The concentration of read the blood stream was measured (g/L) for a sample of 10

Question

The concentration of read the blood stream was measured (g/L) for a sample of 10 children from a large school near a busy road The result are given below Give the following statistical parameters A group of subjects participated in an experiment to study the effectiveness of a certain diet combined with a program of exercise, in redusing serum cholesterol levels The following data shows the serum cholesterol level m mg/100ml from the subjects at the beginning of the program (before) and at the end (after). Has the dit/exercise regime been effective in reducing cholesterol (Assume the usual 95% confidence level) The data on the left are the results of an experiment carried out to investigate the effects of different of irradiation on the percentage inactivation of enzymes in milk. Observations were made on each combination of four doses of irradiation levels a, b, c and D You can assume that the data in normally distributed. Analyse the data using an appropriate technique to establish if there are differences in percentage inactivation between the four radiation levels at the 95% significance level Use the Data Analysis Add-ln of Excel Please fill in your answers below if more space is needed please complete answers on the back identifying which questions if relates to Be accurate and precise by using the correct terminology without being 'wordy'

Explanation / Answer

Q1.

Ordering,

9.7
9.7
9.8
9.9
10
10
10.1
10.3
10.3
10.4
10.5
10.5
10.6
10.7
10.7
10.8
10.9
10.9
11
11.1

Getting the mean, X,          
          
X = Sum(x) / n          
Summing the items, Sum(x) =    207.9      
As n =    20      
Thus,          
X =    10.395   [ANSWER, MEAN]

*********************************************  
          
Setting up tables,          

          
Thus, Sum(x - X)^2 =    3.7695      
          
Thus, as           
          
s^2 = Sum(x - X)^2 / (n - 1)      
          
As n =    20      
          
s^2 =    0.198394737      
          
Thus,          
          
s =    0.445415241   [ANSWER, STANDARD DEVIATION]
******************************************  

The median is the average of the two middle terms,

Median = (10.4+10.5)/2 = 10.45 [ANSWER]

Also,

Q1 (lower quartile) = median of lower half = (10+10)/2 = 10 [ANSWER]

Q3 (upper quartile) = median of upper half = (10.7+10.8)/2 = 10.75 [ANSWER]

Thus,

Interquartile range = Q3 - Q1 = 10.75-10 = 0.75 [ANSWER]

x x - X (x - X)^2 9.7 -0.695 0.483025 9.7 -0.695 0.483025 9.8 -0.595 0.354025 9.9 -0.495 0.245025 10 -0.395 0.156025 10 -0.395 0.156025 10.1 -0.295 0.087025 10.3 -0.095 0.009025 10.3 -0.095 0.009025 10.4 0.005 2.5E-05 10.5 0.105 0.011025 10.5 0.105 0.011025 10.6 0.205 0.042025 10.7 0.305 0.093025 10.7 0.305 0.093025 10.8 0.405 0.164025 10.9 0.505 0.255025 10.9 0.505 0.255025 11 0.605 0.366025 11.1 0.705 0.497025

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.