**Title: Hypothesis Testing for Favorite Color Preference****Background:**A large survey of countries including the United States, China, Russia, France, Turkey, Kenya, and others indicates that most people prefer the color blue. About 24% of the population claim blue as their favorite color.**Scenario:**A random sample of \( n = 57 \) college students was surveyed, and \( x = 17 \) of them said blue is their favorite color. Does this imply that the preference of college students is different from that of the general population?---**Question 1(a):** What is the level of significance?**Options:**- \( \alpha = 0.01 \)- \( \alpha = 0.05 \)- \( \alpha = 0.10 \)**Hypotheses:**- Null Hypothesis \( H_0: p = 0.24 \)- Alternative Hypothesis \( H_a: p \neq 0.24 \)**Question 1(b):** What sampling distribution will you use?Options:1. The standard normal, since \( np > 5 \) and \( nq > 5 \).2. The standard normal, since \( n > 30 \).3. The Student's \( t \), since \( n \leq 30 \) and \( \sigma \) is unknown.4. The Student's \( t \), since \( np \leq 5 \).**What is the value of the sample test statistic?****Question 1(c):** Find the P-value of the test statistic. Round your answer to four decimal places.**Diagrams:**There are four graphs displaying normal distributions with shaded areas under the curve, representative of P-values.1. Graph A (shows left tail)2. Graph B (shows right tail)3. Graph C (symmetrical, both tails)4. Graph D (centered, no clear tails)**Question 1(d):**Based on your answers, should you reject or fail to reject the null hypothesis?Options:- Fail to reject \( H_0 \)- Reject \( H_0 \)---**Conclusion:**This exercise investigates whether the preference for the color blue in college students is statistically different from the general population's preference. Follow the steps of hypothesis testing to determine the outcome.

level of significance α = 0.05.a) Given that test the hypothesis at a 0.05 level of significance level of significance The claim of the test is the color preference of all college students is different (either way) from that of the general population.The null and alternative hypothesis is H0: P = 0.24.H1: P ≠ 0.24.(bnp = 57 ×0.1579 = 9.0003 >5nq = 57 (1 – 0.1579) = 47.9997 >5The standard normal can be used since np >5 and nq>5.The test statistic isc) The p-value associated with the above test statistic is calculated in Excel as below.The above test is a two-sided test, so for a two-sided test statistic 2 is multiplied by the p-valueThe p-value = 0.1471The sketch of the above critical region isThe test is two-sided so the critical region will be on both sizes z<-1.45 and z>1.45So option D) is correct Fig.d) Here p-value (0.1471) > 0.05, fails to reject the null hypothesis.At a 5% significance level, we fail to reject the null hypothesis and conclude that the data is not statistically significant.e) There is insufficient evidence at a 0.05 level to conclude that the true population of college students favoring the color differs from 0.24a) Level of significance = 0.05H0: P = 0.24.H1: P ≠ 0.24.b) The standard normal, since np >5 and nq>5.test statistic -1.45c) P-value= 0.1471option D) is correct Fig.d) D) At a 5% significance level, we fail to reject the null hypothesis and conclude that the data is not statistically significant.e) There is insufficient evidence at a 0.05 level to conclude that the true population of college students favoring the color differs from 0.24