Question

 

Three way catalytic converters have been installed in new vehicles in oder to reduce pollutants from motor vehicles exhaust emissions. However, these converters unintentionally increase the level of ammonia in the air. Environmental Science and Technology published a study on the ammonia levels near the exit ramp of a San Francisco highway tunnel. The data in the next table represent daily ammonia concentrations ( parts per million) on eight randomly selectd days during afternoon drive time in the summer of 1999

X= 1.53, 1.50, 1.37, 1.51, 1.55, 1.42, 1.41, 1.48

1)

a)

Provide descriptive statistics for the data,

specifically mean, median, standard

deviation, variance, interquartile range. Make

sure to report units

b)

Estimate the mean daily ammonia

concentration for the selected exit ramp of a

San Francisco highway in summer 1999.

c)

Compute the standard error of the mean.

d)

What is conceptual difference between the

standard deviation in a) and the standard

error in c)

e)

Estimate the confidence interval of the mean

and interpret it in words

Answer 1

a)

X	(X - X̄)²
1.37	0.0103
1.41	0.004
1.42	0.00
1.48	0.000
1.5	0.00
1.51	0.002
1.53	0.003
1.55	0.0062

	X	(X - X̄)²
total sum	11.77	0.03
n	8	8

mean =    ΣX/n =    11.770   /   8   =   1.4713
                      
sample variance =    Σ(X - X̄)²/(n-1)=   0.0287   /   7   =   0.004
                      
sample std dev =   √ [ Σ(X - X̄)²/(n-1)] =   √   (0.0287/7)   =       0.0640

Median=0.5(n+1)th value =    4.5th   value of sorted data
=   1.490  
      
quartile , Q1 = 0.25(n+1)th value=   2.25th   value of sorted data
=   1.4125  
      
Quartile , Q3 = 0.75(n+1)th value=   6.75th   value of sorted data
=   1.525  
      
IQR = Q3-Q1 =    0.1125  

b)

mean =    ΣX/n =    11.770   /   8   =   1.4713

c)

Standard Error , SE = s/√n =   0.0640   / √    8   =   0.0226  

d)

s =  0.0640 while SE = 0.0226

SE< s

s is amount of variablity of data from mean while SE measures how far the sample mean of the data is likely to be from the true population mean

e)

Level of Significance ,    α =    0.05          
degree of freedom=   DF=n-1=   7          
't value='   tα/2=   2.3646   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   0.0640   / √   8   =   0.022634
margin of error , E=t*SE =   2.3646   *   0.02263   =   0.053520
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    1.47   -   0.053520   =   1.417730
Interval Upper Limit = x̅ + E =    1.47   -   0.053520   =   1.524770
95%   confidence interval is (   1.42   < µ <   1.52   )

it means the mean of ammonia concentration lies with in the range of (   1.42   < µ <   1.52   )