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Alik [6]
3 years ago
5

A test preparation claims that more than​ 50% of the students who take their test prep course improve their scores by at least 1

0 points. Instead of advertising the percentage of customers who improve by at least 10​ points, a manager suggests testing whether the mean score improves at all. For each customer they record the difference in score before and after taking the course.
Required:
a. State the null and alternative hypotheses.
​b. The​ P-value from the test is 0.65. Does this provide any evidence that their course​ works? ​
c. From part​ b, what can you​ tell, if​ anything, about the mean difference in the sample​ scores?
Mathematics
1 answer:
satela [25.4K]3 years ago
3 0

Answer:

Step-by-step explanation:

Corresponding scores before and after taking the course form matched pairs.

The data for the test are the differences between the scores before and after taking the course.

μd = scores before taking the course minus scores before taking the course.

a) For the null hypothesis

H0: μd ≥ 0

For the alternative hypothesis

H1: μd < 0

b) We would assume a significance level of 0.05. The​ P-value from the test is 0.65. The p value is high. It increases the possibility of accepting the null hypothesis.

Since alpha, 0.05 < than the p value, 0.65, then we would fail to reject the null hypothesis. Therefore, it does not provide enough evidence that scores after the course are greater than the scores before the course.

c) The mean difference for the sample scores is greater than or equal to zero

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Answer:

D. The small p value and the small effect size suggest that the observed difference between the drug placebo groups are likely to be due to sampling error. This study should be replicated with a larger sample size.

Step-by-step explanation:

The p value is determined for any test before making a conclusion. If the p-value is smaller than critical value then we reject the null hypothesis which means that drug is not effective for the reducing depression. A very small p-value usually lesser than 0.05 indicates strong evidence against the null hypothesis.

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Question 2 (Essay Worth 10 points)
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Answer:

(a) f(2) = 81

(b) g(2) = 83

(c) Test average for maths class after test 2 is greater

Step-by-step explanation:

Given

f(x) = 0.5x + 80

x \to g(x)

1 \to 81

2 \to 83

3 \to 85

Solving (a): f(2)

We have:

f(x) = 0.5x + 80

f(2) = 0.5*2+80

f(2) = 1 + 80

f(2) = 81

Solving (b): g(2)

From the table:

g(x) = 83 when x = 2

So:

g(2) = 83

Solving (c): Which is greater f(2) or g(2)

In (a) and (b),

f(2) = 81

g(2) = 83

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The number of classes Anna can take so the total cost for the month will be the same is 5.

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<h3>When would the total cost be the same?</h3>

When the monthly cost is equal, both equations would be equal: 7.5x = 5.5x + 10

In order to determine the value of x, take the following steps:

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Monthly cost when the cost is the same: 7.5 x 5 = $37.50

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