Question

A marketing team designed a promotional web page to increase online sales. Visitors to the home page were randomly directed to the old page or the newly designed page. During a day of tests, 160 customers were randomly assigned. The 80 customers who were directed to the old page spent $138 on average, while those who went to the new page spent $166 on average.

a. What are the correct null and alternative hypotheses to answer the question, "DO the data provide evidence that the newly designed webpage increased sales?"

b. If the marketing team makes the appropriate conclusion for a 0.05 and p-value 0.042, but this conclusion ends up being wrong, which error was made?

1. They concluded there was insufficient evidence that the new webpage increased sales, when in reality the new webpage did increase sales.
2. They concluded the new webpage increased sales, when in reality the new webpage did not increase sales.
3. They concluded there was insufficient evidence that the new webpage increased sales, when in reality the new webpage did not increase sales.
4. They concluded the new webpage increased sales, when in reality the new webpage did increase sales.

Answer 1

a) null and alternative hypothesis:

H0: μₙ - μₒ = 0

Ha: μₙ - μₒ > 0

b) 2. They concluded the new webpage increased sales, when in reality the new webpage did not increase sales

Explanation:

The null hypothesis (H0) tries to show that no significant variation exists between variables or that a single variable is no different than its mean. While an alternative Hypothesis (Ha) attempt to prove that a new theory is true rather than the old one. That a variable is significantly different from the mean.

For the case above;

Let μₙ and μₒ represent the average sales on the new and old page.

Null hypothesis is that there is no difference between the average sales on the new and old page.

H0: μₙ - μₒ = 0

Alternative w is that the average sales on the new page is greater than that of the old page.

Ha: μₙ - μₒ > 0

b) Decision Rule

If;

P-value > significance level --- accept Null hypothesis

P-value < significance level --- reject Null hypothesis

Z score > Z(at 95% confidence interval) ---- reject Null hypothesis

Z score < Z(at 95% confidence interval) ------ accept Null hypothesis

Given that, for a 0.05 and p-value 0.042

P-value < 0.05

Therefore, there is enough evidence to reject the null hypothesis. Which implies that, they concluded the new webpage increased sales. Since they are wrong, that means in reality the opposite is true.