Question

3. A random sample of size 21 from a normal distribution with mean u yields observations that are summarized as follows: • ī= 3.5 . 511(2-1) = 0.6156 You are testing the following hypotheses: • H,:=3 • Ha:p #3 Let tay be the critical value of a t random variable with v degrees of freedom. The following table lists values of tay for specific combinations of a and v: v = 20 v=21 a = 0.005 2.845 2.831 a = 0.004 2.945 2.930 a = 0.003 | 3.073 3.056 Calculate the p-value for this test.

Answer 1

Solution:

Here, we have to use one sample t test for the population mean.

H₀: µ = 3 versus H_a: µ ≠ 3

This is a two tailed test.

The test statistic formula is given as below:

t = (Xbar - µ)/[S/sqrt(n)]

We are given

Xbar = 3.5

Sample variance = 0.6156

Sample standard deviation = S = Sqrt(Variance) = Sqrt(0.6156) = 0.784602

S = 0.784602

n = 21

df = ν = n – 1 = 21 – 1 = 20

t = (3.5 – 3)/[ 0.784602/sqrt(21)]

t = 2.9203

p-value = 0.0085

(by using t-table or excel)

Answer: p-value = 0.0085