a. The conditions for performing the test are:
- Randomization: Zenon randomly selected the 50 students to participate in his study.
- Independence: It is reasonable to assume that each student's preference for chips is independent of the other students.
- Sample size: The sample size of 50 students is large enough for the normal approximation to apply, as np = 50 * 0.5 = 25 and n(1-p) = 50 * 0.5 = 25 are both greater than 10.
b. To calculate the standardized test statistic, we first need to find the sample proportion:
p-hat = 32/50 = 0.64
Next, we calculate the standard error of the proportion:
SE = √(p*(1-p)/n) = √(0.5*0.5/50) = 0.071
Now, we can calculate the test statistic z:
z = (p-hat - p) / SE = (0.64 - 0.5) / 0.071 = 1.97
Since the alternative hypothesis is Ha: p > 0.5, we are performing a one-tailed test. Looking at the z-table or using a calculator, we find the critical value for a one-tailed test at α = 0.05 is approximately 1.645.
Since 1.97 > 1.645, we reject the null hypothesis and conclude that there is enough evidence to suggest that the proportion of students who prefer name-brand chips is greater than 0.5 at a significance level of 0.05.