Anorexic People Before And After One-Sample T Test Computations And Please Help Me Understand This?

One-Sample T Test Computations and please help me understand this? - anorexic people before and after

Thirty were treated with anorexic girls cognitive behavioral therapy (CBT). You have to be weighed before and after treatment. Weight gain, which are positive, and weight loss that are negative, are given below. Confirmed an increase of more than zero book, that the treatment works. If the profit of the group to book more than zero?

Dates:

1.7 0.7 -0.1 -0.7 -3.5 14.9 3.5 17.1 -7.6 1.6
11.7 6.1 1.1 -4.0 20.9 -9.1 2.1 -1.4 1.4 -0.3
-3.7 -0.8 2.4 12.6 1.9 3.9 0.1 15.4 -0.7 0

Report of the null hypothesis and the relevant research and the critical value
(alpha = 0.05). Please also provide your complete statistics, including the calculation of test statistics in APA format and a conceptual conclusion of tor two sentences.

My book is not really clear to understand and to know that people here understand thank you explain in simple language for me, that you will

2 comments:

Anonymous said...

H0: μ = 0
H1: μ> 0, α = 0.05

Decision rule: if ((Xmean - μ) SX / (n))> TST sqrt (n - 2, 1 - α), reject H0.

Xmean = Σxi / n = 87.2/30 = 2.91p

S ^ 2 = (Σxi ^ 2 - n * Xmean ^ 2) / (n - 1)

= (1757.80-30 * 2.9 ^ 2) / 29 = 51.91379310
s = 7.21
s / sqrt (30) = 1.315469411

Decision Item:
If (2.91 - 0) / 1.315469411> TST (28, 0.95) Reject H0

If 2.21> 1.699 Reject H0.

Conclusion: The group has won over more than zero pounds. The ITC has not worked.

Anonymous said...

H0: No difference in weight before and after cognitive behavioral therapy
Ha: The weights are greater after cognitive therapy

For the 30 differences
Average 2.9067
Standard deviation = 7.2023

A paired t-test was performed.

H0: μ = 0
HA: μ ≠ 0
Sample mean = 2.9067
Standard deviation = 7.2023
Standard error of mean = σ / √ n
Standard error of mean = 7.2023 / √ 30
SE = 7.2023/5.4772
Average Standard Error 1315
T = (Xbar-μ) / SE
t = (2.9067-0) / 1.315
t = 2.2105

The critical t-value is 1.699, with 29 degrees of freedom with alpha = 0.05 (unilateral test)
Since 2.2105> 1.699, reject the null hypothesis and conclude that the weight is greater after cognitive therapy

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