Respuesta :
Answer:
1) The null and alternative hypothesis are:
[tex]H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2\neq 0[/tex]
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:
[tex]H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2\neq 0[/tex]
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.
[tex]M_d=M_1-M_2=4.3-5.4=-1.1[/tex]
The pooled variance is calculated as:
[tex]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[/tex]
The estimated standard error of the difference between means is computed using the formula:
[tex]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[/tex]
Then, we can calculate the t-statistic as:
[tex]t=\dfrac{M_d-(\mu_1-\mu_2)}{s_{M_d}}=\dfrac{-1.1-0}{0.395}=\dfrac{-1.1}{0.395}=-2.782[/tex]
The degrees of freedom for this test are:
[tex]df=n_1+n_2-1=13+17-2=28[/tex]
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):
[tex]P-value=2\cdot P(t<-2.782)=0.0096[/tex]
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.
