Martin lifted weights in the gym, Monday thru Friday, for 20 weeks. each day he did 25 repetitions. how many repetitions did he
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47 -14i
You can work this out in the straight-forward way, or you can recognize that (6-i) is a common factor. In the latter case, you have ...
... = (6-i)(5 + 3-i)
... = (6 -i)(8 -i)
This product of binomials is found in the usual way. Each term of one factor is multiplied by each term of the other factor and the results summed. Of course, i = √-1, so i² = -1.
... = 6·8 -6i -8i +i²
... = 48 -14i -1
... =
_____
A suitable graphing calculator will work these complex number problems easily.
Answer:
a) For this case the value of the significanceis and
, we need a value on the normal standard distribution thataccumulates 0.025 of the area on each tail and we got:
If the calculated statistic we can reject the null hypothesis at 5% of significance
b) Where
c)
d) Since the calculated value satisfy this condition 2.708>1.96 we have enough evidence at 5% of significance that we have a significant difference between the two proportions analyzed.
Step-by-step explanation:
Data given and notation
represent the number of people with the characteristic 1
represent the number of people with the characteristic 2
sample 1 selected
sample 2 selected
represent the proportion estimated for the sample 1
represent the proportion estimated for the sample 2
represent the pooled estimate of p
z would represent the statistic (variable of interest)
represent the value for the test (variable of interest)
significance level given
Concepts and formulas to use
We need to conduct a hypothesis in order to check if is there is a difference between the two proportions, the system of hypothesis would be:
Null hypothesis:
Alternative hypothesis:
We need to apply a z test to compare proportions, and the statistic is given by:
(1)
a.State the decision rule.
For this case the value of the significanceis and
, we need a value on the normal standard distribution thataccumulates 0.025 of the area on each tail and we got:
If the calculated statistic we can reject the null hypothesis at 5% of significance
b. Compute the pooled proportion.
Where
c. Compute the value of the test statistic.
z-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.
Replacing in formula (1) the values obtained we got this:
d. What is your decision regarding the null hypothesis?
Since the calculated value satisfy this condition 2.708>1.96 we have enough evidence at 5% of significance that we have a significant difference between the two proportions analyzed.