer