Answer to Question #344450 in Statistics and Probability for Ericka

Question

The mean serum level measured in 12 patients twenty-four hours after they received a newly recommended antibiotic was 1.2 mg/dl with a standard deviation of 0.4 mg/dl. If the mean serum level in the general population is 1.0 mg/dl, test whether or not the mean serum level in the sample group is significantly different from that of the general population. Use a = 5%. Compute the test value.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=1$

$H_1:\mu\not=1$

This corresponds to a two-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

Based on the information provided, the significance level is $\alpha = 0.05,$ and the critical value for a two-tailed test is $z_c = 1.96.$

The rejection region for this two-tailed test is $R = \{z:|z|>1.96\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{1.2-1}{0.4/\sqrt{12}}\approx1.732$$

Since it is observed that $|z|=1.732<1.96=z_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value is $p=2P(z>1.732)= 0.083274,$ and since $p= 0.083274>0.05=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$

is different than 1, at the $\alpha = 0.05$ significance level.