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:
47 very easy
Step-by-step explanation:
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Answer:
y = - 10
Step-by-step explanation:
Given that y varies directly with x then the equation relating them is
y = kx ← k is the constant of variation
To find k use the condition y = 2 when x = - 4, that is
2 = - 4k ( divide both sides by - 4 )
= k, that is
k = -
y = - x ← equation of variation
When x = 20, then
y = - × 20 = - 10
First one C second one D sorry if I’m wrong
Answer:
Not sure 737
Step-by-step explanation:
