Question

Below are monthly rents paid by 30 students who live off campus. 730 730 730 930 700 570 690 1,030 740 620 720 670 560 740 650 660 850 930 600 620 760 690 710 500 730 800 820 840 720 700 Click here for the Excel Data File (a) Using Excel, find the mean, median, mode, and standard deviation. (Round your answers to 2 decimal places.) Descriptive Statistics Data Mean Median Mode Standard Deviation (b) Which measure or measures of central tendency are most appropriate for this data set? The mean or the median, because these measures are close in value, which suggests the data set is fairly symmetric. The median because the data set is strongly skewed. The mode because the data is integer valued with a very small range. (c) Do the measures of central tendency agree? Yes No (e) Use Excel or MegaStat to sort and standardize the data. What is the z score for the following rent value? (Round your answer to 2 decimal places.) Rent value z-score 800 (f) Are there unusual data values? No Yes (g) Using the Empirical Rule, do you think the data could be from a normal population? Yes No

Answer 1

Answer:

Step-by-step explanation:

Hello!

a)

To calculate the mean of the monthly rent paid by students living off-campus you have to add al observations and divide it by the sample size:

X[bar]= ∑X/n= 21740 /30= 724.67

The mean always takes values within the range of definition of the variable, but it does not necessarily coincide with an observed value.

To calculate the Median there are two steps to follow, first you calculate its position:

For even samples PosMe= n/2= 30/2= 15 ⇒ This means the median is in the 15th position.

Now you order the sample data from lower to highest and then indentify the observation in the 15th place:

500; 560 ; 570 ; 600 ; 620 ; 620 ; 650 ; 660 ; 670 ; 690 ; 690 ; 700 ; 700 ; 710 ; 720 ; 720 ; 730 ; 730 ; 730 ; 730 ; 740 ; 740 ; 760 ; 800 ; 820 ; 840 ; 850 ; 930 ; 930 ; 1030

Me= 720

Like the mean the median always takes values within the range of definition of the variable, but it does not necesarly coincide with an observed value.

The mode is the value of the variable with more absolute frequency, in other words, the most observed value.

620 is observed 2 times

690 is observed 2 times

700 is observed 2 times

720 is observed 2 times

730 is observed 4 times

740 is observed 2 times

930 is observed 2 times

The rest of the values are observed only one time.

730 is the value with the most absolute frequency so it corresponds to the mode.

Md= 730

The standard deviation is the square root of the variance:

$S^2= \frac{1}{n-1}[sumX^2-\frac{(sumX)^2}{n} ]$

$S^2=\frac{1}{29}[16133000-\frac{(21740)^2}{30} ] = 13060.229$

S= √S²= √13060.23= 114.28

b.

Me= 720 < X[bar]= 724.67 < Md= 730

As you can see the median, mean and mode are similar but not equal, this shows that the distribution of the data set is not exactly symmetrical but slightly skewed to the left. Since the mode is the value with more value with the more absolute frequency you can use it as a starting point to compare the three measurements of central tendency, both the mean and the median are less that it, this shows you the left-skewed tendency of the distribution.

Most correct answer: The mean or the median, because these measures are close in value, which suggests the data set is fairly symmetric.

c.

Yes (check b.)

e.

Z= $\frac{800-724.67}{114.28}$= 0.659

f.

Yes there are unusual data (i.e. outliers)

An outlier is an observation that is significantly distant from the rest of the data set. They usually represent experimental errors (such as a measurement) or atypical observations. Some statistical measurements, such as the sample mean, are severely affected by this type of values and their presence tends to cause misleading results on statistical analysis.

In this data set, there are three outliers: 500, 930 and 1030.

g.

The empirical rule states that

68% of the data is between μ-σ≤0.68≤μ+σ

95% of the data is between μ-2σ≤0.95≤μ+2σ

99% of the data is between μ-3σ≤0.99≤μ+3σ

If this is true then:

1)724.67-114.28 ≤ 0.68 ≤ 724.67+114.28

610.4 ≤ 0.68 ≤ 838.96

Under the standard normal distribution:

-Z ≤ 0.68 ≤ Z

-0.994 ≤ 0.68 ≤ 0.994

Now reversing the standardization with the sample data to check if it is equal to the empirical values:

$-0.994= \frac{x-724.67}{114.28}$

x= 581.25

$0.994= \frac{x-724.67}{114.28}$

x= 838.26

581.25 ≤ 0.68 ≤ 838.26

2)724.67-2*114.28 ≤ 0.95 ≤ 724.67+2*114.28

496.12 ≤ 0.95 ≤ 953.24

Under the standard normal distribution:

-Z ≤ 0.95 ≤ Z

-1.960 ≤ 0.95 ≤ 1.960

Now reversing the standardization with the sample data to check if it is equal to the empirical values:

$-1.96= \frac{x-724.67}{114.28}$

x= 500.68

$1.96= \frac{x-724.67}{114.28}$

x=948.66

500.68 ≤ 0.95 ≤ 948.66

3)724.67-3*114.28 ≤ 0.99 ≤ 724.67+3*114.28

381.84 ≤ 0.99 ≤ 1067.52

Under the standard normal distribution:

-Z ≤ 0.99 ≤ Z

-2.576 ≤ 0.99 ≤ 2.576

Now reversing the standardization with the sample data to check if it is equal to the empirical values:

$-2.576= \frac{x-724.67}{114.28}$

x= 430.28

$2.576= \frac{x-724.67}{114.28}$

x= 1019.055

430.28 ≤ 0.99 ≤ 1019.055

Comparing the empirical intervals with the data intervals, you can say that the data could be from a normal population.

I hope it helps!