Answer:
City A: {2, 3.5, 4, 4, 5, 5.5}
City B: {3.5, 4, 5, 5.5, 6, 6}
Mean is the average.
Average of City A = 4 (2+3.5+4+4+5+5.5 = 24 and 24/6 = 4)
Average of City B = 5 (3.5+4+5+5.5+6+6 = 30 and 30/6 = 5)
Mean Absolute Deviation
Mean Absolute Deviation of City A = 0.8
Mean Absolute Deviation of City B = 0.8
Median is the middle (after putting it in order least to greatest)
Median of City A = 4
Median of City B = 5.25 (5+5.5 = 10.5 and 10.5/2 =5.25)
Step-by-step explanation:
hope this helps!
Answer:
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Step-by-step explanation:
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Answer:
G
Step-by-step explanation:
a relation from a set of inputs to a set of possible outputs where each input is related to exactly one output.
2(9)
How do I solve this and what is the answer
Answer:
Answers are below
Step-by-step explanation:
All you have to do is multiply so,
4(18)
4*18 = 72
2(9)
2*9 = 18
Answer:
4. is x = 9
but I don't know about 5. sorry.
Step-by-step explanation:
5x + 3 = 48
-3. -3
5x = 45
5. 5
x = 9
5(9) + 3 = 48
48 = 48
Reject H0 if tcalc < 1.7960
b. Calculate the Test statistic.
c-1. The null hypothesis should be rejected.
i. TRUE
ii. FALSE
c-2. The average repair time is longer than 5 days.
i. TRUE
ii. FALSE
c-3 At α = .05 is the goal being met?
i. TRUE
ii. FALSE
Answer:
a) Reject H0 if tcalc > 1.7960
b)
c-1) ii. FALSE
c-2) ii.FALSE
c-3)i. TRUE
Step-by-step explanation:
1) Data given and notation
represent the mean time 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)
a) State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean is less than 5 days, the system of hypothesis would be:
Null hypothesis:
Alternative hypothesis:
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".
Rejection zone
On this case we need a critical value that accumulates 0.05 of the area on the right tail. The degrees of freedom are given by 11. And we can use the following excel code to find the critical value : "T.INV(1-0.95,11)" and the critical value would be given by .
And the rejection zone is given by:
Reject H0 if tcalc > 1.7960
b) 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 test the p value would be:
c-1. The null hypothesis should be rejected.
ii. FALSE
c-2. The average repair time is longer than 5 days.
ii. FALSE
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 reject the null hypothesis, and the true mean is not significantly higher than 5.
c-3 At α = .05 is the goal being met?
i. TRUE
We fail to reject the null hypothesis so then the goal is met.
Answer: P/2 - L = W
Step-by-step explanation:
P = 2L + 2W. We are isolating W. Subtract 2L from both sides to isolate the term first
P - 2L = 2W. Divide by 2 on both sides to isolate W.
P/2 - L = W