A company that stocks shelves in supermarkets is considering expanding the supply that it delivers. Items that are not sold must be discarded at the end of the day, so it only wants to schedule additional deliveries if stores regularly sell out. A break-even analysis indicates that an additional delivery cycle will be profitable if items are selling out in more than 59% of markets. A survey during the last week in 48 markets found the shelves bare in 32.

Find the p-value of the test. Do the data reject the null hypothesis if the significance level is 0.05?

(Round to four decimal places as needed.)

The objective is to test if it is profitable to expand supply delivery. The company thinks that if more than 59% (symbolically p > 0.59) of the items are selling out in the markets, then it is profitable to increase the deliveries.

A sample of 48 markets was taken and it was registered that the item was sold out in 32 of them.

The study variable is.

X: Number of markets where the item was sold out in a random sample of 48 markets.

The study parameter is the proportion of "bare shelves"

sample proportion 'p= (32/48) = 0.67

The hypothesis is:

H₀: p ≤ 0.59

H₁: p > 0.59

α: 0.05

Remember: The p-value is defined as the probability corresponding to the calculated statistic if possible under the null hypothesis (i.e. the probability of obtaining a value as extreme as the value of the statistic under the null hypothesis).

So, to calculate the p-value you have to first calculate the statistic under the null hypothesis:

[tex]Z= \frac{'p - p}{\sqrt{\frac{p(1 - p)}{n} } }[/tex]

[tex]Z= \frac{0.67 - 0.59}{\sqrt{\frac{0.59*0.41}{48} } }[/tex]

Z= 1.1269≅ 1.13

Keep in mind that the p-value as the test is one-tailed. Now you can calculate the p-value as:

P(Z ≥ 1.13)= 1 - P(Z < 1.13)= 1 - 0.8706 =0.1292

The decision is to reject the null hypothesis. So at a level of 5% you can say that it is probitable to increase the deliveries.

I hope you have a SUPER day!