Chicken Delight claims that 90% of its orders are delivered within 10 minutes of the time the order is placed. A sample of 100 orders revealed that 82 were delivered within the promised time. At the 0.10 significance level, can we conclude that less than 90% of the orders are delivered in less than 10 minutes?

I think so but wouldn't you need a higher number of orders and at say the same time of day and the same order etc etc to make it fair

Either a delivery is on time (<= 10 minutes), or it isn't, so assume this is a binomial distribution with n=100, p=.9, q=1-p=.1

Then, for a binomial distribution,
mu=np=90, and s=SQRT(npq)=SQRT(100*.9*.1)=3

The sampling distribution of XBAR is a t distribution (because we used a sample standard deviation s), but because n is large we can approximate it with a normal distribution.

We want P[ z = (XBAR-mu)/(s/SQRT(n)) < critical value of z ] = .90

The critical value of z is 1.282 from normal table.

Is (XBAR-mu)/(s/SQRT(n)) < 1.282 ?

XBAR = .82 < 1.282 * (s/SQRT(n)) + mu

.82 < 1.282 * (3/SQRT(100)) + .82 = 1.2046

So since .82 is less than 1.2046, we conclude that less than 90% of orders are delivered in less than 10 minutes.