Answer:
e. The probability of observing a sample mean of 5.11 or less, or of 5.29 or more, is 0.018 if the true mean is 5.2.
Step-by-step explanation:
We have a two-tailed one sample t-test.
The null hypothesis claims that the pH is not significantly different from 5.2.
The alternative hypothesis is that the mean pH is significantly different from 5.2.
The sample mean pH is 5.11, with a sample size of n=50.
The P-value of the test is 0.018.
This P-value corresponds to the probability of observing a sample mean of 5.11 or less, given that the population is defined by the null hypothesis (mean=5.2).
As this test is two-tailed, it also includes the probability of the other tail. That is the probability of observing a sample with mean 5.29 or more (0.09 or more from the population mean).
Then, we can say that, if the true mean is 5.2, there is a probability P=0.018 of observing a sample of size n=50 with a sample mean with a difference bigger than 0.09 from the population mean of the null hypothesis (5.11 or less or 5.29 or more).
The right answer is e.
Answer:
Incomplete question
Step-by-step explanation:
Answer:
1/6
Step-by-step explanation:
Let x represent the proportion of time the third person spends on the project. You want ...
1/2 + 1/3 + x = 1 . . . . full-time equivalents
5/6 + x = 1 . . . . simplify
x = 1/6 . . . . . . . subtract 5/6
The third person should budget 1/6 of their time to the project.
Hello there! The missing y-values are 12, 14, and 16.
Given all our x-values and two additional y-values, we can see that multiplying the x-value by 2 gives us the y-value. This is shown when x is 5 and 9, because multiplying 5 by 2 gave us 10, and multiplying 9 by 2 gave us 18. Because of this rule, we can multiply each given x-value by 2 to receive our y-value. Once solving, we also notice that the y-values all add by 2 to get the next factor as the data number increases. Hope this helps!
