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:
16/3
Step-by-step explanation:
.
Fractions can be converted to decimals and vice versa.
<em>The decimal equivalent of the fraction of months with 31 days is 0.583</em>
Given:
<em />
<em> --- months in a year</em>
<em />
<em> --- months with 31 days</em>
<em />
The fraction (n) of months with 31 days is:
So, we have:
The decimal equivalent is:
Hence, (c) is correct.
Read more about fractions and decimals at:
brainly.com/question/548650
Answer:
5) P (arrive) + P (not arrive) = 100% Complements
80% + P( not arrive) = 100% Substitute thr ralees
P ( not arraive) = 100% - 80% Isolate the needed
P ( not arrive) = 20%
therefor, the probability of not arriving on time is:-
20%, or 0.20, or 1/5
The complements of arriving on time is not arriving on time.
The probability is equal to 100%
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<u>hope it helps..</u>
<u>have a great day!</u>
