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dalvyx [7]
3 years ago
10

Cushing's disease is characterized by muscular weakness due to adrenal or pituitary dysfunction. To provide effective treatment,

it is important to detect childhood Cushing's disease as early as possible. Age at onset of symptoms and age at diagnosis (months) for 15 children suffering from the disease were given in an article "Treatment of Cushing's Disease in Childhood and Adolescence by Transphenoidal Microadenomectomy" (New Engl. J. of Med., 1984: 889). Here are the values of the differences between age at onset of symptoms and age at diagnosis:
-24 -12 -55 -15 -30 -60 -14 -21
-48 -12 -25 -53 -61 -69 -80
Calculate a 95% confidence interval for the population mean difference, and interpret the resulting bound.
Mathematics
1 answer:
Katyanochek1 [597]3 years ago
8 0

Answer:

Since the calculated value of t =-0.427 does not  fall in the critical region so we accept H0 and conclude that there is enough evidence to show that mean difference in the  age of onset of symptoms and age of diagnosis is  25 months  .

Step-by-step explanation:

The given data is

Difference d= -24 -12 -55 -15 -30 -60 -14 -21 -48 -12 -25 -53 -61 -69 -80

∑ d= -579

 ∑d²= 29871

1) Let the hypotheses be

H0: ud= 25 against the claim    Ha: ud ≠25

H0 : mean difference in the  age of onset of symptoms and age of diagnosis is  25 months  .

Ha:   mean difference in the  age of onset of symptoms and age of diagnosis is not 25 months.

 

2) The degrees of freedom = n-1= 15-1= 14

3) The significance level is 0.05

4) The test statistic is

t= d`/sd/√n

The critical region is ║t║≤ t (0.025,14) = ±2.145

d`= ∑di/n= -579/15= -38.6

Sd= 23.178  (using calculators)

Therefore

t= d`/ sd/√n

t= -38.6/ 23.178√15

t= -1.655/3.872= -0.427

5) Since the calculated value of t =-0.427 does not  fall in the critical region so we accept H0 and conclude that there is enough evidence to show that mean difference in the  age of onset of symptoms and age of diagnosis is  25 months  .

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Answer:

[-3, ∞)

Step-by-step explanation:

There are many ways to find the range but I will use the method I find the easiest.

First, find the derivative of the function.

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Great, you have the x value but we need the y value. To find it, plug the x value of 5 back into the original equation.

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