Question

find test statistic and p value and make conclusion for a proportion (using 1-PropZTest)

x = 27, n = 200, H0:p = 10%, Ha:p > 10%, degree of certainty=0.01

Answer 1

Solution :

The null and alternative hypotheses are as follows :

Test statistic :

The test statistic is given as follows :

Where, p̂ is sample proportion, p is hypothesized value of population proportion under H_{​​​​​​0}, q = 1 - p and n is sample size.

Sample proportion is given by,

p̂ = x/n

p̂ = (27/200) = 0.135

n = 200, p = 10% = 0.10 and q = (1 - 0.10) = 0.90

The value of the test statistic is 1.6499.

P-value :

Since, our test is right-tailed test, therefore we shall obtain right-tailed p-value for the test statistic. The right-tailed p-value is given as follows :

P-value = P(Z > value of the test statistic)

P-value = P(Z > 1.6499)

P-value = 0.0495

The p-value is 0.0495.

Conclusion :

Significance level = 0.01

P-value = 0.0495

(0.0495 > 0.01)

Since, p-value is greater than the significance level of 0.01, therefore we shall be fail to reject the null hypothesis at 0.01 significance level.

At 0.01 significance level, there is not sufficient evidence to conclude that p > 10%.