Write an equation of the line that passes through the given point and has the given slope.

An interior automotive supplier places several electrical wires in a harness.Apull test measures the force required to pull spli

oksano4ka [1.4K]

Answer:

a) For this case we can use the following R code to construct the qq plot

> data<-c(28.8, 24.4, 30.1, 25.6, 26.4, 23.9, 22.1, 22.5, 27.6, 28.1, 20.8, 27.7, 24.4, 25.1, 24.6, 26.3, 28.2, 22.2, 26.3, 24.4)

# The above line is in order to store the data in a vector

> qqnorm(data, pch = 1, frame = FALSE)

# The line above is in order to calculate the quantiles from the data assumin Normal distribution

> qqline(data, col = "steelblue", lwd = 2)

# The line above is in order to put a line for the theoretical dsitribution

The result is on the figure attached.

b) For this case as we can see on the figure attached the calculated quantiles are not far from the theorical quantiles given byt the straaigth blue line so then we can conclude that the distribution seems to be normal.

Step-by-step explanation:

For this case we have the following data:

28.8, 24.4, 30.1, 25.6, 26.4, 23.9, 22.1, 22.5, 27.6, 28.1, 20.8, 27.7, 24.4, 25.1, 24.6, 26.3, 28.2, 22.2, 26.3, 24.4

The quantile-quantile or q-q plot is a graphical procedure in order to check the validity of a distributional assumption for a data set. We just need to calculate "the theoretically expected value for each data point based on the distribution in question".

If the values are asusted to the assumed distribution, we will see that "the points on the q-q plot will fall approximately on a straight line"

For this case our distribution assumed is normal.

Part a

For this case we can use the following R code to construct the qq plot

> data<-c(28.8, 24.4, 30.1, 25.6, 26.4, 23.9, 22.1, 22.5, 27.6, 28.1, 20.8, 27.7, 24.4, 25.1, 24.6, 26.3, 28.2, 22.2, 26.3, 24.4)

# The above line is in order to store the data in a vector

> qqnorm(data, pch = 1, frame = FALSE)

# The line above is in order to calculate the quantiles from the data assuming Normal distribution (0,1)

> qqline(data, col = "steelblue", lwd = 2)

# The line above is in order to put a line for the theoretical distribution

The result is on the figure attached.

Part b

For this case as we can see on the figure attached the calculated quantiles are not far from the theorical quantiles given byt the straaigth blue line so then we can conclude that the distribution seems to be normal.

4 0

3 years ago

Can somebody solve this please

blondinia [14]

What’s the question?

4 0

2 years ago

A customer pays $3.27 for oranges and $4.76 for pears. How many pounds of fruit does the customer buy?

ruslelena [56]

Take 3.27 and divide that by 1.09 (for the oranges)
3.27/1.09=3lbs. of oranges
then take 4.76 and divide that by 1.19 (for the pears)
4.76/1.19=4lbs. of pears
add 4+3 to get how many lbs. in all
4+3=7

3 0

2 years ago

A restaurant that can hold 300 people is divided into 20 equal parts. How many people can be put in each section?

alexandr1967 [171]

Answer:

15

Step-by-step explanation:

300/20=15

4 0

2 years ago

A recent survey of 128 high school students indicated the following information about release times from school.

Delicious77 [7]

Answer:

A) Interval estimate for those that want to get out earlier = (35%) ± (4%) = (31%, 39%)

Interval estimate for those that want to get out later = (39%) ± (4%) = (35%, 43%)

B) The group that wants to get out of school earlier can win after all the votes are counted if their true population proportion takes on a value that is higher than the closest true population proportion (for the group that wants to get out of school later)

That is, in the (31%, 39%) and (35%, 43%) obtained in (a), a range of (35.1%, 39%) and (35%, 38.9%) show how possible that the group that wants to get out of school earlier can win after all the votes are counted.

Step-by-step explanation:

The Interval estimate for the proportion is basically an interval of range of values where the true population proportion can be found with a certain level of confidence.

Mathematically,

Interval estimate = (Sample proportion) ± (Margin of error)

Margin of Error is the width of the confidence interval about the proportion.

a) Find the interval estimates

i) for those that want to get out earlier and

ii) those that want to get out later.

i) Sample proportion of those that want to get out earlier = 35%

Margin of Error = 4%

Interval estimate for those that want to get out earlier = (35%) ± (4%) = (31%, 39%)

ii) Sample proportion of those that want to get out later = 39%

Margin of Error = 4%

Interval estimate for those that want to get out later = (39%) ± (4%) = (35%, 43%)

b) Explain how it would be possible for the group that wants to get out of school earlier to win after all the votes are counted.

Since the interval estimates represent the range of values that the true population proportion can take on for each group that prefer a particular option, the group that wants to get out of school earlier van have their proportion take on values between 31% and 39%. If their true population takes on a value that is highest (which is very possible from the interval estimate), and the group with the highest proportion in the sample, (the group that wants to get out of school later, whose true population proportion can take between 35% and 43%) has a true population proportion that is less than that of the group that wants to get out of school earlier, then, the group that wants to get out of school earlier can win after all the votes are counted.

Hope this Helps!!!

3 0

3 years ago