Mathematics, 23.07.2019 16:10 mikaylarichardson96

Na study of speed dating, male subjects were asked to rate the attractiveness of their female dates, and a sample of the results is listed below (1equalsnot attractive; 10equalsextremely attractive). construct a confidence interval using a 95% confidence level. what do the results tell about the mean attractiveness ratings of the population of all adult females? 7, 7, 3, 8, 5, 6, 6, 9, 9, 8, 6, 9 what is the confidence interval for the population mean mu? 5.8less thanmuless than 8.1 (round to one decimal place as needed.) what does the confidence interval tell about the population of all adult females? select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice. a. we are 95% confident that the interval from nothing to nothing actually contains the true mean attractiveness rating of all adult females. (round to one decimal place as needed.) b. the results tell nothing about the population of all adult females, because participants in speed dating are not a representative sample of the population of all adult females. c. we are confident that 95% of all adult females have attractiveness ratings between nothing and nothing. (round to one decimal place as needed.)

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Answer from: kameahtravis

The confidence interval is $\fbox{(5.8, 8.1)}$ and $\fbox{\text{Option A}}$ is correct.

Further Explanation:

Given:

The least value is that is the least attractive.

The highest value is that is the most attractive.

The observations are,

7, 7, 3, 8, 5, 6, 6, 9, 9, 8, 6, 9.

Calculation:

The sum of all observations is .

The population mean is $\mu$ .

The standard deviation is 1.832 .

The sample mean $\bar^{X}$ is 6.92 .

Level of significance $\alpha$ = $5\%$ .

Formula for confidence interval = $\left( \bar{X} \pm t_{n-1, \frac{\alpha}{2}\%} \frac{s}{\sqrt{n}} \right)$

Confidence interval = $\left( 6.92 \pm t_{12-1, \frac{5}{2}\%} \frac{1.832}{\sqrt{12}} \right)$

The value of $t_{11, \frac{5}{2}\%$ = 2.201

Confidence interval = $( 6.92 \pm 2.201}\times \frac{1.832}{\sqrt{12}}) \right)$

Confidence interval = $( 6.92 - 2.201}\times 0.5288 ,6.92 + 2.201}\times \0.5288) \right)$

Confidence interval = (6.92-1.1639,6.92+1.1639)

Confidence interval = $\fbox{(5.8, 8.1)}$

The $95\%$ confidence interval gives us an idea that $95\%$ chances of the true mean or population mean lies in the interval.

A. We are $95\%$ confident that the interval from 5.8 to 8.1 actually contains the true mean of attractiveness rating of all adult females.

B. The results tell nothing about the population of all adult females, because participants in speed dating are not a representative sample of the population of all adult females.

C. We are confident that $95\%$ of all adult females have attractiveness ratings between 5.8 and 8.1 .

$\fbox{\text{Option A}}$ is Correct as we are $95\%$ confident that the interval from 5.8 to 8.1 actually contains the true mean of attractiveness rating of all adult females.

Option B is not correct as the confidence interval tells us about the population mean.

Option C is not correct as the individual rating can be more than 5.8 or less than the 8.1 .

Learn More:

1. Learn more about collinear link

2. Learn more about binomial term link

Answer Details:

Grade: College Statistics

Subject: Mathematics

Chapter: Confidence Interval

Keywords:

Probability, Statistics, Speed dating, Females rating, Confidence interval, t-test, Level of significance , Normal distribution, Central Limit Theorem, t-table, Population mean, Sample mean, Standard deviation, Symmetric, Variance.

Answer from: Quest

step-by-step explanation:

Answer from: Quest

step-by-step explanation:

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