PLEASE HELP ASAP! I have reposted this 2 times and no one will help me.:( please answer all 4 thank you so much!! i need this do
ne in an hour!!!
1 answer:
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Answer:
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 FAIL to reject the null hypothesis, so we can conclude that the mean temperature actually its NOT significant less then 22 at 1% of signficance.
(D) P-val=0.021, fail to reject the null hypothesis
Step-by-step explanation:
1) Data given and notation
represent the mean for the temperatures
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)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean is less than 22C, 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
We can calculate the degrees of freedom like this:
Since is a one left tailed test the p value would be:
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 FAIL to reject the null hypothesis, so we can conclude that the mean temperature actually its NOT significant less then 22 at 1% of signficance.
The best option would be:
(D) P-val=0.021, fail to reject the null hypothesis
Answer:
Step-by-step explanation:
Comparing the triangles in the given questions, the following can be deduced:
5. The scale factor = =
≠
So that;
QR is similar to UT, PR is similar to US, but PQ is NOT similar to TS.
Thus, the triangles are NOT similar.
6. Scale factor = =
=
= 1.5
So that;
QR is similar to UT, PR is similar to US, but PQ is similar to TS.
Thus, the triangles are similar by Side-Side-Side (SSS).
7. Here, an exact scale factor can not be determined. Thus, the triangles are NOT similar.