The chips project Zenon decided to investigate whether students at his school prefer name-brand potato chips to generic potato chips. He randomly selected 50 students and had each student try both types of chips, in random order. Overall, 32 of the 50 students preferred the name-brand chips. Zenon wants to perform a test at the α = 0.05 significance level of H0: p = 0.5 versus Ha: p > 0.5, where = the proportion of all students at Zenon’s school who prefer name-brand chips. a. Check that the conditions for performing the test are met. b. Calculate the standardized test statistic.
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.