Question

For each of the following Hypotheses (including test statistic and other information), find de p-value (or its bounds) and draw basic conclusions with 99% confidence.

1) Ho : p = po, H1: P <> po n = 314 test statistics = 2.03 P

value =

Conclusion w.r.t. Ho:

2) Ho : σ1^2<= σ2^2 , H1: σ1^2>σ2^2 n1= 8, n2=4 test statistics = 3.02

P value =

Conclusion w.r.t. Ho:

3) Ho : µd = D0, , H1: µd <> D0, n=18 test statistics = -1.47

P value =

Conclusion w.r.t. Ho:

Answer 1

1)

Since hypotheses are about proportion so test statistics must be z test statistics. Test is two tailed so p-value of the test using z table is

p-value = 2 P(z > 2.03) = 2 * [ 1 - P(z <= 2.03)] = 2* [1 - 0.9788] = 0.0424

Conclusion: Since p-value is greater than 0.01 so we fail to reject the null hypothesis.

2)

Here hypotheses are comparing variances so test statistics must be F test statistics.

Degree of freedom of numerator: df1 = 8-1 = 7

Degree of freedom of denominator: df2 = 4-1 = 3

The test statistics is

F = 3.02

Using F table, the p-value is:

p-value > 0.10

Conclusion: Since p-value is greater than 0.01 so we fail to reject the null hypothesis.

3)

Here hypotheses are about paired population mean so test statistics must be t test statistics. Test is two tailed.

Degree of freedom: df = 18-1 = 17

The test statistics is

t = -1.47

Using t table, the p-value is:

0.10 < p-value < 0.20

Conclusion: Since p-value is greater than 0.01 so we fail to reject the null hypothesis.