Question

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

Answer 1

Using a one-tailed t-test with the given sample mean, standard deviation, and sample size, the calculated t-statistic exceeds the critical t-value from the t-distribution table for α = .05. Thus, we reject the null hypothesis, supporting that the attraction rating with humor is significantly higher than neutral.

Step-by-step explanation:

To assess whether the average attractiveness rating of 4.53 is significantly higher than the neutral rating of 4, we must conduct a one-tailed t-test. The null hypothesis (H₀) asserts that there is no difference between the sample mean and the neutral rating, meaning μ = 4. The alternative hypothesis (H₁) is that the sample mean is greater than the neutral rating, which translates to μ > 4.

We are given the following information for the sample: sample mean (Μ) = 4.53, standard deviation (s) = 1.04, and sample size (n) = 16. Utilizing these values, we can calculate the t-statistic using the formula

t = (Μ - μ) / (s / √n)

Substituting the given values into the formula, we obtain:

t = (4.53 - 4) / (1.04 / √16) = 2.019

To determine the critical t-value, we refer to the t-distribution table for a one-tailed test with α = .05 and df = n - 1 = 15. From the table, the critical t-value is approximately 1.753.

Since the calculated t-statistic (2.019) is greater than the critical t-value (1.753), we reject the null hypothesis and conclude that the average attractiveness rating of the man with a great sense of humor is significantly higher than neutral at the α = .05 level of significance.

Answer 2

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

Null hypothesis is accepted

There is the average rate is less than μ ≤ 4

Step-by-step explanation:

Step(i):-

The Population of the mean 'μ' =4

sample size 'n' = 16

sample mean 'x⁻' = 4.53

given sample standard deviation 's' = 1.04

level of significance α = 0.05

Step(ii):-

Null hypothesis:H₀ : There is no significance difference between two means

Alternative hypothesis : H₁: There is significance difference between two means

Test statistic

`t = (x^(-) - mean)/((S)/(√(n) ) )`

`t = (4.53-4)/((1.04)/(√(16) ) )`

t = 2.038

Degrees of freedom ν = n-1 = 16-1 =15

t₀.₀₂₅ = 2.145

Conclusion:-

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

Null hypothesis is accepted

There is the average rate is less than μ ≤ 4