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1 answer:
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Answer:
h =
Step-by-step explanation:
Given
A = (a + c)h
Multiply both sides by 2 to clear the fraction
2A = (a + c)h ( divide both sides by (a + c) )
= h
Answer:
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, and the group CTS, NOT have a significant higher mean compared to the Normal group at 1% of significance.
Step-by-step explanation:
1) Data given and notation
represent the mean for the sample CTS
represent the mean for the sample Normal
represent the sample standard deviation for the sample of CTS
represent the sample standard deviation for the sample of Normal
sample size selected for the CTS
sample size selected for the Normal
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 for the group CTS is higher than the mean for the Normal, the system of hypothesis would be:
Null hypothesis:
Alternative hypothesis:
If we analyze the size for the samples both are less than 30 so for this case is better apply a t test to compare means, 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 whether the means of two groups are equal to each other".
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 right tailed test the p value would be:
We can use the following excel code to calculate the p value in Excel:"=1-T.DIST(1.507,15,TRUE)"
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, and the group CTS, NOT have a significant higher mean compared to the Normal group at 1% of significance.