A small hair salon in Denver, Colorado, averages about 62 customers on weekdays with a standard deviation of 16. It is safe to assume that the underlying distribution is normal. In an attempt to increase the number of weekday customers, the manager offers a $3 discount on 9 consecutive weekdays. She reports that her strategy has worked because the sample mean of customers during this 9-weekday period jumps to 74.

a) What is the null hypothesis in this scenario?
b) What is the alternative hypothesis in this scenario?
c) Calculate the z-score for the sample mean of 74 customers.
d) Based on the z-score calculated, can we reject the null hypothesis at a significance level of 0.05?