Respuesta :
Answer:
(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:
[tex]\bar x=\frac{1}{n}\sum X_{i}[/tex]
[tex]=\frac{1}{12}\times [35+36+37+33+34+35+33+35+37+36+34+35]\\=35[/tex]
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:
[tex]z_{\alpha}=z_{0.05}=-1.645[/tex]
*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:
[tex]z=\frac{\bar x-\mu}{\sigma/\sqrt{n}}=\frac{35-31}{4.12/\sqrt{12}}=3.36[/tex]
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.
