One math student John can solve 6 math problems in 20 minutes while another student Jessica can solve the same6 problems at a ra
1 answer:
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Answer:
a) Null hypothesis:
Alternative hypothesis:
b)
For this case the p value is given
If we compare the p value and the significance level given we see that
so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 15 at 5% of signficance.
c) Type I error, also known as a “false positive” is the error of rejecting a null hypothesis when it is actually true. Can be interpreted as the error of no reject an alternative hypothesis when the results can be attributed not to the reality.
So for this case a type I of error would be reject the hypothesis that the true mean is less or equal than 15 and is actually true.
Step-by-step explanation:
Data given and notation
represent the sample mean
represent the sample standard deviation
sample size
represent the value that we want to test
represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
represent the p value for the test (variable of interest)
Part a: State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the true mean for vacation days is higher than 15, the system of hypothesis would be:
Null hypothesis:
Alternative hypothesis:
If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:
(1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
Part b: P-value and conclusion
The first step is calculate the degrees of freedom, on this case:
For this case the p value is given
Conclusion
If we compare the p value and the significance level given we see that
so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 15 at 5% of signficance.
Part c
Type I error, also known as a “false positive” is the error of rejecting a null hypothesis when it is actually true. Can be interpreted as the error of no reject an alternative hypothesis when the results can be attributed not to the reality.
So for this case a type I of error would be reject the hypothesis that the true mean is less or equal than 15 and is actually true.
X² - 3x - 10
x² = x * x
10 can be factored:
1 x 10 ⇒⇒⇒ difference between factors = 9
2 x 5 ⇒⇒⇒ difference between factors = 3 ⇒ (The correct pair)
(x-5)(x+2) = x(x+2) - 5(x+2)
= x² + 2x - 5x - 10
= x² - 3x - 10
∴ x - 5 = 0 ⇒⇒⇒ x = 5
x + 2 = 0 ⇒⇒⇒ x = -2
I have attached <span>algebra tiles for the given equation.</span>
x² = x * x
10 can be factored:
1 x 10 ⇒⇒⇒ difference between factors = 9
2 x 5 ⇒⇒⇒ difference between factors = 3 ⇒ (The correct pair)
(x-5)(x+2) = x(x+2) - 5(x+2)
= x² + 2x - 5x - 10
= x² - 3x - 10
∴ x - 5 = 0 ⇒⇒⇒ x = 5
x + 2 = 0 ⇒⇒⇒ x = -2
I have attached <span>algebra tiles for the given equation.</span>