beiachapman122
05.05.2020 •
Mathematics
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).
Solved
Show answers
More tips
- G Goods and services How to Make Soap at Home: Useful Tips for Beginners...
- C Computers and Internet How to Reinstall Windows: A Detailed Guide for Beginners...
- S Style and Beauty How to Get Rid of a Bruise: Tips and Tricks...
- F Food and Cooking Лечо: вкусное и простое блюдо для любой кухни...
- H Health and Medicine Relieving Swelling in Legs: Causes and Ways to Alleviate the Symptom...
Answers on questions: Mathematics
- M Mathematics HOLY Fu*ck PLEASE HELP ME!...
- M Mathematics How many do u wanna bet nobody on this app is not going to help me on this question...
- M Mathematics PLEASE HELP! I’ll give brainliest for the correct answer :) An undefined term is a term that has been given no formal definition. In geometry, there are three undefined terms that...
- E English Augmentive communication tools children do what? ( parenting skills on virtual school)...
- E English Reescreva o período: A cidade avisou que crianças, idosos e pessoas com sistema imunológico debilitado usando o Present Perfect, Em seguida, descreva a diferença semântica (de sentido)...
- E English A supporting text may be the opening paragraph of a case study. the concluding paragraph of a case study. comparing and contrasting a case study. a quote providing evidence in a...
- M Mathematics 10 - 7b 10and5b-6 -11solve each compound inequality...
Ответ:
1) The null and alternative hypothesis are:
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:
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.
The pooled variance is calculated as:
The estimated standard error of the difference between means is computed using the formula:
Then, we can calculate the t-statistic as:
The degrees of freedom for this test are:
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):
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.
Ответ:
Rate of change in elevation = 0.6 in/year
Step-by-step explanation:
Note:
Current elevation (Missing) = 7,602 feet
Given:
Old elevation = 7,602 feet
Number of year = 7,600
Find:
Rate of change in elevation
Computation:
Change in elevation = 7,602 - 7,600
Change in elevation = 2 ft
Change in elevation = 2 x 12 = 24 inches
Rate of change in elevation = 24 / 40
Rate of change in elevation = 0.6 in/year