Question

The average age at which adolescent girls reach their adult height is 16 years. Suppose you have a sample of 27 adolescent girls who are developmentally delayed, and who have an average age at which they reached their adult height of 17.1 years and a sample variance of 36.0 years. You want to test the hypothesis that adolescent girls who are developmentally delayed have a different age at which they reached their adult height than all adolescent girls.
Calculate the t statistic. To do this, you first need to calculate the estimated standard error. The estimated standard error is SM=_________ .The t statistic is _________. Now suppose you have a larger sample size n 81. Calculate the estimated standard error and the t statistic for this sample with the same sample average and the same standard deviation as above, but with the larger sample size. The new estimated standard error is __________ .The new t statistic is___________

Answer 1

Answer:

a) t-statistic  t = 0.9532

b) The standard error  S.E = 1.2

c) The new  t-statistic = 1.95

d)  The new estimated standard error =0.666

Step-by-step explanation:

Step(i):-

Given that the mean of the Population = 16

The sample size n=27

The mean of the sample = 17.1

Given that the variance of the sample (S²) = 36.0

The standard deviation of the sample (S) = √36 = 6

Test statistic

  $t = \frac{x^{-} -mean}{\frac{S.D}{\sqrt{n} } }$

 $t = \frac{17.1 -16}{\frac{6}{\sqrt{27} } }$

t = 0.9532

b)

   The standard error is defined by

  $S.E = \frac{S.D}{\sqrt{n} } = \frac{6}{\sqrt{27} } =1.154$

Step(ii):-  

c) given that the sample size n = 81

Test statistic

 $t = \frac{x^{-} -mean}{\frac{S.D}{\sqrt{n} } }$

 $t = \frac{17.1 -16}{\frac{6}{\sqrt{81} } }$

t = 1.65

d)

The new standard error is defined by

  $S.E = \frac{S.D}{\sqrt{n} } = \frac{6}{\sqrt{81} } = 0.66$