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Dr. Vegapunk thinks that watching anime (Japanese animated shows) decreases social skills in college students. To test this, Dr. Vegapunk randomly selected 30 brooklyn college students and assigned them to watch an episode of anime everyday for a week. After the week, each of the students answered a questionnaire about their social skills. The results showed that the sample had a mean social skills score of 7.3 and a standard deviation of 2.4. A previous study showed that the overall population of brooklyn college students had a mean social skills score of 6.2, but the standard deviation was not reported. Dr. Vegapunk decides to use an alpha level of 0.05. e) what is the obtained statistic

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Answer:

The test statistic = 2.51

Explanation:

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, Dr. Vegapunk wants to test the claim that watching anime decreases social skills in college students.

So, the null hypothesis would be that there isn't significant evidence to conclude that watching anime decreases social skills in college students. That is, watching anime does not decrease social skills in college students.

And the alternative hypothesis is that there is significant evidence to conclude that watching anime decreases social skills in college students.

If the previous known mean social skills of college students is 6.2 and the new population mean social skills of college students who watch anime is μ,

Mathematically,

The null hypothesis is represented as

H₀: μ ≥ 6.2

The alternative hypothesis is given as

Hₐ: μ < 6.2

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 mean = 7.3

μ₀ = Standard to be compared against = 6.2

σₓ = standard error of sample mean = [σ/√n]

where n = Sample size = 30

σ = Sample standard deviation = 2.4

σₓ = (2.4/√30) = 0.4382

t = (7.3 - 6.2) ÷ 0.4382

t = 2.51

Hope this Helps!!!

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