A^2+ 12^2 = 15^2
a^2 + 144 = 225
225 - 144 = 81
square root of 81 = 9
a = 9
a^2 + 144 = 225
225 - 144 = 81
square root of 81 = 9
a = 9
Simplify!! 2y^2 • 3x
2 answers:
You might be interested in
Answer:
If we compare the p value and the significance level for example we see that
so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can conclude that the mean annual premium in Pennsylvania is significantly lower than the national mean annual premium of 1503 at 5% of significance.
Step-by-step explanation:
1) Data given and notation
represent the mean annual premium value for the sample
represent the sample standard deviation for the sample
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)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean annual premium in Pennsylvania is lower than the national mean annual premium, 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".
Calculate the statistic
We can replace in formula (1) the info given like this:
P-value
The first step is calculate the degrees of freedom, on this case:
Since is a one side left tailed test the p value would be:
Conclusion
If we compare the p value and the significance level for example we see that
so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can conclude that the mean annual premium in Pennsylvania is significantly lower than the national mean annual premium of 1503 at 5% of significance.