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lesya692 [45]
2 years ago
5

The Sport Journal (Winter 2004) reported on a study of a speed-training program for high school football players. Each participa

nt was timed in a 40-yard sprint both before and after training. The researchers measured two variables: (1) the difference between the before and after sprint times (in seconds), and (2) the category of improvement ("improved," "no change," and "worse") for each player. What is the type for the variable "improvement"?
A. Ordinal
B. Interval
C. Ratio
D. Nominal
Mathematics
1 answer:
ruslelena [56]2 years ago
7 0

Answer:

Ordinal

Step-by-step explanation:

The definition of ordinal type of variables states that "if the variable has meaningful ordering or the data values of variable can be arranged in ranking then the variable is ordinal"

Here are some examples of ordinal variable to have a clear idea about it.

i) Ratings of product i.e. Excellent,good,fair, worse.

ii) Grades in exam i.e. A,B,C,D and F.

iii) Ranks in army.

iv) Ranks in sports team.

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375,000 no se sisera esa

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The realtor and her clients do not know the average home sale price for all of Guelph (500 was actually just a guess). However,
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Answer:

a) The 99% confidence interval would be given by (346.708;453.292)

b) The 99% confidence interval would be given by (338.445;461.555)

Step-by-step explanation:

1) Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X=400 represent the sample mean for the sample  

\mu population mean (variable of interest)

\sigma=80 represent the population standard deviation

n=15 represent the sample size  

2) Part a

The confidence interval for the mean is given by the following formula:

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}   (1)

Since the Confidence is 0.99 or 99%, the value of \alpha=0.01 and \alpha/2 =0.005, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.005,0,1)".And we see that z_{\alpha/2}=2.58

Now we have everything in order to replace into formula (1):

400-2.58\frac{80}{\sqrt{15}}=346.708    

400+2.58\frac{80}{\sqrt{15}}=453.292

So on this case the 99% confidence interval would be given by (346.708;453.292)    

3) Part b

For this case we don't know the population deviation so we need to use the t distribution instead the normal standard distribution.

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

We need to find the degrees of freedom first df=n-1=15-1=14

Since the Confidence is 0.99 or 99%, the value of \alpha=0.01 and \alpha/2 =0.005, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,14)".And we see that t_{\alpha/2}=2.98

Now we have everything in order to replace into formula (1):

400-2.98\frac{80}{\sqrt{15}}=338.445    

400+2.98\frac{80}{\sqrt{15}}=461.555

So on this case the 99% confidence interval would be given by (338.445;461.555)    

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