Using the following sample data; 6, 7, 11, 6, 11, 5, 15, 11, 5;

Compute the sample standard deviation using either the computing formula or the defining formula.

Given are five observations for two variables,  and .

The estimated regression equation for these data is y = 70.84 - 2.83x

a. Compute SST, SSE , and SSR.

b. Compute the coefficient of determination r^2. Comment on the goodness of fit.

(to 3 decimals)

The least squares line provided an - Select your answer -goodbadItem 5 fit;   of the variability in  has been explained by the estimated regression equation (to 1 decimal).

c. Compute the sample correlation coefficient. Enter negative value as negative number.

(to 3 decimals)

El documento de Excel anexo presenta las estadísticas de criminalidad en una ciudad. También se presentan otros datos importantes a cerca de educación. El propósito de este ejercicio es crear dos modelos de regresión lineal múltiple donde se trate de predecir: a) Y1 usando como predictores X3,X5,X6 b) Y2 usando como predictores X3,X4,X7 En cada caso se necesita: 1. El modelo (todos los coeficientes beta) y la interpretación de cada coeficiente. 2. Cuan significativos son cada uno de los coeficientes 3. El coeficiente de determinación del modelo (R cuadrado) 4. La interpretación de R cuadrado 5. En el caso (a) prediga: Cuál será la tasa de crímenes totales reportados por milón de habitantes si se asignan 50 dólares anuales por habitante a la policía, hay un 10% de jóvenes entre 16 y 19 años que no asisten a la escuela superior (ni la han finalizado) y hay un 50% de jóvenes entre 18 y 24 años que asisten a la universidad. 6. En el caso (b) prediga: Cuántos crímenes de violencia se reportarán si se asignan 20 dólares anuales por habitante a la policía, hay 60% de personas de más de 25 años que finalizaron la escuela superior y hay un 5% de personas de 25 años o más que lograron una carrera universitaria de 4 años. 7. Luego de hacer todo este análisis arroje conclusiones prácticas acerca de los hallazgos hechos en esta ciudad. 8. Si usted es un consejero para las autoridades de esa ciudad, por favor escriba un parrafo de recomendaciones a seguir para tratar de reducir la criminalidad. ABAJO APARECEN CIERTAS FÓRMULAS QUE LE PUEDE SER DE UTILIDAD, AUNQUE LA RECOMENDACIÓN QUE RESUELVA TODO EL PROBLEMA USANDO R Y/O EXCEL PARA EL MISMO.

Describe in 175 words please type response:

Describe statistical inference and how it corralates with hypothesis testing for single populations.

Describe in 175 words please type response:

Describe how decision making is done using one sample hypothesis testing.

Suppose that you decide to randomly sample people ages 18-24 in your county to determine whether or not they are registered to vote. In your sample of 50 people, 35 said they were registered to vote. a) (2 points) Find a 95% confidence interval for the true proportion of the county population ages 18-24 who are registered to vote. Make sure to check any necessary conditions and to state a conclusion in the context of the problem. Also, explain what 95% confidence means in this context. b) (1 point) What is the probability that the true proportion of people ages 18-24 who registered to vote in your county is in your particular confidence interval? (Note: Be careful). c) (1 point) According to a separate news report, about 73% of 18- to 24-year-olds in the same county said that they were registered to vote. Does the 73% figure seem reasonable with your own poll? Explain. d) (1 point) Assume you have not done your poll yet, but you knew the news report poll results. In designing your poll now, you want separately estimate the same percentage to within ±4 percentage points with 95% confidence, how many people should you poll?

Consider a sample with data values of 26, 25, 20, 15, 31, 33, 29, and 25. Compute the 20th, 25th, 65th, and 75th percentiles.

20th percentile

25th percentile

65th percentile

75th percentile

4. One thousand cars were stopped at random for a roadside test of tyres and lights: 115 had unroadworthy tyres; and 46 had a lighting fault. These figures include 22 cars with both defects. If drivers had been fined $100 if their car had one defect, and $500 if their car had both defects, how much revenue would have been raised per car?

According to the latest financial reports from a sporting goods store, the mean sales per customer was $75 with a population standard deviation of $6. The store manager believes 39 randomly selected customers spent more per transaction.

What is the probability that the sample mean of sales per customer is between $76 and $77 dollars?

You may use a calculator or the portion of the z -table given below. Round your answer to two decimal places if necessary.

$\mu_{\overline{x}}=$ $

sigma sub line segment x is equal to $\sigma_{\overline{x}}=$

$

cap p times open paren 76 is less than or equal to line segment x comma line segment x is less than or equal to 77 close paren is equal to $P\left(76\le\overline{x}\le77\right)=$

Assume adult IQ scores are normally distributed with a mean of 100 and a standard deviation of 15

a) What is the probability that a randomly selected adult has an IQ that is less than 115

b) Find the probability that an adult has an IQ greater than 131.5 (requirement to join MENSA)

c) Find the probability that a randomly selected adult has an IQ between 110 and 120

d) Find the IQ separating the top 15% from the others e) Find the IQ score separating the bottom 10% from the others

In a statistic class, 11 scores were randomly selected with the following results were obtained: 68, 74, 66, 37, 52, 71, 90, 65, 76, 73, 22. What are the outer fences?

A)-6.0, 140.0

B)-10.0, 92.0

C)2.0, 128.0

D)2.0, 162.0

E)37.0, 107.0

This question will be perfomed entirely in R. Consider the following sample from an unkown distribution:

sample_data <- c(1.15, 0.5, 28.03 , 0.085, 1.82, 25.30, 0.7, 0.02, 0.01 ,13.23)

1.a) Calculating the p-value using the bootstrap hypothesis testing method to determine whether the mean is greater than 1 (one-sided test). Use 10,000 bootstrap samples. Set the seed to 248 before beginning the bootstrap process, i.e. include this line of code at the beginning of your script.

set.seed(248)
Include a histogram of the generated bootstrap samples of X̄⋆, does it appear to be symmetric? Do you

reject the null at 0.05 significance level? (10 points)
1.b) Find the rejection region (for X̄) based on your bootstrap samples at a 0.05 significance level. (5 points)

1.c) Perform a t-test for the same hypothesis as in 1.a) using the t.test function in R. Make sure that you apply the correct arguments.

Do you reject the null at a 0.05 significance level based on the t-test?

Compare the p-value from 1.a) to the p-value from the t-test, do they imply different conlcusion? Which p-value would you trust more? Support your anwser. (10 points)

According to the college board, scores by women on the SAT-I test were normally distributed with a mean of 998 and standard deviation of 202. Score by women on the ACT test are normally distributed with a mean of 20.9 and a standard deviation of 4.6. Assume that the two tests use different scales to measure the same aptitude

a) If a woman gets an SAT-I score in the 67thh percentile, find her actual SAT-I score and her equivalent ACT score

b) If a woman gets an SAT score of 1220, find her equivalent ACT score

Answer True or False for the following:

1. Chi-square requires assumptions about the shape of the population distribution from which a sample is drawn.

2. Goodness of fit refers to how close the observed data are to those predicted from a hypothesis.

3. The null hypothesis (H0) states that no association exists between the two cross-tabulated variables in the population, and therefore the variables are statistically independent.

4. High chi square values indicate a high probability that the observed deviations are due to random chance alone

5. A scatter plot shows the direction of a relationship between the variables.