A sample of n 16 individuals is selected from a population

A sample of n = 16 individuals is selected from a population with a mean of = 78. A treatment is administered to the individuals in the sample and, after treatment, the sample variance is found to be 2 = 144.

a. If the treatment produces a sample mean of M = 81, is this sufficient to conclude that there is a significant treatment effect using a two-tailed test with = .05

b. If the treatment produces a sample mean of M = 87, is this sufficient to conclude that there is a significant treatment effect using a two-tailed test with = .05?

Solution

mu =78 and n =16

Sample variance = 144

a) Sample mean = 82

Let us create hypohteses as

H0: mu = 78

H0: mu not equals 78

Two tailed test

Sample size = 16 <30 hence t test can be done.

alpha = 0.05

t = 1.3333
  df = 15
  standard error of difference = 3.000

The   95% confidence interval of this difference:
From -10.39 to 2.39

The P-Value is 0.202329. The result is not significant at p < 0.05.

Hence there is not sufficient evidence to show that sample mean differs from population mean at 5%

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b) Here M =86

H0: mu = 78

H0: mu not equals 78

Two tailed test

The difference between these two values is 8.00
The   95% confidence interval of this difference:
From -68.73 to 84.73

t = 0.2222
  df = 15
  standard error of difference = 36.000

The two-tailed P value equals 0.8271
  By conventional criteria, this difference is considered to be not statistically significant.