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3 years ago

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the claim that 20% of americans know their credit score is tested. suppose that the sample evidence causes us to fail to reject

the null hypothesis claim that 20% of americans know their credit score. state the final conclusion in simple non technical terms

1 answer:

kipiarov [429]3 years ago

5 0

Answer:

Conclusion : People ≠ 20% don't know about their credit score.

Step-by-step explanation:

Hypothesis is testing a statement for its statistical significance.

Null Hypothesis (H0) implies 'no difference from tested value', Alternate hypothesis (H1) implies 'significant difference from tested value'

Let % of people knowing their credit score = CS

H0 : CS = 20

H1 : CS ≠ 20

If the null hypothesis is rejected, it implies that we reject the claim that CS i.e '% of americans knowing their credit score = 20%'. So, the alternate hypothesis is accepted, i.e we conclude that '% americans knowing their credit score ≠ 20%'.

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The function in Exercise represents the rate of flow of money in dollars per year. Assume a 10-year period at 8% compounded cont

anzhelika [568]

Answer:

a) The present value is 688.64 $

b) The accumulated amount is 1532.60 $

Step-by-step explanation:

<u>a)</u><u> The preset value equation is given by this formula:</u>

$P=\int^{T}_{0}f(t)e^{-rt}dt$

where:

T is the period in years (T = 10 years)
r is the annual interest rate (r=0.08)

So we have:

$P=\int^{T}_{0}(0.01t+100)e^{-rt}dt$

Now we just need to solve this integral.

$P=\int^{T}_{0}0.01te^{-rt}dt+\int^{T}_{0}100e^{-rt}dt$

$P=e^{-0.08t}(-1.56-0.13t)|^{10}_{0}+1250e^{-0.08t}|^{10}_{0}$

$P=0.30+688.34=688.64 $$

The present value is 688.64 $

<u>b)</u><u> The accumulated amount of money flow formula is:</u>

$A=e^{r\tau}\int^{T}_{0}f(t)e^{-rt}dt$

We have the same equation but whit a term that depends of τ, in our case it is 10.

So we have:

$A=e^{r\tau}\int^{T}_{0}(0.01t+100)e^{-rt}dt=e^{0.08\cdot 10}P$

$A=e^{0.08\cdot 10}688.64=1532.60 $$

The accumulated amount is 1532.60 $

Have a nice day!

6 0

3 years ago

Please help me asap<br><br><br> ty :)

Art [367]

Answer:

y = 5x+7

Step-by-step explanation:

Parallel lines have the same slope so y = 5x+9 is in slope intercept form

y = mx+b where m is the slope and b is the y intercept

The slope is 5

The new line will have a slope of 5

y = 5x+b

We know a point on the line (-2,-3)

Substitute this point into the equation

-3 = 5(-2)+b

-3 = -10+b

-3+10 = b

7 = b

y = 5x+7

3 0

2 years ago

Read 2 more answers

How do i solve this problem ?

Usimov [2.4K]

N is 7.5 because prq is 3x the size of abc

8 0

3 years ago

PLLLZ HELP I'LL GIVE BRAINLISET

Sholpan [36]

Answer:

Step-by-step explanation:

Sandy only because the 2 prisms overlap

5 0

3 years ago

Dannette and Alphonso work for a computer repair company. They must include the time it takes to complete each repair in their r

olga55 [171]

Answer:

(a):

Dannette Alphonso

$\bar x_D = 4.33$ $\bar x_A = 5.17$

$M_D = 2.5$ $M_A = 5$

$\sigma_D = 3.350$ $\sigma_A = 1.951$

$IQR_D = 7$ $IQR_A = 1.5$

(b):

Measure of center: Median

Measure of spread: Interquartile range

(c):

There are no outliers in Dannette's dataset

There are outliers in Alphonso's dataset

Step-by-step explanation:

Given

See attachment for the appropriate data presentation

Solving (a): Mean, Median, Standard deviation and IQR of each

From the attached plots, we have:

IQR_A = 1.5 ---- Dannette

$A = \{3,4,4,4,4,5,5,5,5,6,6,11\}$ ---- Alphonso

n = 12 --- number of dataset

Mean

The mean is calculated

$\bar x = \frac{\sum x}{n}$

So, we have:

$\bar x_D = \frac{1+1+1+1+2+2+3+7+8+8+9+9}{12}$

$\bar x_D = \frac{52}{12}$

$\bar x_D = 4.33$ --- Dannette

$\bar x_A = \frac{3+4+4+4+4+5+5+5+5+6+6+11}{12}$

$\bar x_A = \frac{62}{12}$

$\bar x_A = 5.17$ --- Alphonso

Median

The median is calculated as:

$M = \frac{n + 1}{2}th$

$M = \frac{12 + 1}{2}th$

$M = \frac{13}{2}th$

$M = 6.5th$

This implies that the median is the mean of the 6th and the 7th item.

So, we have:

$M_D = \frac{2+3}{2}$

$M_D = \frac{5}{2}$

$M_D = 2.5$ ---- Dannette

$M_A = \frac{5+5}{2}$

$M_A = \frac{10}{2}$

$M_A = 5$ ---- Alphonso

Standard Deviation

This is calculated as:

$\sigma = \sqrt{\frac{\sum(x - \bar x)^2}{n}}$

So, we have:

$\sigma_D = \sqrt{\frac{(1 - 4.33)^2 +.............+(9- 4.33)^2}{12}}$

$\sigma_D = \sqrt{\frac{134.6668}{12}}$

$\sigma_D = 3.350$ ---- Dannette

$\sigma_A = \sqrt{\frac{(3-5.17)^2+............+(11-5.17)^2}{12}}$

$\sigma_A = \sqrt{\frac{45.6668}{12}}$

$\sigma_A = 1.951$ --- Alphonso

The Interquartile Range (IQR)

This is calculated as:

$IQR =Q_3 - Q_1$

Where

$Q_3 \to$ Upper Quartile and $Q_1 \to$ Lower Quartile

$Q_3$ is calculated as:

$Q_3 = \frac{3}{4}*({n + 1})th$

$Q_3 = \frac{3}{4}*(12 + 1})th$

$Q_3 = \frac{3}{4}*13th$

$Q_3 = 9.75th$

This means that $Q_3$ is the mean of the 9th and 7th item. So, we have:

$Q_3 = \frac{1}{2} * (8+8) = \frac{1}{2} * 16$ $Q_3 = \frac{1}{2} * (5+6) = \frac{1}{2} * 11$

$Q_3 = 8$ ---- Dannette $Q_3 = 5.5$ --- Alphonso

$Q_1$ is calculated as:

$Q_1 = \frac{1}{4}*({n + 1})th$

$Q_1 = \frac{1}{4}*({12 + 1})th$

$Q_1 = \frac{1}{4}*13th$

$Q_1 = 3.25th$

This means that $Q_1$ is the mean of the 3rd and 4th item. So, we have:

$Q_1 = \frac{1}{2}(1+1) = \frac{1}{2} * 2$ $Q_1 = \frac{1}{2}(4+4) = \frac{1}{2} * 8$

$Q_1 = 1$ --- Dannette $Q_1 = 4$ ---- Alphonso

So, the IQR is:

$IQR = Q_3 - Q_1$

$IQR_D = 8 - 1$ $IQR_A = 5.5 - 4$

$IQR_D = 7$ --- Dannette $IQR_A = 1.5$ --- Alphonso

Solving (b): The measures to compare

Measure of center

By observation, we can see that there are outliers is the plot of Alphonso (because 11 is far from the other dataset) while there are no outliers in Dannette plot (as all data are close).

Since, the above is the case; we simply compare the median of both because it is not affected by outliers

Measure of spread

Compare the interquartile range of both, as it is arguably the best measure of spread, because it is also not affected by outliers.

Solving (c): Check for outlier

To check for outlier, we make use of the following formulas:

$Lower =Q_1 - 1.5 * IQR$

$Upper =Q_3 + 1.5 * IQR$

For Dannette:

$Lower = 1 - 1.5 * 7 = -9.5$

$Upper = 8 + 1.5 * 7 = 18.5$

Since, the dataset are all positive, we change the lower outlier to 0.

So, the valid data range are:

$Valid = 0 \to 18.5$

From the question, the range of Dannette's dataset is: 1 to 9. Hence, there are no outliers in Dannette's dataset

For Alphonso:

$Lower = 4 - 1.5 * 1.5 =1.75$

$Upper = 5.5 + 1.5 * 1.5 =7.75$

So, the valid data range are:

$Valid = 1.75\to 7.75$

From the question, the range of Alphonso's dataset is: 3 to 11. Hence, there are outliers in Alphonso's dataset

4 0

3 years ago