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Sophie [7]
3 years ago
13

Which function is graphed below?

Mathematics
1 answer:
kupik [55]3 years ago
7 0

Answer:

f(x)=cos(x)

The amplitude is equal to 1. The period is 2π.

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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
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