If I have this sequence 2, 4, 6, ... I can describe the rule as

(1 point) Suppose we want a 95% confidence interval for the average amount spent on books by freshmen in their first year at col

Ne4ueva [31]

Answer:

a) $n=(\frac{1.960(16)}{5})^2 =39.33 \approx 40$

So the answer for this case would be n=40 rounded up to the nearest integer

b) For this case if we see the formula for the margin of error

$ME=z_{\alpha/2}\frac{s}{\sqrt{n}}$ (a)

We can see that the margin of error is inversely proportional to the sample size so if we want a samller margin of error we need a LARGER sample

Answer: LARGER

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X$ represent the sample mean for the sample

$\mu$ population mean

$\sigma=16$ represent the population standard deviation

n represent the sample size

Solution to the problem

Part a

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{s}{\sqrt{n}}$ (a)

And on this case we have that ME =5 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025;0;1)", and we got $z_{\alpha/2}=1.960$ , replacing into formula (b) we got:

$n=(\frac{1.960(16)}{5})^2 =39.33 \approx 40$

So the answer for this case would be n=40 rounded up to the nearest integer

Part b

For this case if we see the formula for the margin of error

$ME=z_{\alpha/2}\frac{s}{\sqrt{n}}$ (a)

We can see that the margin of error is inversely proportional to the sample size so if we want a samller margin of error we need a LARGER sample

Answer: LARGER

5 0

2 years ago

I need answers I’ll cash app 10 dollars

posledela

1. 0 > 3x - 3 - 6

0 > 3x -9

9 > 3x

3 > x

It would be a closed circle starting from 3 going to the right.

2. 4x + 1 - 1 >/ -8

4x >/ -8

x >/ -2

It would be an open circle starting from -2 going to the right.

3. -1 </ 2n + 4 - 5

-1 </ 2n -1

0 </ 2n

Undefined.

4. -6 > 5n + 5 + 4

-6 > 5n + 9

-15 > 5n

-3 > n

It would be a closed circle going to the right

4 0

2 years ago

Some scientists believe alcoholism is linked to social isolation. One measure of social isolation is marital status. A study of

frez [133]

Answer:

1) H0: There is independence between the marital status and the diagnostic of alcoholic

H1: There is association between the marital status and the diagnostic of alcoholic

2) The statistic to check the hypothesis is given by:

$\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

3) $\chi^2 = \frac{(21-33.143)^2}{33.143}+\frac{(37-41.429)^2}{41.429}+\frac{(58-41.429)^2}{41.429}+\frac{(59-46.857)^2}{46.857}+\frac{(63-58.571)^2}{58.571}+\frac{(42-58.571)^2}{58.571} =19.72$

4) $df=(rows-1)(cols-1)=(3-1)(2-1)=2$

And we can calculate the p value given by:

$p_v = P(\chi^2_{2} >19.72)=5.22x10^{-5}$

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(19.72,2,TRUE)"

Since the p value is lower than the significance level so then we can reject the null hypothesis at 5% of significance, and we can conclude that we have association between the two variables analyzed.

Step-by-step explanation:

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Assume the following dataset:

Diag. Alcoholic Undiagnosed Alcoholic Not alcoholic Total

Married 21 37 58 116

Not Married 59 63 42 164

Total 80 100 100 280

Part 1

We need to conduct a chi square test in order to check the following hypothesis:

H0: There is independence between the marital status and the diagnostic of alcoholic

H1: There is association between the marital status and the diagnostic of alcoholic

The level os significance assumed for this case is $\alpha=0.05$

Part 2

The statistic to check the hypothesis is given by:

$\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

Part 3

The table given represent the observed values, we just need to calculate the expected values with the following formula $E_i = \frac{total col * total row}{grand total}$

And the calculations are given by:

$E_{1} =\frac{80*116}{280}=33.143$

$E_{2} =\frac{100*116}{280}=41.429$

$E_{3} =\frac{100*116}{280}=41.429$

$E_{4} =\frac{80*164}{280}=46.857$

$E_{5} =\frac{100*164}{280}=58.571$

$E_{6} =\frac{100*164}{280}=58.571$

And the expected values are given by:

Diag. Alcoholic Undiagnosed Alcoholic Not alcoholic Total

Married 33.143 41.429 41.429 116

Not Married 46.857 58.571 58.571 164

Total 80 100 100 280

And now we can calculate the statistic:

$\chi^2 = \frac{(21-33.143)^2}{33.143}+\frac{(37-41.429)^2}{41.429}+\frac{(58-41.429)^2}{41.429}+\frac{(59-46.857)^2}{46.857}+\frac{(63-58.571)^2}{58.571}+\frac{(42-58.571)^2}{58.571} =19.72$

Part 4

Now we can calculate the degrees of freedom for the statistic given by:

$df=(rows-1)(cols-1)=(3-1)(2-1)=2$

And we can calculate the p value given by:

$p_v = P(\chi^2_{2} >19.72)=5.22x10^{-5}$

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(19.72,2,TRUE)"

Since the p value is lower than the significance level so then we can reject the null hypothesis at 5% of significance, and we can conclude that we have association between the two variables analyzed.

7 0

3 years ago

Photo attached please help

aniked [119]

C and E that is the answer

4 0

2 years ago

Use the Distributive Property to expand 5(3x-8y)-4

lubasha [3.4K]

Answer:

it should be 60x-160y

Step-by-step explanation: