Write a decimal in tenths that is less than 2.42 but greater than 2.0

Health insurance benefits vary by the size of the company (the Henry J. Kaiser Family Foundation website, June 23, 2016). The sa

xxMikexx [17]

Answer:

$\chi^2 = \frac{(32-42)^2}{42}+\frac{(18-8)^2}{8}+\frac{(68-63)^2}{63}+\frac{(7-12)^2}{12}+\frac{(89-84)^2}{84}+\frac{(11-16)^2}{16}=19.221$

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:

$p_v = P(\chi^2_{2} >19.221)=0.000067$

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

"=1-CHISQ.DIST(19.221,2,TRUE)"

Since the p values is higher than a significance level for example $\alpha=0.05$ , we can reject the null hypothesis at 5% of significance, and we can conclude that the two variables are dependent at 5% of significance.

Step-by-step explanation:

Previous concepts

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".

Solution to the problem

Assume the following dataset:

Size Company/ Heal. Ins. Yes No Total

Small 32 18 50

Medium 68 7 75

Large 89 11 100

_____________________________________

Total 189 36 225

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

H0: independence between heath insurance coverage and size of the company

H1: NO independence between heath insurance coverage and size of the company

The statistic to check the hypothesis is given by:

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

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:

$E_{1} =\frac{50*189}{225}=42$

$E_{2} =\frac{50*36}{225}=8$

$E_{3} =\frac{75*189}{225}=63$

$E_{4} =\frac{75*36}{225}=12$

$E_{5} =\frac{100*189}{225}=84$

$E_{6} =\frac{100*36}{225}=16$

And the expected values are given by:

Size Company/ Heal. Ins. Yes No Total

Small 42 8 50

Medium 63 12 75

Large 84 16 100

_____________________________________

Total 189 36 225

And now we can calculate the statistic:

$\chi^2 = \frac{(32-42)^2}{42}+\frac{(18-8)^2}{8}+\frac{(68-63)^2}{63}+\frac{(7-12)^2}{12}+\frac{(89-84)^2}{84}+\frac{(11-16)^2}{16}=19.221$

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:

$p_v = P(\chi^2_{2} >19.221)=0.000067$

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

"=1-CHISQ.DIST(19.221,2,TRUE)"

Since the p values is higher than a significance level for example $\alpha=0.05$ , we can reject the null hypothesis at 5% of significance, and we can conclude that the two variables are dependent at 5% of significance.

3 0

3 years ago

A worker has four different job offers, each with a contract for six years. Assuming the job descriptions are identical,

Phoenix [80]

9514 1404 393

Answer:

$48,000 to start, with 6% raises

Step-by-step explanation:

The starting salaries are within a few percentage points of each other, so the majority of the difference in earnings will come because of the raises. The offer with the largest raise percentage is likely the best.

The earnings total can be figured as the sum of a geometric series with a first term of "starting salary" and a growth factor of (1+raise percentage). For starting salary 's' and raise percentage 'r', the total earnings in 6 years will be ...

S = s((1+r)^6 -1)/r

Here are the total earnings, in thousands, for each of the offers:

a) s = 49, r = .03, S = 316.95

b) s = 51, r = .01, S = 313.75

c) s = 48, r = .06, S = 334.82 . . . . best offer

d) s = 50, r = .02, S = 315.41

The worker can earn the most from a starting salary of $48,000 and 6% increases each year.

7 0

2 years ago

Read 2 more answers

State the domain and range of the following relation<br> x^2+y^2=16

Vlada [557]

The domain and range is [-4, 4] and [0, 4]

<h3 /><h3>What is Domain and range?</h3>

The domain of a function is the set of values that we are allowed to plug into our function.

The range of a function is the set of values that the function assumes.

x² + y² = 16

y = √16 - x²

For domain under root should not be negative quantity,

16 - x²≥0

16≥x²

So, x≤4 or x≥4

Thus, the domain is [-4, 4]

Range:

y is maximum at x=0, y=4

y is minimum at x=4, y=0

Thus, range = [0, 4]

Learn more about domain and range here:

brainly.com/question/12751831

#SPJ1

8 0

1 year ago

2/5da80% como ahora la cuenta

andreev551 [17]

I don't speak Spanish but I think that this is what you meant:
80 * 2/5 = 32
so 2/5 of 80 % is 32 % (80/5 * 2)

4 0

3 years ago

Write an equation for the line that is parallel to the given line and that passes through the given point. y = –6x + 2; (–1, 2)

kap26 [50]

Parallel=same slope
Y=mx+b
Sub in the slope and point to solve for b
2=-6(-1)+b
2=6+b
2-6=b
-4=b
The equation is y=-6x-4; the answer is B
Hope this helps!

6 0

3 years ago