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.
9514 1404 393
Answer:
2/3
Step-by-step explanation:
The ratio of any corresponding coordinates will give the scale factor.
Using the x-coordinates of C' and C, we have ...
scale factor = C'/C = 4/6 = 2/3
The dilation scale factor is 2/3.
Answer:
The cost of each cavity filling was $ 134.60.
Step-by-step explanation:
Given that the total cost of Anfa's trip to the dentist was $ 628.35, and she paid a flat fee of $ 89.95 which included the checkup: cleaning and then had 4 cavities filled, each of which cost the same amount, to determine which shows the correct equation and value of x, the cost of each cavity filling, the following calculation must be performed:
(628.35 - 89.95) / 4 = X
538.4 / 4 = X
134.6 = X
Therefore, the cost of each cavity filling was $ 134.60.
Answer:
2 out of 100, 2, 0.02 out of 100, and 0.02
Step-by-step explanation:
