Question

You wish to test the following claim (H) at a significance level of or 0.005.

H.:p=0.36
H: p<0.36
You obtain a sample of size n 679 in which there are 224 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.
The test statistic's value is (224/679-0.36)/sqrt(0.36 (1-0.36)/679)--1.634. What is the test statistic for this sample?
A. z
B. t

Answer 1

The test statistic for this sample is a z-test statistic of -1.634. Fail to reject the null hypothesis.

Step-by-step explanation:

The test statistic for this sample is a z-test statistic. It is calculated using the formula (sample proportion - population proportion) divided by the standard deviation of the sampling distribution. In this case, the formula is (224/679 - 0.36) / sqrt(0.36 * (1 - 0.36) / 679), which evaluates to -1.634. Since this is a one-tailed test with the alternative hypothesis p < 0.36, the critical value for a significance level of 0.005 is approximately -2.58 for the standard normal distribution. Since the test statistic (-1.634) is not less than the critical value (-2.58), we would fail to reject the null hypothesis and do not find strong evidence to support the claim that p < 0.36.