Question

A theater arts department enlisted the help of a social psychology professor to design a study to see if being surrounded by highly attractive people affected the performance of young actors. Thirty student actors were randomly assigned to act a particular scene under one of two conditions: For 13 of the actors, the other performers were dressed and made up to look very attractive; for the other 17 actors, the other performers were dressed and made up to look very unattractive. A panel of judges rated the performance of the two groups, with higher scores indicating a better performance, and the mean performance ratings for the actors with attractive co-performers was 4.3 (SD = .86) and with unattractive co-performers, the mean was 5.4 (SD = 1.30). Using the .05 significance level, did acting with attractive co-performers affect performance?Step 1: State Hypotheses (2 points)Null:Research:Step 2: Determine Comparison Distribution (1 point)What is the mean and standard deviation of the comparison distribution?Step 3: Set the Criteria for a Decision (3 points)One-tailed or two-tailed?df =Critical value(s):Step 4: Compute the Test Statistic (6 points)Compute the pooled estimate of the population variance (round to 2 decimal places)Compute the estimated standard error (also called the standard deviation of the distribution of differences between means). (round to 2 decimal places)Compute the t-statistic (round to 2 decimal places)Step 5: Make a Decision (3 points)Reject/Fail to Reject the null?Write your results as they would appear in a research journal (one sentence).

Answer 1

Answer:

1) The null and alternative hypothesis are:

$H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2\neq 0$

being μ1: mean performance score for actors with attractive partners, and μ2: mean performance score for actors with unattractive partners.

- The mean (difference of sample means) is M_d=-1.1.

- The standard error is S_md=0.395.

- This is a two-tailed test, as we are claiming a significative difference (higher or lower).

- Degrees of freedom df=28.

- Pooled estimation of the population variance s_p=1.05.

- Test statistic t=-2.78.

Decision: Reject the null hypothesis (P-value=0.0096).

Conclusion: There is enough evidence to support the claim that being surrounded by highly attractive people affects the performance of young actors.

Step-by-step explanation:

This is a hypothesis test for the difference between populations means.

The claim is that being surrounded by highly attractive people affects the performance of young actors.

Then, the null and alternative hypothesis are:

$H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2\neq 0$

The significance level is 0.05.

The sample 1 (attractive partners), of size n1=13 has a mean of 4.3 and a standard deviation of 0.86.

The sample 2 (unattractive partners), of size n2=17 has a mean of 5.4 and a standard deviation of 1.3.

The difference between sample means is Md=-1.1.

$M_d=M_1-M_2=4.3-5.4=-1.1$

The pooled variance is calculated as:

$s_p=\sqrt{\dfrac{n_1s_1+n_2s_2}{n_1+n_2}}=\sqrt{\dfrac{13*0.86+17*1.3}{30}}=\sqrt{\dfrac{33.28}{30}}=\sqrt{1.1093}=1.05$

The estimated standard error of the difference between means is computed using the formula:

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{0.86^2}{13}+\dfrac{1.3^2}{17}}\\\\\\s_{M_d}=\sqrt{0.057+0.099}=\sqrt{0.156}=0.395$

Then, we can calculate the t-statistic as:

$t=\dfrac{M_d-(\mu_1-\mu_2)}{s_{M_d}}=\dfrac{-1.1-0}{0.395}=\dfrac{-1.1}{0.395}=-2.782$

The degrees of freedom for this test are:

$df=n_1+n_2-1=13+17-2=28$

This test is a two-tailed test, with 28 degrees of freedom and t=-2.782, so the P-value for this test is calculated as (using a t-table):

$P-value=2\cdot P(t<-2.782)=0.0096$

As the P-value (0.0096) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that being surrounded by highly attractive people affects the performance of young actors.