A physiologist believes he has discovered a new steroid hormone that can control the blood glucose in mammals. In an experiment he injects laboratory rats with a small amount of this putative steroid hormone and, after a suitable waiting period measures the blood glucose level of the rats. In his twelve experimental rats he obtains the following levels of blood glucose {35,36,37,33,34,35,33,35,37,36,34,3 5}. From previous studies of thousands of normal, un-manipulated rats he knows that the mean (?) blood glucose level in this species of rat (his laboratory population) is 31 and that the standard deviation (?) is 4.12. Use a Z-test to evaluate the directional hypothesis that this new steroid hormone decreases blood glucose levels. Use an alevel of 0.05 for this analysis.

A) what is the null hypothesis?
B) what is the alternative hypothesis?
C) what is the critical value with and alpha of .05?
D) what is Z obt?
E) what should the cognitive psychologist conclude?

(a) H₀: μ = 31.

(b) Hₐ: μ < 31.

(c) The critical value of z at α = 0.05 is -1.645.

(d) The test statistic is 3.36.

(e) The cognitive psychologist should conclude that the new steroid hormone does not decreases the blood glucose levels.

Step-by-step explanation:

In this scenario, we need to determine whether the new steroid hormone decreases blood glucose levels.

A physiologist  injects 12 laboratory rats with a small amount of this putative steroid hormone and, after a suitable waiting period measures the blood glucose level of the rats.

The data provided is:

S = {35, 36, 37, 33, 34, 35, 33, 35, 37, 36, 34, 35}

Compute the sample mean as follows:

Also it is given that:

μ = 31

σ = 4.12

(a)

The null hypothesis for this test can be defined as:

H₀: The new steroid hormone does not decreases the blood glucose levels, i.e. μ = 31.

(b)

The alternate hypothesis for this test can be defined as:

Hₐ: The new steroid hormone decreases the blood glucose levels, i.e. μ < 31.

(c)

The critical value of the distribution is the value beyond which the rejection region lies.

In this case we will use the z-test for single, since the population standard deviation is known.

The critical value of z at α = 0.05 is:

*Use a z-table.

The negative value is because of the alternate hypothesis, i.e. since we need the mean to be less than 31, the rejection region will be at μ < 31.

Thus, the critical value of z at α = 0.05 is -1.645.

(d)

Compute the value of the test statistic as follows:

The test statistic is 3.36.

(e)

The test statistic value, z = 3.36 > z₀.₀₅ = -1.645.

The null hypothesis was failed to be rejected.

Thus, the cognitive psychologist should conclude that the new steroid hormone does not decreases the blood glucose levels.

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