Question

A nationwide poll suggest that 60% of Americans have consumed fast-food in the last week. In a small poll conducted in West Virginia (WV), one of the states in the US with the highest obesity rate, 36 out of 50 WV residents said they consumed fast-food in the last week. Which of the following is the correct set-up for calculating the p-value for a test evaluating whether the percentage of WV residents who consumed fast-food in the last week is higher than the national average.

A. Randomly sample 50 non-WV residents, and record how many consumed fast-food in the last week. Repeat this many times and calculate the proportion of samples where at least 72% of the respondents consumed fast-food in the last week.

B. Roll a 10-sided die 50 times and record the proportion of times you get a 6 or lower. Repeat this many times, and calculate the number of simulations where the sample proportion is 60% or more.

C. Roll a 10-sided die 50 times and record the proportion of times you get a 6 or lower. Repeat this many times, and calculate the number of simulations where the sample proportion is 72% or more.

D. In a bag place 50 chips, 36 red 14 white. Randomly sample 50 chips, with replacement, and record the proportion of red chips in the sample. Repeat this many times, and calculate the proportion of samples where at least 60% of the chips are red.

Answer 1

The hypotheses to test are

This is a left tailed test.

The test statistic is

Let the critical value be

The rejection region is  . That is

The P-value of the test is

The above probability is equivalent to calculating, the proportion of simulations (where the base probability of favourable even is 0.6 ) where the sample proportion is 72% or more.

So the correct set up is

C. Roll a 10-sided die 50 times and record the proportion of times you get a 6 or lower. Repeat this many times, and calculate the number of simulations where the sample proportion is 72% or more.

Here as required.