What is the median?<br> 8 1 69 3 6 3

A restaurant owner is interested in the proportion of his customers who order dessert. He looks at 65 randomly selected receipts

xxTIMURxx [149]

Answer:

c. Population

Step-by-step explanation:

All customers who come to the restaurant are the population.

The 65 randomly selected customers are the sample.

The correct answer is:

c. Population

4 0

3 years ago

Find the greatest common factor of x^2 and 15xy. A. 1 B. 15 C. x D. 15xy

Sindrei [870]

C. x is the greatest common factor because it is the only term from these answer choices that is a factor of both terms.

5 0

3 years ago

Read 2 more answers

A farmer is going to divide her 30 acre farm between two crops. Seed for corn costs $15 per acre. Seed for wheat costs $30 per a

Deffense [45]

Answer:

$3200

Step-by-step explanation:

The total area of land is 30 acre. Let x represent the area in acre that corn is planted and y represent the area in acre that wheat is planted. Therefore:

x + y = 30 (1)

Also, the farmer can spend at most $750 on seeds. Corn cost $15 per acre, maize cost $30 per acre. Hence:

15x + 30y ≤ 750 (2)

Also, x and y > 0

The profit is given by the function:

Profit = 100x + 110y

Graphing equation 1 and equation 2 and solving, we get the points:

(10, 20)

Hence the maximum profit is at:

Max profit = 100(10) + 110(20) = $3200

8 0

3 years ago

What’s the correct answer

SVETLANKA909090 [29]

Answer: C. 1998, 2000, 2002

7 0

3 years ago

Read 2 more answers

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

What is the median? 8 1 69 3 6 3