2. in a random sample of 50 creamery customers, 30 preferred chocolate over vanilla. is there evidence at…

Explanation:

Step1: State the hypotheses

Let (p) be the proportion of customers who prefer chocolate ice - cream over vanilla. (H_0:p = 0.8) (null hypothesis) (H_1:p<0.8) (alternative hypothesis)

Step2: Calculate the sample proportion (\hat{p})

(\hat{p}=\frac{30}{50}=0.6)

Step3: Calculate the test statistic (z)

The formula for the test statistic in a one - sample proportion test is (z=\frac{\hat{p}-p}{\sqrt{\frac{p(1 - p)}{n}}}) Substitute (p = 0.8), (\hat{p}=0.6), and (n = 50) [ \begin{align*} z&=\frac{0.6 - 0.8}{\sqrt{\frac{0.8\times(1 - 0.8)}{50}}}\ &=\frac{- 0.2}{\sqrt{\frac{0.8\times0.2}{50}}}\ &=\frac{-0.2}{\sqrt{\frac{0.16}{50}}}\ &=\frac{-0.2}{\sqrt{0.0032}}\ &=\frac{-0.2}{0.0566}\ &\approx - 3.53 \end{align*} ]

Step4: Find the critical value

For a one - tailed test with (\alpha=0.01), the critical value (z_{\alpha}) is (z_{0.01}=- 2.33) (from the standard normal distribution table)

Step5: Make a decision

Since the test statistic (z=-3.53<z_{0.01}=-2.33), we reject the null hypothesis.

Answer:

Yes, there is evidence at the (0.01) level of significance that the percentage of Creamery customers who prefer chocolate ice - cream over vanilla is less than (80%)