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The proportion of blue M&Ms in a sample of 100 M&Ms is equal to 27%.

The null hypothesis is written as follows:

The proportion of blue M&Ms is equal to 24%.

$H_0:p=0.24$

The alternative hypothesis is written as follows:

The proportion of blue M&Ms is not equal to 24%.

$H_1:p鈮�0.24$

The test is two-tailed.

The sample size is equal to n=100.

The sample proportion of blue M&Ms isas follows:

$$\begin{gathered} \hat{p}=27\% \\ =\frac{27}{100} \\ =0.27 \end{gathered}$$

The population proportion of blue M&Ms is equal to 0.27.

The value of the test statistic is computed below:

$$\begin{gathered} z=\frac{\hat{p}-p}{\sqrt{\frac{pq}{n}}} \\ =\frac{0.27-0.24}{\sqrt{\frac{0.24\left(1-0.24\right)}{100}}} \\ =0.702 \end{gathered}$$

Thus, z=0.702.

Referring to the standard normal distribution table, the critical value of z at $伪=0.05$ for a two-tailed test is equal to 1.96.

Referring to the standard normal distribution table, the p-value for the test statistic value of 2.694 is equal to 0.4827.

Since the p-value is greater than 0.05, the null hypothesis is failed to reject.

There is not enough evidence to reject the claim that the proportion of blue M&Ms is equal to 0.24.

Since the proportion of blue M&Ms is equal to 24% per the claim of the company, the company does not need to take any corrective measure.