I need in an hour pls help Thankyou. ### Hypothesis Testing and Standard Normal Distribution#### Hypothesis Testing:1. **Test Statistic**: A z-value of 2.79 is obtained when testing the claim that \( p \neq 0.487 \). - **Identify** the type of test: two-tailed, left-tailed, or right-tailed. - Find the P-value. - Using a significance level of \(\alpha = 0.05\), decide whether to reject or fail to reject \( H_0 \). - [View page 1 of the standard normal distribution table.](#) - [View page 2 of the standard normal distribution table.](#)#### Steps:**a. Type of Test:** - This is a \(\text{two-tailed}\) test.**b. P-value:** - Calculated from the standard normal distribution table (round to three decimal places).**c. Conclusion:** - Choose the correct conclusion below: - A. Fail to reject \( H_0 \): There is not sufficient evidence to support the claim that \( p \neq 0.487 \). - B. Reject \( H_0 \): There is not sufficient evidence to support the claim that \( p \neq 0.487 \). - C. Fail to reject \( H_0 \): There is sufficient evidence to support the claim that \( p \neq 0.487 \). - D. Reject \( H_0 \): There is sufficient evidence to support the claim that \( p \neq 0.487 \).---#### Standard Normal Distribution Table:**Graph**:- The graph is a bell-shaped curve representing the standard normal distribution. It is symmetrical about the mean (z=0).**Table:**- **Title**: NEGATIVE z Scores- **Axis**: - X-axis: z-values - Y-axis: Cumulative area from the LEFT up to the specific z-value.**Values**:- The table provides cumulative probabilities for negative z-scores. For example: - For \( z = -3.50 \), the cumulative area is 0.0002. - For \( z = -2.9 \), column 0.00 lists 0.0019, indicating the cumulative probability up to that z-score.This table is essential for finding probabilities and critical values associated **Hypothesis Testing with Z-Scores**The problem involves testing the claim \( p \neq 0.487 \) using a test statistic of \( z = 2.79 \).**Steps to Follow:**a. Determine the type of test:- Select from right-tailed, two-tailed, or left-tailed.b. Identify the hypothesis test:- Use a significance level of \( \alpha = 0.05 \) and determine whether to reject \( H_0 \).c. Choose the correct conclusion:- Options range from failing to reject \( H_0 \) to rejecting \( H_0 \) based on the evidence.**Options:**A. Fail to reject \( H_0 \). There is not sufficient evidence to support the claim that \( p \neq 0.487 \).B. Reject \( H_0 \). There is sufficient evidence to support the claim that \( p \neq 0.487 \).C. Fail to reject \( H_0 \). There is not sufficient evidence to support the claim that \( p < 0.487 \).D. Reject \( H_0 \). There is sufficient evidence to support the claim that \( p < 0.487 \).**Standard Normal Distribution Table (Page 2):**- This table provides cumulative areas from the left for positive \( z \) scores.- A bell curve diagram represents the normal distribution, centered at zero.- The table contains z-scores ranging from 0.0 to 0.8 in increments of 0.1, with respective cumulative areas (e.g., \( z = 0.01 \) corresponds to an area of 0.5040).The table is essential for determining the probability associated with specific z-scores and making decisions about hypothesis testing.

Given that Test statistic Z = 2.79 And the claim is p is not equal to 0.487 Level of significance = 0.05a) To test the claim p ≠ 0.487, the null and alternative hypotheses are Ho : p = 0.487 Ha : p ≠ 0.487 Since the alternative hypothesis contains "≠" symbol then the test is two-tailed. b) The p-value for z = 2.79 is P-value = 2 × P(Z > 2.79) = 0.0053 (from Std. Normal table) c) Since p-value is less than 0.05 level of significance then D. Reject Ho. There is sufficient evidence to support the claim that p ≠ 0.487