Question

A researcher obtains a sample of n= 16 adults who are between the ages of 65 and 75. The researcher measures cognitive performance for each individual before and after a two-month program in which participants receive daily doses of a blueberry supplement. The results show an average increase in performance of MD= 7.4, with SS = 1215. Does the result support the conclusion that the blueberry supplement significantly increase cognitive performance? Use a one-tailed test with α = .05

(A) The alternative hypothesis in symbols is:

H1: µ1= µ2

H1: µ1> µ2

H1: µD= 0

H1: µD≠ 0

H1: µD> 0

H1: MD = 7.4

H1: MD ≠ 7.4

H1: µD≤ 0

(B) The critical t-value is (Include all the numbers after the decimal point.)

(C) Construct a 95% confidence interval to estimate the average cognitive performance improvement for the population of older adults.

The lower boundary is

(D) The upper boundary of the confidence interval you constructed in the previous question is

Answer 1

Answer:

a)

Given,

sample n = 16

MD= 7.4

SS = 1215

Alternative hypothesis H1 : µD > 0

b)

standard deviation = sqrt(SS/(n-1))

= sqrt(1215/(16-1))

= 9

Degree of freedom = n - 1

= 16 - 1

= 15

alpha = 0.05

Critical value corresponding to df = 15, alpha = 0.05 is 1.753

Test statistic t = xbar/(s/sqrt(n))

substitute values

= 7.4/(9/sqrt(16))

= 3.29

c)

At 95% confidence interval , t = 2.131

Interval = xbar +/- t*s/sqrt(n)

substitute values

= 7.4 +/- 2.131*9/sqrt(16)

= 7.4 +/- 4.795

= (7.4 - 4.795 , 7.4 + 4.795)

= (2.605 , 12.195)

Lower limit = 2.605

d)

Upper limit = 12.195