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 49 markets found the shelves bare in 34. Required:
a. State the null and alternative hypotheses.
b. Describe a Type I error and a Type Il error in this context.
c. Find the p-value of the test. Do the data supply enough evidence to reject the null hypothesis if the α-level is 0.05?

a) If p represents the proportion of markets in which the items sell out.

The null hypothesis is represented as

H₀: p ≤ 0.59

The alternative hypothesis is represented as

Hₐ: p > 0.59

b) In this context, a type I error would be concluding that the items are selling out in more than 59% of the markets when they are not in more than 59% of the markets.

While, a type II error would be concluding that the items aren't selling out in more than 59% of the markets when they are actually selling out in more than 59% of the markets

c) p-value = 0.060337

The p-value obtained is greater than the significance level at which the test was performed, hence, we fail to reject the null hypothesis & say that there is enough evidence to suggest that the items are selling out in more than 59% of markets.

Step-by-step explanation:

a) For hypothesis testing, the first thing to define is the null and alternative hypothesis.

The null hypothesis plays the devil's advocate and usually takes the form of the opposite of the theory to be tested. It usually contains the signs =, ≤ and ≥ depending on the directions of the test.

While, the alternative hypothesis usually confirms the the theory being tested by the experimental setup. It usually contains the signs ≠, < and > depending on the directions of the test.

For this question, we want to find out if items are selling out in more than 59% of markets.

So, the null hypothesis would be that there isn't significant evidence to suggest that the items are selling out in more than 59% of markets. That is, the items aren't selling out in more than 59% of the markets.

The alternative hypothesis is that there is significant evidence to suggest that the items are selling out in more than 59% of markets.

Mathematically, if p represents the proportion of markets in which the items sell out.

The null hypothesis is represented as

H₀: p ≤ 0.59

The alternative hypothesis is represented as

Hₐ: p > 0.59

b) In Hypothesis testing, a type I error involves rejecting the null hypothesis and accepting the alternative hypothesis when in reality, the null hypothesis is true.

WHILE

A type II error involves failing to reject the null hypothesis when in reality it should have been rejected because the null hypothesis is false.

For this question, a type I error would be concluding that the items are selling out in more than 59% of the markets when they are not in more than 59% of the markets.

While, a type II error would be concluding that the items aren't selling out in more than 59% of the markets when they are actually selling out in more than 59% of the markets.

c) To do this test, we will use the t-distribution because no information on the population standard deviation is known

So, we compute the t-test statistic

t = (x - μ)/σₓ

x = sample proportion of markets where the items are selling out = (34/49) = 0.694

μ = p₀ = The standard we are comparing against = 0.59

σₓ = standard error = √[p(1-p)/n]

p = sample proportion = 0.694

n = Sample size = 49

σₓ = √[0.694×0.306/49] = 0.0658328125 = 0.06583

t = (0.694 - 0.59) ÷ 0.06583

t = 1.5798268267 = 1.58

checking the tables for the p-value of this t-statistic

Degree of freedom = df = n - 1 = 49 - 1 = 48

Significance level = 0.05

The hypothesis test uses a one-tailed condition because we're testing only in one direction (whether the true proportion is greater than 0.59)

p-value (for t = 1.58, at 0.05 significance level, df = 48, with a one tailed condition) = 0.060337

The interpretation of p-values is that

When the (p-value > significance level), we fail to reject the null hypothesis and when the (p-value < significance level), we reject the null hypothesis and accept the alternative hypothesis.

So, for this question, significance level = 0.05

p-value = 0.060337

0.060337 > 0.05

Hence,

p-value > significance level

This means that we fail to reject the null hypothesis & say that there is enough evidence to suggest that the items are selling out in more than 59% of markets.

Hope this Helps!!!

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