We' supposed to indicate which statement is true/false.
Note that, if a sample size is 40 or over, we can use the t distribution even with skewed data. So it's not highly sensitive to non-normality of the population from which samples are taken. So statement A is false.
It's true that the t-distribution assumes that the population from which samples are drawn is normally distributed. So B is true.
For skewed data or with extreme outliers, we can't use the t distribution. We only use t distribution as long as we believe that the population from which samples are drawn is closed to a bell-shape. So C is true.
Lastly, statement D is against statement C. So D is false.
Answer:
there are no solutions
Step-by-step explanation:
Answer:
10.04 × 8.8 =?, ? = 88.352 :)
Using a calculator, it is found that for the two-tailed test of significance, the p-value is of 0.9195.
The correlation coefficient is also called <u>Pearson's r-score</u>, and is used for two-tailed tests. To find the p-value, the information needed is:
- The value of the Pearson's r-score, that is, the value of the correlation coefficients.
- The sample size.
In this problem, we have that the correlation coefficient is of r = 0.02, with a sample size of n = 28.
- Using it as the input for a r-score calculator, the p-value is of 0.9195.
A similar problem is given at brainly.com/question/13873630
Answer:
7^12
Step-by-step explanation:
7^14
-----------
7^2
When we divide with the same base, we subtract the exponents
7^(14-2)
7^12
