Answer:
1). Z score = 1.464733
Mean = 38.87917
Standard deviation = 3.609405
2). Quartiles:
Q1 = 37.025
Q2 = 38.450
Q3 = 40.800
3). IQR = 3.775
4).
CI (lower fence) = 37.4351
CI (upper fence) = 40.32323
5). There is outlier in the data set. Please see attached box plot for evidence.
Step-by-step explanation:
1). By Z score, we mean:
Z = [tex]\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex],
where:
x-bar ==> mean(g) = 38.87917
[tex]\mu[/tex] = 37.80
standard deviation = 3.609405
sample size (n) = 24.
2). By quantile, we mean:
Q = L + (i*(n/4) - Cf)*c
Where L is the lower class limit of the quartile class
Cf is the cumulative frequency before the quartile class
c is the class size.
3) . IQR = Q3 - Q1
4). CI = [tex]\mu \pm *Z_{\alpha/2} \frac{\sigma}{\sqrt{n}}[/tex],
Where [tex]Z_{\alpha/2}[/tex] ==> 1.96
In order to replicate and obtain the result, please use the R code below:
g = c(31.5, 36.0, 37.8, 38.5, 40.1, 42.2,34.2, 36.2, 38.1, 38.7, 40.6, 42.5,34.7, 37.3, 38.2, 39.5,
41.4, 43.4,35.6, 37.6, 38.4, 39.6, 41.7, 49.3)
boxplot(g)
Z = (mean(g) - 37.8)/(sd(g)/sqrt(length(g)))
mean(g)
quantile(g)
IQR(g)
CIl = mean(g) - 1.96*(sd(g)/sqrt(length(g)))
CIU = mean(g) + 1.96*(sd(g)/sqrt(length(g)))
From the dataset, we have:
First, we calculate the mean of the dataset
[tex]\mu = \frac{\sum x}{n}[/tex]
So, we have:
[tex]\mu = \frac{31.5 +36.0 +37.8 +38.5 +40.1+......+ 41.7+ 49.3}{24}[/tex]
[tex]\mu = \frac{933.1}{24}[/tex]
[tex]\mu = 38.88[/tex]
Next, the we calculate the standard deviation:
[tex]\sigma = \sqrt{\frac{\sum(x - \mu)^2}{n-1}}[/tex]
So, we have:
[tex]\sigma= \sqrt{\frac{(31.5 - 38.9)^2 +(36.0 - 38.9)^2 +(37.8 - 38.9)^2 +(38.5 - 38.9)^2 +(40.1 - 38.9)^2 +......+ (41.7 - 38.9)^2 +(49.3 - 38.9)^2}{24-1}[/tex]
[tex]\sigma= 3.61[/tex]
A. Calculate z score
Given that:
[tex]x = 37.8[/tex]
The z-score is calculated using:
[tex]z =\frac{x - \mu}{\sigma}[/tex]
So, we have:
[tex]z =\frac{37.8 - 38.88}{3.61}[/tex]
[tex]z =-0.30[/tex]
Interpretation: The value of x (i.e. 37.8) is 0.30 standard deviations below the mean.
B. Quartiles
Quartile is calculated using:
[tex]Q_i = \frac{i}{4}(n + 1)th[/tex]
For the lower quartile, we have:
[tex]Q_1 = \frac{1}{4}(24 + 1)th[/tex]
[tex]Q_1 = \frac{1}{4}(25)th[/tex]
[tex]Q_1 = 6.25th[/tex]
This implies that the lower quartile is the mean of the 6th and 7th data element.
From the sorted data (see attachment), we have:
[tex]Q_1 = \frac{36.2+ 37.3}{2}[/tex]
[tex]Q_1 = 36.75[/tex]
For the upper quartile, we have:
[tex]Q_3 = \frac{3}{4}(24 + 1)th[/tex]
[tex]Q_3 = \frac{3}{4}(25)th[/tex]
[tex]Q_3 = 18.75th[/tex]
This implies that the upper quartile is the mean of the 18th and 19th data element.
So, we have:
[tex]Q_3 = \frac{1}{2}(40.6+ 41.4)[/tex]
[tex]Q_3 = 41.00[/tex]
C. The interquartile range (IQR)
This is calculated using:
[tex]IQR =Q_3 - Q_1[/tex]
So, we have:
[tex]IQR =41.00 - 36.75[/tex]
[tex]IQR =4.25[/tex]
Interpretation of the IQR
The range of the middle half of the data is 4.25
D. Outliers
The lower fence of the outlier is calculated using:
[tex]L = Q_1 - 1.5 \times IQR[/tex]
[tex]L = 36.75 - 1.5 \times 4.25[/tex]
[tex]L= 30.375[/tex]
The upper fence of the outlier is calculated using:
[tex]U = Q_3 + 1.5 \times IQR[/tex]
[tex]U = 41.00 + 1.5 \times 4.25[/tex]
[tex]U = 47.375[/tex]
This means that, data elements outside the range 30.375 to 47.475 are outliers.
So, we can conclude that, there is an outlier.
And the outlier is: 49.3
Learn more about quartiles and z-score at:
https://brainly.com/question/13649404
