This study is based on a health agency that is taking a survey in order to evaluate all the hospitals in Melbourne. Statistical calculation and analyzing the attributes of the data will be done. The hospital agency seeks to obtain a reportgiving the number of admissions, the type of control, and the type of service. The report will focus onthe provision of services by each hospital. Hospital categories; non-government, non-federal, hospitals for profitand for federal government will be examined.
In order to facilitate the analytical part of the study, a sample of 60 admissions will be randomly selected out of the total population of 300.Landau & Shefler (2013 p.30) portend that random sampling gives an equal opportunity for participants in the study. The researcher assumes that a sample size of 60 will result in rich information. Notably, the study will investigate the effect of varied variables on the number of hospital admissions at the Melbourne hospitals.Descriptive analyses will help in the formulation of the conclusion.
- a) Based a series of random samples of 60, are the mean values of these random samples normally distributed? Explain
The distribution is not normal. According to central limit theorem, the sampling distribution of mean tends quite closer to the normal distribution, provided the number of sample items is large. When a distribution of the mean values of a random sample population is said to be normal, the distribution of the sample mean is also normal; the sample distribution mean ( ) = the population mean (µ). For a normal population distribution with mean and standard deviation , the distribution of the sample mean is normal, with mean and standard deviation .
- b) Calculate the standard error of the mean and explain the meaning of this value
= = 903.12
The standard error of the mean is 903.12. This value is not significantly far from the mean. Thus, the standard error of the mean cannot affect the mean of the population.
- c) Determine the 95% confidence interval and explain its meaning in the context of the overall problem
To obtain the 95% confidence interval, multiply the SEM by 1.96 and add the result to the sample mean to obtain the upper limit of the interval in which the population parameter will fall.
Upper limit = 8729
Lower limit = 5189
The populations mean lies between 5189 admissions and 8729 admissions. Given that the population mean may be 5189, it is possible that 5189×60 admissions made were are outliers.
- 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?
The probability that a sample of 60 hospitals taken randomly from the Melbourne region is 0.05
- e) If the admissions times were more variable, what effect would this have on the confidence interval?
If the admissions time variables increase, the confidence interval and its width decrease. This will provide a more exact estimate of the population value. Larger samples result in smaller standard errors, and therefore, in sampling distributions that are more clustered around the population mean. A more closely clustered sampling distribution indicates that our confidence intervals will be narrower and more precise. Thus, if the admissions times are more variable, confidence intervals will not be more precise.
- Write down both the null and alternative hypotheses
7500 admissions are made in Melbourne hospitals.
7500 admissions are made in Melbourne hospitals.
- Carry out the t test and report the p-value, and the test statistic
The P-value, 0.168, tells us it is unlikely that we would observe such an extreme test statistic t* in the direction of HA if the null hypothesis were true. Therefore, our null hypothesisis rejected. That is, since the P-value, 0.168, is more than α = 6995.56, we reject the null hypothesis H0: μ = 7500 in favor of the alternative hypothesis HA: μ> 6959.
The average admissions are less than the assumed mean of 7500 (t=0.6, df=533, p<.05).
- Write an appropriate conclusion in the context of the problem
The services provided in the hospitals at Melbourne have a big impact on the number of admission since the descriptive analyses provided depict a not normal distribution of variables.
Based on answers to questions 1 and 2: A report of the effect of various variables on the number of admissions in Melbourne hospitals
The standard error variable provides a wide range of a class in which the exact mean should fall. Thus, it shows how unequal the sampled data is distributed.
The t test value
The t test value failed to accept the null hypothesis provided hence indicating that the relationship of variables shown in the table might have happened by chance.
This section is based on a survey that was conducted by ComeReal Estate Agency on house prices. The prices depended on the conditions of the local market.
A sample of 211 was randomly selected for the July 2015 sales in Melbourne.
- Write down the regression equation.
y = -137.88+178.63x
- State the R-squared value and the standard error and explain what they mean with respect to the data.
Adjusted R2 is the coefficient of determination and tells us how price of the house varies with bedroom number in a house.From the table provided, the value of R2 is 0.6680. This implies that, there was a variation of 66.8% of between the bedroom and the house price. This simply means that, a bedroom is accountingfor a, 66.8% of the regression equation.
- Write down the value of the gradient of the regression line and explain what it means for this data.
The gradient shows that an addition of one bedroom causes an increase of house price by 178.63.
- Are the values for the constant and the gradient (slope) significant (i.e. significantly different from zero) in this case? Justify your answer.
The constant not significant
This is because the constant term is in part estimated by the omission of predictors from a regression analysis. Typically, it serves as a garbage bin for any error that is not taken into account by the terms in the model. Also the constant in this case is a negative.
The gradient is significant since it indicates the value rate by which the bedroom number impacts on the price of the house.
- Conduct a hypothesis test on the slope coefficient and test whether there is a linear relationship that there between number of bedrooms and prices of the houses.
Include the null and alternative hypotheses; key test results and an appropriate conclusion.
H0: The slope of the regression line is equal to zero.
Ha: The slope of the regression line is not equal to zero.
Y = Β0 + Β1X
y = -137.88+178.63(1.542) =137.65
There is a linear relationship since the gradient of the regression greatly differs from the zero.
- Does the linear regression provide a good model? Give statistical reasons based on the scatterplot, p-values, the standard error and coefficient of determination.
The model illustrates that when all variables are held at zero (constant), the value of employee performance would be -137.88. However, holding other factors constant, a unit increase in bedroom, would lead to a 178.62 increase in house price.
- 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?
I would factor in variables such as security, income, accessibility of the area, and infrastructural development for instance drainage system and lighting.
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