Answer to Question #266773 in Statistics and Probability for Rubeka

Question

Blood glucose levels for obese patients have a mean of 100 with a standard deviation of 15. A researcher thinks that a diet high in protein will have a positive or negative effect on blood glucose levels. A sample of 30 obese patients who have tried a high in protein diet have a mean glucose level of 94. Test the hypothesis that a diet high in protein has an effect on blood glucose using a z-test with a level of significance =.04. Z-table

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=100$

$H_1:\mu\not=100$

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 $\alpha = 0.04,$ and the critical value for a two-tailed test is $z_c = 2.0537.$

The rejection region for this two-tailed test is$R = \{z: |z| > 2.0537\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{94-100}{15/\sqrt{30}}=-2.19089$$

Since it is observed that $|z| = 2.19089 >2.0537= z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is $p=2P(Z<-2.19089)=0.02846,$ and since $p = 0.02846 < 0.04=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$ is different than $100,$ at the $\alpha = 0.04$ significance level.

Therefore, there is enough evidence to claim that a diet high in protein has an effect on blood glucose, at the $\alpha = 0.04$ significance level.