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evablogger [386]
3 years ago
11

The highway mileages of 147 cars are in a file named CARS2.† These cars are 2003 models. Find the mean, standard deviation, and

median for the mileages. (Round your answers to two decimal places.)
Mathematics
1 answer:
grigory [225]3 years ago
6 0

In statistics, mean is the average in a set of data collected. To calculate the mean, you have to add up all the numbers in the table and divide it by the number of data that exists. For example, if the data set is 2 4 1 5 2 8, the mean will be:

mean = \frac{(2+4+1+5+2+8)}{6}= 3.6

So, for this data set, the mean is 3.6.

The median is also a average number, however, the median is the middle value in the list. For example, with the data set above, the median is 3, because, as the set has an even number of data, the middle term would be the mean of the two middle number, in other words:

median = \frac{1+5}{2} = 3

The standard deviation (σ) is the spread of data distribution, which means it's how "far" a number is from the mean of the set. The formula is

σ = √∑ (x - μ)²/N, where ∑ is the total sum; μ is the mean; and N is the number of data points in the sample. So, to calculate the standard deviation, using the example:

1) Calculate the mean of the set: μ=3.6

2) Find the difference between each data point and the mean and then, the square of each one:

(2-3.6)² = 2.56;

(4-3.6)² = 0.16

(1-3.6)² = 6.76

(5-3.6)² = 1.96

(2-3.6)² = 2.56

(8-3.6)² = 19.36

3) Add up all the squares: ∑= 33.36

4) Divide by the number of data points: \frac{33.36}{6}=5.56

5) Take the square root: σ = 2.36

For the set of numbers above (2,4,1,5,2,8), the standard deviation is

σ = 2.36.

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Answer:

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Step-by-step explanation:

Assuming this info from R

hist(gifted$count)

##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max.  

##   21.00   28.00   31.00   30.69   34.25   39.00

## Sd

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Data given and notation  

\bar X=30.69 represent the mean  

s=4.3149 represent the sample standard deviation

n=36 sample size  

\mu_o =32 represent the value that we want to test

\alpha represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is different than 32, the system of hypothesis would be:  

Null hypothesis:\mu = 32  

Alternative hypothesis:\mu \neq 32  

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

t=\frac{30.69-32}{\frac{4.3149}{\sqrt{36}}}=-1.822    

P-value

The first step is calculate the degrees of freedom, on this case:  

df=n-1=36-1=35  

Since is a two sided test the p value would be:  

p_v =2*P(t_{(35)}  

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