Question

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.)

Answer 1

Answer:

p-value: 0 .1292

Step-by-step explanation:

Hello!

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:

$Z= \frac{'p - p}{\sqrt{\frac{p(1 - p)}{n} } }$

$Z= \frac{0.67 - 0.59}{\sqrt{\frac{0.59*0.41}{48} } }$

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!