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2 years ago

The charts show the effect of an interest rate adjustment on an adjustable-rate mortgage.

Mathematics

1 answer:

Tcecarenko [31]2 years ago

3 0

Answer:

C. a higher monthly payment.

Step-by-step explanation:

<h3>Answer options:</h3>

Rate adjustment resulted in

<u>A. a lower interest rate. </u>

No, interest rate increased from 0.4 to 0.5 percent

<u>B. a higher principal. </u>

No, principal decreased from $200000 to $187000

<u>C. a higher monthly payment. </u>

Yes, monthly payment increased from $1059.85 to $1162.56

<u>D. a higher number of payments.</u>

No, number of payments remained as 336

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A shoe manufacturer compared material A and material B for the soles of shoes. Twelve volunteers each got two shoes. The left wa

sveta [45]

Answer:

a) Are dependent since we are mesuring at the same individuals but on different times and with a different method

b) If we see the qq plot we don't have any significant deviation for the values and we don't have any heavy tail so we can conclude that we can approximate the differences with the normal distribution.

c) $p_v =P(t_{(12)}>0.969) =0.353$

So the p value is higher than the significance level given, so then we can conclude that we FAIL to reject the null hypothesis. So we can conclude that the mean differences is NOT significantly different from 0 .

Step-by-step explanation:

A paired t-test is used to compare two population means where you have two samples in which observations in one sample can be paired with observations in the other sample. For example if we have Before-and-after observations (This problem) we can use it.

The Q-Q plot, or quantile-quantile plot, "is a graphical tool to help us assess if a set of data plausibly came from some theoretical distribution such as a Normal or exponential".

Let put some notation

x=value for A , y = value for B

A: 379, 378, 328, 372, 325, 304, 356, 309, 354, 318, 355, 392

B: 372, 376, 328, 368, 283, 252, 369, 321, 379, 303, 328, 411

(a) Are the two samples paired or independent? Explain your answer.

Are dependent since we are mesuring at the same individuals but on different times and with a different method

(b) Make a normal QQ plot of the differences within each pair. Is it reasonable to assume a normal population of differences?

The first step is calculate the difference $d_i=A_i-B_i$ and we obtain this:

d: 7,2,0,4,42,52,-13,-12,-25,15,27,-19

In order to do the qqplot we can use the following R code:

d<-c(7,2,0,4,42,52,-13,-12,-25,15,27,-19)

qqnorm(d)

And the graph obtained is attached.

If we see the qq plot we don't have any significant deviation for the values and we don't have any heavy tail so we can conclude that we can approximate the differences with the normal distribution.

(c) Choose a test appropriate for the hypotheses above and justify your choice based on your answers to parts (a) and (b). Perform the test by computing a p-value, make a test decision, and state your conclusion in the context of the problem

The system of hypothesis for this case are:

Null hypothesis: $\mu_A- \mu_B = 0$

Alternative hypothesis: $\mu_A -\mu_B \neq 0$

The second step is calculate the mean difference

$\bar d= \frac{\sum_{i=1}^n d_i}{n}= \frac{80}{12}=6.67$

The third step would be calculate the standard deviation for the differences, and we got:

$s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =23.849$

The 4 step is calculate the statistic given by :

$t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{6.67 -0}{\frac{23.849}{\sqrt{12}}}=0.969$

The next step is calculate the degrees of freedom given by:

$df=n-1=12-1=11$

Now we can calculate the p value, since we have a two tailed test the p value is given by:

$p_v =P(t_{(12)}>0.969) =0.353$

4 0

2 years ago

PLEASE HELP ME!

algol13

Step-by-step explanation:

$1.\sum_{i=1}^{5}3i$

The simplest method is "brute force". Calculate each term and add them up.

