see attached The test statistic of \( z = 1.19 \) is obtained when testing the claim that \( p > 0.3 \).a. Identify the hypothesis test as being two-tailed, left-tailed, or right-tailed.**Answer:**This is a **right-tailed** test.b. Find the P-value.**P-value =** \(\_\_\_\) (Round to three decimal places as needed.)c. Using a significance level of \( \alpha = 0.05 \), should we reject \( H_0 \) or should we fail to reject \( H_0 \)?- To find the P-value, refer to the standard normal distribution table linked below:[Click here to view page 1 of the standard normal distribution table.](#)[Click here to view page 2 of the standard normal distribution table.](#)For guidance, the P-value is the probability of observing a test statistic as extreme as, or more extreme than, the observed value under the null hypothesis. It can be found using the standard normal distribution table based on the calculated \( z \)-score.

Name: see attached The test statistic of \( z = 1.19 \) is obtained when tes
Uploaded: 2024-03-20T06:46:14.000Z
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Description: see attached The test statistic of \( z = 1.19 \) is obtained when testing the claim that \( p > 0.3 \).a. Identify the hypothesis test as being two-tailed, left-tailed, or right-tailed.**Answer:**This is a **right-tailed** test.b. Find the P-value.*

a. Different null and alternative hypotheses can be defined as given below Two tailed test: Left tail test: Right tail test: From the given information, the claim of the test is p > 0.3. Here, the claim is p > 0.3. The symbol of greater than (>) represent the right tailed test. Thus, the hypothesis test as being right tailed test.b. P-Value: It is given that, the level of significance is α = 0.05, test statistic (z) is 1.19. The excel formula for calculate the P-value is = 1- NORMDIST (x, mean, standard_dev, cumulative) The P-value is: = 1- NORMDIST (1.19, 0, 1, 1) Thus, the P-value is 0.1170.c. Decision: Use the level of significance α = 0.05. The P-value (0.1170) is greater than level of significance (0.05). By the decision rule, we should fail to reject H0. Therefore, we fail to reject H0.