I need help in math please I am not able to figure these out, I will give you the brainlest and thank you.

Suppose each of the following data sets is a simple random sample from some population. For each dataset, make a normal QQ plot.

adell [148]

Answer:

a) For this case the histogram is not too skewed and we can say that is approximately symmetrical so then we can conclude that this dataset is similar to a normal distribution

b) For this case the data is skewed to the left and we can't assume that we have the normality assumption.

c) This last case the histogram is not symmetrical and the data seems to be skewed.

Step-by-step explanation:

For this case we have the following data:

(a)data = c(7,13.2,8.1,8.2,6,9.5,9.4,8.7,9.8,10.9,8.4,7.4,8.4,10,9.7,8.6,12.4,10.7,11,9.4)

We can use the following R code to get the histogram

> x1<-c(7,13.2,8.1,8.2,6,9.5,9.4,8.7,9.8,10.9,8.4,7.4,8.4,10,9.7,8.6,12.4,10.7,11,9.4)

> hist(x1,main="Histogram a)")

The result is on the first figure attached.

For this case the histogram is not too skewed and we can say that is approximately symmetrical so then we can conclude that this dataset is similar to a normal distribution

(b)data = c(2.5,1.8,2.6,-1.9,1.6,2.6,1.4,0.9,1.2,2.3,-1.5,1.5,2.5,2.9,-0.1)

> x2<- c(2.5,1.8,2.6,-1.9,1.6,2.6,1.4,0.9,1.2,2.3,-1.5,1.5,2.5,2.9,-0.1)

> hist(x2,main="Histogram b)")

The result is on the first figure attached.

For this case the data is skewed to the left and we can't assume that we have the normality assumption.

(c)data = c(3.3,1.7,3.3,3.3,2.4,0.5,1.1,1.7,12,14.4,12.8,11.2,10.9,11.7,11.7,11.6)

> x3<-c(3.3,1.7,3.3,3.3,2.4,0.5,1.1,1.7,12,14.4,12.8,11.2,10.9,11.7,11.7,11.6)

> hist(x3,main="Histogram c)")

The result is on the first figure attached.

This last case the histogram is not symmetrical and the data seems to be skewed.

7 0

3 years ago

Write an equation to represent the direct proportional relationship between the amount of gratuity (9) received by a room servic

AleksandrR [38]

Oh wait this is confusing.. i think somebody asked this question and got it answered

so search it up on here

3 0

3 years ago

Read 2 more answers

Evaluate 5y + 4x when y=10 and x= -5

salantis [7]

5y+4x

5(10) + 4(-5)

50 - 20

30

4 0

3 years ago

Read 2 more answers

Emily's bill for dinner at a restaurant was $61. She left an 18% tip. What was the amount of the tip?

UkoKoshka [18]

You cross multiply. So 61 over 18 = x over 100. You multiply 61 and get the answer of 1600. Now with what you have left, you divide that with 18 and you should get your answer of 338.888889. Make sure to round it to whatever it tells you to round to!

7 0

3 years ago

Assume that a cross is made between a heterozygous tall pea plant and a homozygous short pea plant. Fifty offspring are produced

Butoxors [25]

Answer:

1) $\chi^2 = \frac{(30-25)^2}{25} +\frac{(20-25)^2}{25} =1 +1 =2$

2) $df = c-1 = 2-1$

Where c represent the number of categories c=2

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".

Assume the following dataset:

Tall =30 , Short =20

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

H0: The deviations from a 1:1 ratio (25 tall and 25 short) are due to chance.

H1: The deviations from a 1:1 ratio (25 tall and 25 short) are NOT due to chance.

Part 1

So then we know that the expected values would be 25 for each case

The statistic to check the hypothesis is given by:

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

And if we replace we got:

$\chi^2 = \frac{(30-25)^2}{25} +\frac{(20-25)^2}{25} =1 +1 =2$

Part 2

For this case the degreed of freedom are given by:

$df = c-1 = 2-1$

Where c represent the number of categories c=2

And we can calculate the p value given by:

$p_v = P(\chi^2_{1} >2)=0.157$

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

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

7 0

3 years ago