ECO502: Decision Making - Regression Statistics - Assessment Answer

Decision Making Assignment

Assignment Task

Question 1

1. a) For a series of random samples of 60, are the mean values of these randomsamples normally distributed? Explain

b) Calculate the standard error of the mean and explain the meaning of this value.

c) Determine the 95% confidence interval and explain its meaning in thecontext of the overall problem.

d) What is the probability that a sample of 60 hospitals selected at random in the Melbourne area will have a mean greater than 7000.00 admissions?

e) If the admissions times were more variable, what effect would this have on the confidence interval?

Question 2

Assume that the average admission for all hospitals in Melbourne is 7500. Conduct a statistical hypothesis test to determine if the admission of hospitals in Melbourne is significantly different from the average admission 6959. Mention any assumptions and include relevant hypotheses and report the results and conclusion in the conventional manner.

a) Write down both the null and alternative hypotheses

b) Carry out the t test and report the p-value, and the test statistic

c) Write an appropriate conclusion in the context of the problem.

Based on your answers to questions 1 and 2 please write a report of the effect of various variables on the number of admissions in Melbourne hospitals

Question 3

Most of the time houses prices depend on the local market conditions. In addition one of the factors is the number of bedrooms (as bedrooms increase prices increases). Recently Come Real Estate Agency has conducted a survey and selected a random sample of 211 for July 2015 sale in Melbourne and the data analyzed is summarized as follows.

a) Write down the regression equation.

b) State the R-squared value and the standard error and explain what they mean with respect to the data.

c) Write down the value of the gradient of the regression line and explain what it means for this data

d) Are the values for the constant and the gradient (slope) significant (i.e. significantly different from zero) in this case? Justify your answer.

e) Conduct a hypothesis test on the slope coefficient to test whether there is a linear relationship between number of bedrooms and prices of the houses. Include the null and alternative hypotheses; key test results and an appropriate conclusion.

f) Does the linear regression provide a good model? Give statistical reasons based on the scatterplot, p-values, the standard error and coefficient of determination.

g) If you were developing a model to predict the prices of the houses on the number of bedrooms, what other factors would you like to be able to include?

Question 1:

As per the Central Limit Theorem, the mean of the iterates of equal number of samples from the same population will be approximately normally distributed. Thus, for a series of random samples of 60, the mean values if these random samples normally distributed.

Mean; x? = 6959.00

Sample size; n = 60

Standard deviation; s = 6995.56

Standard error; SE =sn= 6995.5660=903.12

The standard error of 903.12 signifies that the variability between sample means that we will obtain if we take multiple samples from the same population is 906.12.

Since, the population standard deviation is unknown, we will use t-distribution to calculate the confidence interval for the mean admissions. The 95% confidence interval is:

Lower Confidence Limit = x? – SE * Z?/2 = 6959.00 – 903.12 * 2.00 = 5151.86

Upper Confidence Limit = x? + SE * Z?/2 = 6959.00 + 903.12 * 2.00 = 8766.14

We are 95% confident that the mean admissions in hospitals in the Melbourne area lies somewhere between 5151.86 and 8766.14.

The probability that a sample of 60 hospitals selected at random in the Melbourne area will have a mean greater than 7000.00 admissions is:

P(X > 7000) = 1 – P(X ? 7000)

P(X > 7000) = 1 – P(Z ? (7000 – 6959) / 903.12)

P(X > 7000) = 1 – P(Z ? 0.0454)

P(X > 7000) = 1 – 0.5180

P(X > 7000) = 0.4820

Thus, there is 48.20% probability that a sample of 60 hospitals selected at random in the Melbourne area will have a mean greater than 7000.00 admissions.

As the variance in the data increases, the width of the confidence interval also increases. Thus, if the admissions times were more variable, the confidence interval would be wider.

Question 2:

Null hypothesis: The average admissions of hospitals in Melbourne is equal to 7500.

Alternate hypothesis: The average admissions of hospitals in Melbourne is not equal to 7500.

Using t-distribution to test the above hypothesis with degrees of freedom = 59, at ? = 0.05.

Calculation of test statistics:

t-statistic=X-MeanStandard Error= 7500.00-6959.00903.12

t-statistic = –0.60

Using t-distribution table to find the probability for the above t-value:

p-value = 0.5514

Since, p-value is greater than the level of significance (0.05), we fail to reject the null hypothesis. Thus, there is insufficient evidence to conclude that the average admissions of hospitals in Melbourne is different from 7500.

Based on the above analysis, we are 95% confident the mean admissions of hospitals in Melbourne lies somewhere between 5151.86 and 8766.14. Further, there is high probability (48.20%) that a sample of 60 hospitals selected at random in the Melbourne area will have a mean greater than 7000.00 admissions. Since, we did not include and any variable in our analysis, we cannot make any significant conclusion regarding its effect on the number of admissions in Melbourne hospitals. However, we conducted a hypothesis test, based on which the average admissions in Melbourne hospitals does not appear to be significantly different from 7500.

Question 3:

The regression equation which can help in estimating the house prices in Melbourne (Y) based on the number of bedrooms (X) is:

Y = 178.63 X – 137.88

The R-squared of 0.6680 signifies that around 66.80% of variation in house prices in Melbourne can be estimated by the number of bedrooms. Further, the standard error of 115.81 signifies that the average distance of the observed values from the regression line.

The gradient of the regression line of 178.63 signifies that with each unit increase in the number of bedrooms, the house prices in Melbourne increase by $178,630.

In the regression equation used to predict the house prices in Melbourne based on the number of bedroom, the values for the constant is –137.88 and the gradient is 178.63. Further, the p-values for the constant and gradient mentioned in the regression analysis output is very low, which signifies that their values are significantly different from zero.

Null hypothesis: There is no relationship between the number of bedrooms and house prices.

Alternate hypothesis: There is a significant relationship between the number of bedrooms and house prices.

We will conduct a t-test on the slope coefficient of the regression equation in order to test the above hypothesis at ? = 0.05.

The appropriate test statistic and p-value is:

t stat = 18.93

p-value = 0.0000

Since, p-value is less than the level of significance (0.05), we will reject the null hypothesis. Thus, there is sufficient evidence to conclude that there is a significant relationship between the number of bedrooms and house prices.

The following points signifies that the linear regression provides a good model:

• The hypothesis test on the slope coefficient of the regression equation help us determine that there is a significant relationship between the number of bedrooms and house prices.
• Further, the correlation coefficient of 0.8173 signifies a strong positive correlation between the two variables.
• The moderate coefficient of determination (0.6680) and standard error (115.81) also supports the fitness of the regression model in this situation.

Some of the other factors which can help determine the price of houses along with the number of bedrooms are:

• The total area (in square feet) of the house;
• Parking facility;
• Location