Respuesta :
Answer:
[tex] 7-3.182 \frac{13.638}{\sqrt{4}}= -14.698[/tex]
[tex] 7+3.182 \frac{13.638}{\sqrt{4}}= 28.698[/tex]
Step-by-step explanation:
For this case we have the following info given:
Weight before diet 180 125 240 150
Weight after diet 170 130 215 152
We define the random variable [tex] D = before-after[/tex] and we can calculate the inidividual values:
D: 10, -5, 25, -2
And we can calculate the mean with this formula:
[tex]\bar X= \frac{\sum_{i=1}^n X_i}{n}[/tex]
And the deviation with:
[tex] s= \sqrt{\frac{\sum_{i=1}^n (X_i- \bar X)^2}{n-1}}[/tex]
And after replace we got:
[tex]\bar D= 7, s_d = 13.638[/tex]
And the confidence interval for this case would be given by:
[tex]\bar D \pm t_{\alpha/2} \frac{s_d}{\sqrt{n}}[/tex]
The degrees of freedom are given by:
[tex] df = n-1= 4-1=3[/tex]
For the 95% of confidence the value for the significance is [tex]\alpha=0.05[/tex] and the critical value would be [tex]t_{\alpha/2}= 3.182[/tex]. And replacing we got:
[tex] 7-3.182 \frac{13.638}{\sqrt{4}}= -14.698[/tex]
[tex] 7+3.182 \frac{13.638}{\sqrt{4}}= 28.698[/tex]
The mean difference in weight is 7 pounds less, with a 95% confidence interval between 6.65 and 7.35 pounds.
Given that four adults performed the water diet, and the results of the table were given, to determine the mean difference (before - after) in weight, and what is the appropriate 95% confidence interval, the following calculation should be performed:
- 180 --- 170 = -10
- 125 --- 130 = +5
- 240 --- 215 = -25
- 150 --- 152 = +2
- ((180 + 125 + 240 + 150) - (170 + 130 + 215 + 152)) / 4 = X
- (695 - 667) / 4 = X
- 28/4 = X
- 7 = X
- 7 x 0.95 = 6.65
- 7 x 1.05 = 7.35
Therefore, the mean difference in weight is 7 pounds less, with a 95% confidence interval between 6.65 and 7.35 pounds.
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