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maxonik [38]
3 years ago
5

In studies examining the effect of humor on interpersonal attractions, McGee and Shevlin (2009) found that an individual’s sense

of humor had a significant effect on how the individual was perceived by others. In one part of the study, female college students were given brief descriptions of a potential romantic partner. The fictitious male was described positively as being single and ambitious and having good job prospects. For one group of participants, the description also said that he had a great sense of humor. For another group, it said that he has no sense of humor. After reading the description, each participant was asked to rate the attractiveness of the man on a seven-point scale from 1 (very unattractive) to 7 (very attractive). A score of 4 indicates a neutral rating. The females who read the "great sense of humor" description gave the potential partner an average attractiveness score of M = 4.53 with a standard deviation of s = 1.04. If the sample consisted of n = 16 participants, is the average rating significantly higher than neutral (μ = 4)? Use a one-tailed test with α = .05
Mathematics
1 answer:
son4ous [18]3 years ago
6 0

Answer:

<em>The calculated value t = 2.038< 2.145 at 0.05 level of significance</em>

<em>Null hypothesis is accepted </em>

<em>There is the average rate is less than  μ ≤ 4</em>

Step-by-step explanation:

<u><em>Step(i):-</em></u>

<em>The Population of the mean 'μ' =4</em>

<em>sample size   'n' = 16</em>

<em>sample mean 'x⁻' = 4.53</em>

<em>given sample standard deviation 's' = 1.04</em>

<em>level of significance  α = 0.05</em>

<u><em>Step(ii)</em></u><em>:-</em>

<u><em>Null hypothesis:H₀</em></u> : There is no significance difference between two means

<u>Alternative hypothesis : H₁</u><em>: There is significance difference between two means</em>

Test statistic

<em>                     </em>t = \frac{x^{-} - mean}{\frac{S}{\sqrt{n} } }<em />

<em>                    </em>t = \frac{4.53-4}{\frac{1.04}{\sqrt{16} } }

                t = 2.038

<em>Degrees of freedom ν = n-1 = 16-1 =15</em>

t₀.₀₂₅ = 2.145

<u><em>Conclusion</em></u>:-

<em>The calculated value t = 2.038< 2.145 at 0.05 level of significance</em>

<em>Null hypothesis is accepted </em>

<em>There is the average rate is less than  μ ≤ 4</em>

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