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Joe Mama is the Chief Marketing Officer of a major investment firm and is looking to provide several new investment opportunities to his customers. Since this will be a major project for his firm, he wants to be sure that at least 70% of his customers will be interested in buying. He presents the investments to a random sample of 150 of his customers and determines whether they would be interested in buying. He coded their response as either 1 = "likely to buy" or 0 = "unlikely to buy." 117 customers indicated they would be interested in buying, 33 said they would not. "What should Joe Mama conclude about these investment opportunities"?

User Funmi
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Answer:

Joe Mama can conclude that there is significant evidence to conclude that more than 70% of the total population of customers will be interested in buying into the new investment opportunity.

Explanation:

We can check for the right conclusion from the results of this sample with an hypohesis test or finding the confidence interval of the population proportion that would be willing to buy into the new investment opportunity.

Using the hypothesis test method

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, the null hypothesis is that there is no significant evidence to conclude that more than 70% of the total population of customers will be interested in buying into the new investment opportunity.

The alternative hypothesis will now be that there is significant evidence to conclude that more than 70% of the total population of customers will be interested in buying into the new investment opportunity.

Mathematically,

The null hypothesis is represented as

H₀: p ≤ 0.70

The alternative hypothesis is given as

Hₐ: p > 0.70

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 = p = sample proportion that are interested in buying into.the new investment opportunity = (117/150) = 0.78

μ = p₀ = the benchmark we're comparing against = 0.70

σₓ = standard error = √[p(1-p)/n]

where n = Sample size = 150

σₓ = √[0.78×0.22/150] = 0.0338

t = (0.78 - 0.70) ÷ 0.0338

t = 2.37

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

Degree of freedom = df = n - 1 = 150 - 1 = 149

Significance level = 0.05 (most tests are performed at this significance level)

The hypothesis test uses a one-tailed condition because we're testing only in one direction.

p-value (for t = 2.37, at 0.05 significance level, df = 149, with a one tailed condition) = 0.009534

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 = 0.009534

0.009534 < 0.05

Hence,

p-value < significance level

This means that we reject the null hypothesis, accept the alternative hypothesis & say that there is significant evidence to conclude that more than 70% of the total population of customers will be interested in buying into the new investment opportunity.

Hope this Helps!!!

User Etpinard
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