∑ = 3(1) + 3(2) + 3(3) + 3(4) + 3(5)

∑ = 3 + 6 + 9 + 12 + 15

∑ = 45

$2.\sum_{k=1}^{4}(2k)^{2}$

∑ = (2×1)² + (2×2)² + (2×3)² + (2×4)²

∑ = 4 + 16 + 36 + 64

∑ = 120

$3.\sum_{k=3}^{6}(2k-10)$

∑ = (2×3−10) + (2×4−10) + (2×5−10) + (2×6−10)

∑ = -4 + -2 + 0 + 2

∑ = -4

4. 1 + 1/4 + 1/16 + 1/64 + 1/256

This is a geometric sequence where the first term is 1 and the common ratio is 1/4. The nth term is:

a = 1 (1/4)ⁿ⁻¹

So the series is:

$\sum_{j=1}^{7}(\frac{1}{4})^{j-1}$

5. -5 + -1 + 3 + 7 + 11

This is an arithmetic sequence where the first term is -5 and the common difference is 4. The nth term is:

a = -5 + 4(n−1)

a = -5 + 4n − 4

a = 4n − 9

So the series is:

$\sum_{j=1}^{5}(4j-9)$

5 0

2 years ago

the area of a Blackboard is 1 1/3 square yards posters area is 8/9 square yards find the unit rate and explain in words with uni

Rzqust [24]

The answer is 143 square feet

3 0

2 years ago

Read 2 more answers

Let Y denote a geometric random variable with probability of success p. a Show that for a positive integer a, P(Y > a) = qa .

lakkis [162]

Answer:

a) For this case we can find the cumulative distribution function first:

$F(k) = P(Y \leq k) = \sum_{k'=1}^k P(Y =k')= \sum_{k'=1}^k p(1-p)^{k'-1}= 1-(1-p)^k$

So then by the complement rule we have this:

$P(Y>a) = 1-F(a)= 1- [1-(1-p)^a]= 1-1 +(1-p)^a = (1-p)^a = q^a$

b) $P(Y>a)= q^a$

$P(Y>b) = q^b$

So then we have this using independence:

$P(Y> a+b) = q^{a+b}$

We want to find the following probability:

$P(Y> a+b |Y>a)$

Using the definition of conditional probability we got:

$P(Y> a+b |Y>a)= \frac{P(Y> a+b \cap Y>a)}{P(Y>a)} = \frac{P(Y>a+b)}{P(Y>a)} = \frac{q^{a+b}}{q^a} = q^b = P(Y>b)$

And we see that if a = 2 and b=5 we have:

$P(Y> 2+5 | Y>2) = P(Y>5)$

c) For this case we use independent identical and with the same distribution experiments.

And the result for part b makes sense since we are interest in find the probability that the random variable of interest would be higher than an specified value given another condition with a value lower or equal.

Step-by-step explanation:

Previous concepts

The geometric distribution represents "the number of failures before you get a success in a series of Bernoulli trials. This discrete probability distribution is represented by the probability density function:"

$P(X=x)=(1-p)^{x-1} p$

If we define the random of variable Y we know that:

$Y\sim Geo (1-p)$

Part a

For this case we can find the cumulative distribution function first:

$F(k) = P(Y \leq k) = \sum_{k'=1}^k P(Y =k')= \sum_{k'=1}^k p(1-p)^{k'-1}= 1-(1-p)^k$

So then by the complement rule we have this:

$P(Y>a) = 1-F(a)= 1- [1-(1-p)^a]= 1-1 +(1-p)^a = (1-p)^a = q^a$

Part b

For this case we can use the result from part a to conclude that:

$P(Y>a)= q^a$

$P(Y>b) = q^b$

So then we have this assuming independence:

$P(Y> a+b) = q^{a+b}$

We want to find the following probability:

$P(Y> a+b |Y>a)$

Using the definition of conditional probability we got:

$P(Y> a+b |Y>a)= \frac{P(Y> a+b \cap Y>a)}{P(Y>a)} = \frac{P(Y>a+b)}{P(Y>a)} = \frac{q^{a+b}}{q^a} = q^b = P(Y>b)$

And we see that if a = 2 and b=5 we have:

$P(Y> 2+5 | Y>2) = P(Y>5)$

Part c

For this case we use independent identical and with the same distribution experiments.

8 0

2 years ago

A man bought a mobile phone for $800 and sold it for $1000. What was his profit as a percentage of the cost price

bija089 [108]

Answer:

1000/800 = 125% profit

5 0

1 year ago

Read 2 more answers