A teacher wants to see if a new unit on taking square roots is helping students learn. She has five randomly selected students t
2 answers:
Answer:
The P-value for this test is P=0.00006.
Step-by-step explanation:
We have a matched-pair test, with a test statistic t=-14.9.
The degrees of freedom in a sample of 5 students is:
For a t=-14.9 and 4 degrees of freedom, a left-tail test will have a P-value of:
The claim is that the new unit on taking square roots is helping students to learn. This test concludes that there is statistical evidence to support the claim that the new unit is helping students to learn.
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First, lets determine 
(the number of elements in 
).
We can see that the negative even integers that are between -27 and 0 are: -24, -22, -20, -18, -16, -14, -12, -10, -8, -6, -4, and -2.
This means
.
A proper subset of is can't be itself, this means the a proper subset of can't have 13 elements.
Given that re-arranging the elements of any set doesn't yield a different set, we use combinations (binomial coefficients) to get the number of subsets in a particular set.
The number of subsets of with 1 element are given by 
(which means 13 combine 1), in the same way the number of subsets with 2 elements is given by 
and so forth until reaching the number of subsets with 12 elements.
The previous statement can be written as:
In the previous summation we also included the empty subset, which is always a proper subset of.
So is an acceptable answer, but we can simplify it.
The following statement is true (it can be easily proven):
From our previous discussion, this means that the number of subsets of any arbitrary set is equal to 
.
This means that the number of proper subsets (which is the number of subset minus the set itself, for finite subsets) is equal to
.
So, the answer is:
There are 8191 proper subsets of.
We can see that the negative even integers that are between -27 and 0 are: -24, -22, -20, -18, -16, -14, -12, -10, -8, -6, -4, and -2.
This means
A proper subset of
Given that re-arranging the elements of any set doesn't yield a different set, we use combinations (binomial coefficients) to get the number of subsets in a particular set.
The number of subsets of
The previous statement can be written as:
In the previous summation we also included the empty subset, which is always a proper subset of
So
The following statement is true (it can be easily proven):
From our previous discussion, this means that the number of subsets of any arbitrary set
This means that the number of proper subsets (which is the number of subset minus the set itself, for finite subsets) is equal to
So, the answer is:
There are 8191 proper subsets of