Use the given information to find the P-value. Also, use a 0.05 significance level and state the conclusion about the null hypothesis (reject the null hypothesis or fail to reject the null hypothesis).

The test statistic in a left-tailed test is z = –1.83.

A) 0.0336; reject the null hypothesis
B) 0.0672; reject the null hypothesis
C) 0.9664; fail to reject the null hypothesis
D) 0.0672; fail to reject the null hypothesis

1 answer

To find the P-value for a left-tailed test with a test statistic \( z = -1.83 \), we can refer to a standard normal (Z) distribution table or use a calculator that provides cumulative distribution function (CDF) values for the standard normal distribution.

The P-value in a left-tailed test is given by the area to the left of the test statistic. For \( z = -1.83 \):

  1. Look up the cumulative probability for \( z = -1.83 \) in the Z-table, or use a calculator. The cumulative probability for \( z = -1.83 \) is approximately \( 0.0336 \).

This means that:

  • P-value ≈ 0.0336.

Now, we compare the P-value with the significance level \( \alpha = 0.05 \):

  • Since \( 0.0336 < 0.05 \), we reject the null hypothesis.

Thus, the correct answer is:

A) 0.0336; reject the null hypothesis.