Question

H0: p ? 0.75 Ha: p < 0.75 A sample of 300 items was selected. Compute the p-value and state your conclusion for each of the following sample results. Use standard deviation = .05. Round your answers to four decimal places. a. = 0.66 p-value is Conclusion: p-value is H0 b. = 0.72 p-value is Conclusion: p-value is H0 c. = 0.7 p-value is Conclusion: p-value is H0 d. = 0.77 p-value is Conclusion: p-value is H0

Answer 1

The test statistic here is computed as:

$z^* = \frac{p-P}{\sqrt{\frac{P(1-P)}{n}}} = \frac{0.66 - 0.75}{\sqrt{\frac{0.75*0.25}{300}}} =- 3.6$

As this is a one tailed test, the p-value here is obtained from standard normal tables as:
p = P(Z < -3.6) = 0.0002

Therefore 0.0002 is the required p-value here.

As the p-value here is 0.0002 < 0.05 which is the level of significance, therefore the test is significant here and we can reject the null hypothesis here. Therefore we have sufficient evidence here that the proportion is less than 0.75 here.