Question

You wish to test the following claim (HaHa) at a significance level of α=0.01α=0.01.

      Ho:p=0.39Ho:p=0.39
      Ha:p>0.39Ha:p>0.39

You obtain a sample of size n=708n=708 in which there are 303 successful observations. For this test, you should NOT use the continuity correction, and you should use the normal distribution as an approximation for the binomial distribution.

What is the test statistic for this sample? (Report answer accurate to three decimal places.)
test statistic =

What is the p-value for this sample?

Answer 1

Solution :

Given that,

= 0.39

1 - = 0.61

n = 708

x = 303

Level of significance = = 0.01

Point estimate = sample proportion = = x / n = 0.428

This a right (One) tailed test.

Test statistics

z = ( - ) / *(1-) / n

= ( 0.428 - 0.39) / (0.39*0.61) / 708

= 2.071

Test statistic = 2.071

P-value = P(Z>z)

= 1 - P(Z <z )

= 1- P(Z < 2.071)

= 1 - 0.9808

= 0.0192

P-value = 0.192

The p-value is p = 0.0192, and since p = 0.0192 > 0.01, it is concluded that the null hypothesis is fails to reject.