Is this congruent by SSS, SAS, AAS, ASA? :)

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

Crop researchers plant 15 plots with a new variety of corn. The yields in bushels per acre are: 138.0 139.1 113.0 132.5 140.7 10

son4ous [18]

There is a relationship between confidence interval and standard deviation:
$\theta=\overline{x} \pm \frac{z\sigma}{\sqrt{n}}$
Where $\overline{x}$ is the mean, $\sigma$ is standard deviation, and n is number of data points.
Every confidence interval has associated z value. This can be found online.
We need to find the standard deviation first:
$\sigma=\sqrt{\frac{\sum(x-\overline{x})^2}{n}$
When we do all the calculations we find that:
$\overline{x}=123.8\\ \sigma=11.84$
Now we can find confidence intervals:
$($90\%,z=1.645): \theta=123.8 \pm \frac{1.645\cdot 11.84}{\sqrt{15}}=123.8 \pm5.0\\($95\%,z=1.960): \theta=123.8 \pm \frac{1.960\cdot 11.84}{\sqrt{15}}=123.8 \pm 5.99\\ ($99\%,z=2.576): \theta=123.8 \pm \frac{2.576\cdot 11.84}{\sqrt{15}}=123.8 \pm 7.87\\$
We can see that as confidence interval increases so does the error margin. Z values accociated with each confidence intreval also get bigger as confidence interval increases.
Here is the link to the spreadsheet with standard deviation calculation:
https://docs.google.com/spreadsheets/d/1pnsJIrM_lmQKAGRJvduiHzjg9mYvLgpsCqCoGYvR5Us/edit?usp=sharing

6 0

3 years ago

The diameter of a circle is 10 cm. Find its area in terms of \piπ.

Salsk061 [2.6K]

Answer:

I probably need like a picture or example

7 0

2 years ago

PLS HELP FAST. CORRECT ANSWER WILL BE MARKED BRAINLIEST

Marianna [84]

Answer:

it a or the first one or whatevr u wanna call it jsut #1

Step-by-step explanation:

7 0

3 years ago

a study was to be undertaken to determine if a particular training program would improve physical fitness. an srs of 31 universi

Nataliya [291]

Step-by-step explanation:

A study was to be undertaken to determine if a particular training program would improve physical fitness. A sample of 31 university students was selected to be enrolled in the fitness program. The researchers wished to determine if there was evidence that their sample of students differed from the general population of untrained subjects. The sample mean is 47.4 and a standard deviation of 5.3. The 98% confidence interval is determined and is given as, (45.2, 49.6) .

If the level of confidence is changed to 95%, then the confidence interval will become shorter but the p-value will not change because it is calculated using the test statistic. So the correct answer is (a).

3 0

1 year ago