Answer to Question #337957 in Statistics and Probability for kyasu

Question

Blood glucose levels for those obese teenagers have a mean of 120. A researcher thinks that a diet high in raw corn starch will have a positive or negative effect on blood glucose levels. A sample of 25 students who have tried the raw corn starch diet has a mean glucose level of 135 with a standard deviation of 38. Test the hypothesis using the level of significance of 0.10 that the raw corn starch diet had an effect.

Answer 1

1. The following null and alternative hypotheses need to be tested:

$H_0:\mu=120$

$H_a:\mu\not=120$

This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation, using the sample standard deviation, will be used.

2. Based on the information provided, the significance level is $\alpha = 0.10,$ $df=n-1=24$ degrees of freedom, and the critical value for a two-tailed test is $t_c =1.710882.$

The rejection region for this two-tailed test is $R = \{t: |t|>1.710882\}$

3.The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{135-120}{38/\sqrt{25}}=1.973684$$

4. Since it is observed that $|t| =1.973684>1.710882= t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for two-tailed, $df=24$ degrees of freedom, $t=1.973684$ is $p =0.056589,$ and since $p= 0.056589<0.10=\alpha,$ it is concluded that the null hypothesis is rejected.

5. Therefore, there is enough evidence to claim that the population mean $\mu$ is different than 120, at the $\alpha = 0.10$ significance level.