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).
1) Vvith H1: p * 0.733, the test statistic is z = 2.09.
A) 0.0183; fail to reject the null hypothesis B) 0.0183; reject the null hypothesis
C) 0.0366; reject the null hypothesis D) 0.0366; fail to reject the null hypothesis

Here's a hint:

The null will be rejected using the data given. Use a z-table to determine the p-value, which is the actual level of the z-test statistic.

To find the p-value, we need to use the given test statistic (z = 2.09) and a standard normal distribution table (also known as a z-table).

The p-value represents the probability of obtaining a test statistic as extreme or more extreme than the one observed, assuming the null hypothesis is true.

Since the null hypothesis is stated as H1: p * 0.733, we have a one-tailed test (either greater than or less than). We need to find the probability of observing a z-value as extreme or more extreme than 2.09 in the distribution.

Using a standard normal distribution table (z-table), we can find that the cumulative probability corresponding to a z-value of 2.09 is approximately 0.9818.

However, since we are interested in finding the probability of obtaining a test statistic as extreme or more extreme than the observed one, we need to consider both tails of the distribution. Therefore, we need to subtract this cumulative probability from 1.

1 - 0.9818 = 0.0182.

Hence, the p-value is approximately 0.0182 or 0.0183 (rounded to four decimal places).

To make a conclusion about the null hypothesis, we compare the p-value with the significance level (0.05 in this case).

If the p-value is less than the significance level (p-value < 0.05), we would reject the null hypothesis. However, if the p-value is greater than or equal to the significance level (p-value >= 0.05), we would fail to reject the null hypothesis.

In this case, the p-value is approximately 0.0183, which is less than the significance level of 0.05.

Therefore, the conclusion is:
B) 0.0183; reject the null hypothesis