Question

In: Statistics and Probability

A deficiency of the trace element selenium in the diet can negatively impact growth, immunity, muscle...

A deficiency of the trace element selenium in the diet can negatively impact growth, immunity, muscle and neuromuscular function, and fertility. The introduction of selenium supplements to dairy cows is justified when pastures have low selenium levels. The following data on the initial milk selenium concentration (mg/L) for a sample of cows given a selenium supplement (treatment group) and a control sample given no supplement is given below:

Treatment-11.4, 9.6, 10.1, 8.5, 10.3, 10.6, 11.8, 9.8, 10.9, 10.3, 10.2, 11.4, 9.2, 10.6, 10.8, 8.2

control- 9.1, 8.7, 9.7, 10.8, 10.9, 10.6, 10.1, 12.3, 8.8, 10.4, 10.9, 10.4, 11.6, 10.9

treatment-128.3, 104.0, 96.4, 89, 88, 103.8, 147.3, 97.1, 172.6, 146.3, 99, 122.3, 103, 117.8, 121.5, 93

control-9.3, 8.8, 8.8, 10,1, 9.6, 8.6, 10.4, 12.4, 9.3, 9.5, 8.4, 8.7, 12.5, 9.1

a.) Carry out a complete hypothesis testing to determine if there is a significant different in the initial mean selenium concentration for the control group and the treatment group. Use significance level of α = 0.10 (Hint: You will be using the first two columns to conduct hypothesis testing. Treatment and control do not affect each so it will be two-sample hypothesis testing).

b.) Carry out a complete hypothesis testing to determine if for the treatment group there is a significant evidence to conclude that the mean selenium concentration is greater after 9 days of the selenium supplementation compared to the initial measurement. Use significance level of α = 0.05 (Hint: You will be using the column one and three to conduct hypothesis testing. We are looking at two observations for treatment columns – initial and after 9 days so it will be paired-sample hypothesis testing).

Solutions

Expert Solution

A.

Let the group 1 be treatment group and group 2 be control group.

1 = 10.23

s1 = 1

n1 = 16

2 = 10.37

s2 = 1.03

n2 = 14

H0: 1 =  2

H1: 1   2

Assumptions: Population variances are equal and populations are normally distibuted.

df = n1 + n2 - 2

= 28

p-value = 0.7088 > 0.1

i.e. H0 can't be rejected and hence we can say that there is no significant difference between the initial mean selenium concentration for the control group and the treatment group.

B.

Data:

Treatment2 Treatment1 D (2-1)
128.3 11.4 116.9
104 9.6 94.4
96.4 10.1 86.3
89 8.5 80.5
88 10.3 77.7
103.8 10.6 93.2
147.3 11.8 135.5
97.1 9.8 87.3
172.6 10.9 161.7
146.3 10.3 136
99 10.2 88.8
122.3 11.4 110.9
103 9.2 93.8
117.8 10.6 107.2
121.5 10.8 110.7
93 8.2 84.8

= 104.11

SD = 23.61

SD/ = 5.9 ( n = 16)

H0: D = 0

H1: D > 0

t = /SD/

= 104.11/5.9

= 17.646

df = n-1

= 15

p-value ~ 0 < 0.1

i.e. H0 can be rejected and hence it can be claimed that there is a significant evidence that the mean selenium concentration is greater after 9 days of the selenium supplementation compared to the initial measurement.

Please upvote if you have liked my answer, would be of great help. Thank you.


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