Answer:
We conclude that cheddar popcorn weighed less than 5.5 ounces.
Step-by-step explanation:
We are given the following in the question:
Population mean, μ =5.5 ounces
Sample mean,
= 5.23 ounces
Sample size, n = 64
Alpha, α = 0.05
Sample standard deviation, σ = 0.24 ounce.
First, we design the null and the alternate hypothesis

We use One-tailed z test to perform this hypothesis.
Formula:

Putting all the values, we have

Now, 
Since,

We reject the null hypothesis and accept the alternate hypothesis. Thus, we conclude that cheddar popcorn weighed less than 5.5 ounces.