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AnnyKZ [126]
3 years ago
6

__________We can be 99% confident that the true percentage of U.S. adults Twitter users who get some news through Twitter is bet

ween the upper and lower bounds of the confidence interval.
Mathematics
1 answer:
irakobra [83]3 years ago
7 0

First part of question states;

A poll conducted in 2013 found that the proportion of of U.S. adult Twitter users who get at least some news on Twitter was 0.52 The standard error for this estimate was 0.024, and a normal distribution may be used to model the sample proportion.

Answer:

<u>True</u>

<u>Step-by-step explanation:</u>

From the statement above and after constructing the 99% confidence interval for the proportion of U.S. adult Twitter users we can infer that the 95% confidence upper and lower bounds are 0.52 and 0.79, which means that there's a 99% confident that the true percentage of U.S. adults Twitter users who get some news through Twitter is between the upper and lower bounds of the confidence interval.

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Answer:

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Step-by-step explanation:

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cross-multiply:

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6.81 / 11 = x

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2 years ago
The independent-samples t testResearch Scenario: A forensic psychologist has developed a new program that teaches empathy and po
8_murik_8 [283]

Answer:

Step-by-step explanation:

Hello!

The researcher developed a treatment to teach social skills to youth offenders. To test if the treatment is effective in increasing empathy compared to the standard treatment she randomly selected a group of 9 offenders and applied the new treatment and to another group of 9 randomly selected youth offenders, she applied the standard treatment. (Note: the data corresponds to two samples of 9 units each, so I've used those sizes to conduct the test)

At the end of the treatment, she administers BES to measure their empathy levels. Her claim is that the offenders that received the new treatment will have higher BES scores than those who received the standard treatment.

1) Using the records obtained for both groups, she  intends to conduct an independent t-test to analyze her claim.

X₁: BES results of a youth offender treated with the new treatment.

X₂: BES results of a youth offender treated with the standard treatment.

H₀: μ₁ = μ₂

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α:0.05

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t_{H_0}= -0.32

p-value: 0.7517

The p-value is greater than the significance level so the decision is to not reject the null hypothesis. This means that at a 5% significance level you can conclude that there is no difference between the mean BES scores of the youth offenders treated with the new treatment and the mean BES score of  the youth offenders treated with the standard treatment. The new treatment doesn't increase the levels of social empathy of the youth offenders.

I hope this helps

(Box plot in attachment)

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3 years ago
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Step-by-step explanation:

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2 years ago
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