Answer to Question #170899 in Statistics and Probability for john lester alisuag

Question

1. Blood glucose levels for obese patients have a mean of 100 with a <span style="font-size: 12pt; color: rgb(0, 0, 0); background-color: transparent; font-variant-numeric: normal; font-variant-east-asian: normal; vertical-align: baseline; white-space: pre-wrap;">standard deviation</span> of 15. A researcher thinks that a diet high in raw cornstarch will have a positive or negative effect on blood glucose levels. A sample of 30 patients who have tried the raw cornstarch diet have a mean glucose level of 140.
   

Answer 1

We have that

$\mu=100$

$\sigma=15$

$n=30$

$\bar x=140$

$H_0:\mu=100$

$H_1:\mu\ne100$

The hypothesis test is two-tailed.

Since the population standard deviation is known and the sample size is large (≥30) we use Z-test.

Let $\alpha=0.05$ thus the critical value is Z_0.025 = ±1.96

The rejection region is Z < –1.96 and Z > 1.96

Test statistic:

$$Z_{test}=\frac{\bar x-\mu}{\frac{\sigma}{\sqrt n}}=\frac{140-100}{\frac{15}{\sqrt 30}}=14.61$$

Since 14.61 > 1.96 thus the Z_test falls in the rejection region we reject the null hypothesis.

At the 5% significance

level the data do provide sufficient evidence to support the claim. We are 95% confident to conclude that a diet high in raw cornstarch has an effect on blood glucose levels.