Ima just give all my points away. P.S I don't like you brainly.
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Answer:
1. The 99% confidence interval for the difference in average is -6.47377 < μ₁ - μ₂ < -11.34623
2. The possible issues in the calculations includes;
a. The confidence level used in the confidence interval can influence the result of the confidence interval observed
b. The sample size is small
Step-by-step explanation:
1. The number of adults with insomnia in the sample = 45
The number of adults that participated in the therapy, n₁ = 22
The number of candidates that served as control group, n₂ = 23
The average score for the for the 22 participants of the program, = 6.59
The standard deviation for the 22 participants of the program, s₁ = 4.10
The average score for the for the 23 subjects in the control group, = 15.50
The standard deviation for the 23 subjects in the control group, s₂ = 5.34
The confidence interval for unknown standard deviation, σ, is given by the following expression;
α = 1 - 0.99 = 0.01
α/2 = 0.005
The degrees of freedom, df = 22 - 1 = 21
=
= 1.721
Therefore, we have;
The 99% confidence interval for the difference in average is therefore given as follows;
-6.47377 < μ₁ - μ₂ < -11.34623
Therefore, there is considerable evidence that the participants in the survey had lower average score than the subjects in the control group
2. The possible issues in the calculations are;
a. The confidence level used in the confidence interval can influence the result of the confidence interval observed
b. The sample size is small