To solve this problem, I’ll guide you through the steps:

(a) Significance Level and Hypotheses:

• Level of significance: $\alpha = 0.05$
• Null hypothesis ($H_0$): The true proportion of college students who favor the color blue is equal to 0.24.
  $H_0: p = 0.24$
• Alternative hypothesis ($H_1$): The true proportion of college students who favor the color blue is different from 0.24 (two-tailed test).
  $H_1: p \neq 0.24$

(b) Sampling Distribution and Test Statistic:

1. Sampling Distribution:
   Since the sample size is sufficiently large ($np > 5$ and $nq > 5$, where $q = 1 - p$), we use the standard normal distribution.

2. Test Statistic Formula: $z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}$ where:

   • $\hat{p}$ = sample proportion (provided in the question),
   • $p_0 = 0.24$ (null hypothesis proportion),
   • $n$ = sample size (provided in the question).

3. Plug the values into the formula and calculate $z$-value. Round your $z$-value to two decimal places.

(c) P-Value of the Test Statistic:

1. P-value Calculation:
   Use the $z$-value obtained from part (b) to calculate the P-value from a standard normal table or a statistical calculator. Since this is a two-tailed test, multiply the one-tail probability by 2.

2. Round your P-value to four decimal places.

(d) Decision on Null Hypothesis:

• If the P-value is less than $\alpha = 0.05$, reject $H_0$.
• If the P-value is greater than or equal to $\alpha = 0.05$, fail to reject $H_0$.

(e) Conclusion:

Interpret the result in the context of the problem:

• If $H_0$ is rejected, there is sufficient evidence to conclude that the true proportion of college students favoring blue differs from 0.24.
• If $H_0$ is not rejected, there is insufficient evidence to conclude that the true proportion of college students favoring blue differs from 0.24.

Let me know the sample proportion ($\hat{p}$) and the sample size ($n$) from the problem so I can compute the detailed calculations for you.