I have his task:
In the past a supermarket has sold an average of 50 pounds of macaroni and cheese per day. Each day of the week, the same average amount of macaroni and cheese is sold. Recently, the store has put a mouthwatering picture of macaroni and cheese on all shopping carts. In the 15 days since putting the picture on the carts, they have found $\bar{x}=56\,\mathrm{lb}$ and $s = 10\,\mathrm{lb}$. Let $\mu$ represent the average pounds sold per day since putting the picture on carts. The store wishes to test $H_o$: $\mu \leq 50$ against $H_a$: $\mu > 50$. For $\alpha = 0.05$, what do you conclude? Assume the relevant population follows a normal random variable.
I tried to get the p-value by calculating:
$$Z=\frac{50-56}{\frac{10}{\sqrt{15}}}$$
But that seems to be wrong? What's the correct calculation?
You almost have it but the numerator is backwards. I always tell my students "observed minus expected." You observe $56$ and expect $50$.
The calculations are the same. The test to use is a T test since $\sigma$, the population standard deviation, is unknown.
The test statistic is $2.32379$. On $14$ degrees of freedom, we reject the null hypothesis.