Answer:
a. P-value = 0.039.
The null hypothesis is rejected.
At a significance level of 0.05, there is enough evidence to support the claim that the population mean significantly differs from 100.
b. P-value = 0.013.
The null hypothesis is rejected.
At a significance level of 0.05, there is enough evidence to support the claim that the population mean significantly differs from 100.
c. P-value = 0.130.
The null hypothesis failed to be rejected.
At a significance level of 0.05, there is not enough evidence to support the claim that the population mean significantly differs from 100.
Step-by-step explanation:
This is a hypothesis test for the population mean.
The claim is that the population mean significantly differs from 100.
Then, the null and alternative hypothesis are:
The significance level is 0.05.
The sample has a size n=65.
The degrees of freedom for this sample size are:
a. The sample mean is M=103.
As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=11.5.
The estimated standard error of the mean is computed using the formula:
Then, we can calculate the t-statistic as:
This test is a two-tailed test, with 64 degrees of freedom and t=2.103, so the P-value for this test is calculated as (using a t-table):
As the P-value (0.039) is smaller than the significance level (0.05), the effect is significant.
The null hypothesis is rejected.
At a significance level of 0.05, there is enough evidence to support the claim that the population mean significantly differs from 100.
b. The sample mean is M=96.5.
As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=11.
The estimated standard error of the mean is computed using the formula:
Then, we can calculate the t-statistic as:
This test is a two-tailed test, with 64 degrees of freedom and t=-2.565, so the P-value for this test is calculated as (using a t-table):
As the P-value (0.013) is smaller than the significance level (0.05), the effect is significant.
The null hypothesis is rejected.
At a significance level of 0.05, there is enough evidence to support the claim that the population mean significantly differs from 100.
c. The sample mean is M=102.
As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=10.5.
The estimated standard error of the mean is computed using the formula:
Then, we can calculate the t-statistic as:
This test is a two-tailed test, with 64 degrees of freedom and t=1.536, so the P-value for this test is calculated as (using a t-table):
As the P-value (0.13) is bigger than the significance level (0.05), the effect is not significant.
The null hypothesis failed to be rejected.
At a significance level of 0.05, there is not enough evidence to support the claim that the population mean significantly differs from 100.
with its inverse 
you'll get the identity function:![\mathsf{(f\circ f^{-1})(x)=x}\\\\ \mathsf{f\big[f^{-1}(x)\big]=x}\\\\](https://tex.z-dn.net/?f=%5Cmathsf%7B%28f%5Ccirc%20f%5E%7B-1%7D%29%28x%29%3Dx%7D%5C%5C%5C%5C%20%5Cmathsf%7Bf%5Cbig%5Bf%5E%7B-1%7D%28x%29%5Cbig%5D%3Dx%7D%5C%5C%5C%5C)
![\mathsf{f\big[f^{-1}(x)\big]=10-[f^{-1}(x)]^2}\\\\ \mathsf{x=10-[f^{-1}(x)]^2}\\\\ \mathsf{[f^{-1}(x)]^2=10-x}\\\\ \mathsf{f^{-1}(x)=\pm \sqrt{10-x}}\\\\ \mathsf{f^{-1}(x)=-\sqrt{10-x}~~~or~~~f^{-1}(x)=\sqrt{10-x}}](https://tex.z-dn.net/?f=%5Cmathsf%7Bf%5Cbig%5Bf%5E%7B-1%7D%28x%29%5Cbig%5D%3D10-%5Bf%5E%7B-1%7D%28x%29%5D%5E2%7D%5C%5C%5C%5C%20%5Cmathsf%7Bx%3D10-%5Bf%5E%7B-1%7D%28x%29%5D%5E2%7D%5C%5C%5C%5C%20%5Cmathsf%7B%5Bf%5E%7B-1%7D%28x%29%5D%5E2%3D10-x%7D%5C%5C%5C%5C%20%5Cmathsf%7Bf%5E%7B-1%7D%28x%29%3D%5Cpm%20%5Csqrt%7B10-x%7D%7D%5C%5C%5C%5C%20%5Cmathsf%7Bf%5E%7B-1%7D%28x%29%3D-%5Csqrt%7B10-x%7D~~~or~~~f%5E%7B-1%7D%28x%29%3D%5Csqrt%7B10-x%7D%7D)
itself, and where it's invertible.