### Need Proportion Test Assessment Help And Assignment Help Services To Accomplish Your Academic Goals?

Home   Course
Previous << || >> Next

Proportion Test

SAVE TOP GRADE USING PROPORTION TEST ASSIGNMENT HELP SERVICE OF EXPERTSMINDS.COM!

Question 1: What is the purpose of the test?

Introducing Proportion Test: The purpose of this function is to test the equality of population proportions using a chi-square test statistics. The idea behind the test can be better discussed by first discussing the sampling distribution of proportions. According to central limit theorem the sampling distribution of a population mean is normal with if the sample size is sufficiently large. The rule also applies for small samples when the population being sampled is normal. Researchers prove this rule by experimental.

According to the above theory the specification of the normal distribution is mean µ and standard deviation σ/√ni.eµ ~N(µ,σ/√n ). If the variable measured has two outcomes, we can assign a binary scale, where 0 is the failure outcome and 1 I the success outcome. With such a scale, the proportion of successes is simply the mean. This means that taking the mean of the sample is the same as counting the number of successes and dividing by the total count observed. The mean of such a variable is therefore p with standard deviation of p is√(p(1-p).)

A Proportion being a mean obeys central limit theorem too, as said before if we replace µ and σ on the specified sampling distribution, we get that for proportions p N~(p,√((p(1-p).)/n)) . Now in this case where the central limit theorem applies, a z test can be used to compare any two proportions or a proportion against a hypothetical value.

Why chi-square test statistic is used in prop.test()

One main limitation of Z test is that it cannot be used when we want to compare more than two proportions, also the sample size has to be large, and this now introduces the need for the chi-square test statistic which is used in prop. test() function.

A chi-square goodness of fit test is a method used to compare two given sets of probabilities (proportions). It is most common in comparing probability distribution. Researchers use the method to see how two probability distributions differ. Also when if one wants to establish if a given sample follows a specified distribution. The distribution of the sample is sample is the observed while the hypothesized distribution is what we expect by theory. For instance, if one wishes to establish if the package color of a product really matters, and then collects a sample of 80 shoppers to see what color they purchased. We expect that if color has no contribution, then the number of shoppers who purchase blue =20, green =20 and red =20, yellow=20. This means the probability of purchasing any color is equal and it is 20/80. But suppose we get, blue =25, green =15 and red =22, yellow=18. We use chi-square test to compare the following two distributions and decide if they are statistically different

Table 1: Example 1

 Colour Expected distribution(probabilities) Observed distribution (probabilities) blue 20/80 25/80 green 20/80 15/80 red 20/80 22/80 yellow 20/80 18/80

For our problem we are comparing the following two distributions. The expected probability is half because we have only two out comes, we therefore expect 50 heads and 50 tails

Table 2: Example 2

 Expected distribution(probabilities) Observed distribution (probabilities) Heads 50/100 48/100

The prop.test() function uses this method to compare any number of proportions, the expected distribution makes the null hypothesis where we assume that all the probabilities of success(probabilities) are equal. The observed probabilities (sample proportions) help in construction of the observed probabilities.

DON'T MISS YOUR CHANCE TO EXCEL IN ASSIGNMENT! HIRE TUTOR OF EXPERTSMINDS.COM FOR PERFECTLY WRITTEN PROPORTION TEST ASSIGNMENT SOLUTIONS!

Question 2: Define the meaning of each input parameter.

Answer: X - this is used to input the observed distribution, a vector or a matrix is required here, only counts of for each group is needed then the function can calculate the observed probabilities. The purpose of this input is to supply the values from which observed probabilities (proportions) are calculated from.

n - this input supplies the number of trials, it is also used in calculation of the observed probabilities, as shown on the previous table the observed probabilities(proportions) is x/n. so for example 1, n is 80, and for example 2 n=100.

P - This argument is used to supply the hypothesized probabilities (expected probabilities), It therefore accepts probabilities (values between 0 and 1) only. In case the input x is a matrix, this input can be ignored, also whenever the input is ignored. The default is to assume that all proportions are equal.

Conf level - this argument defines the levels of confidence to be assumed, it also defines the significance level because, and alpha is equal to 1-confidence level. A confidence interval is also printed on the results based on the specified cl specified. If the argument is ignored the function assumes a 95% confidence level.

Alternative - the argument tells whether the test is upper sided test, lower sided test or two sided test. If ignored the function assumes a two tailed test hypothesis.

Correct - the argument specifies whether we are correcting the test statistic using Yates correction factor, the factor corrects the impact of the skewness of chi-square distribution on our test statistic. Is set to TRUE. The correction is applied, if ignored the default is not to correct the test statistic (FALSE).

From the problem presented

X = 48 which are the observed counts of success (the success event is an head)

n= 100, this is the number of trials

p = 0.5, it is hypothesized that the probability of a heads is 0.5. it forms the expected distribution to be compared with x/n
conf level =.95, the argument is ignored therefore the function uses the default .95 and will output a 95% confidence level on the output.

Alternative - this argument is not provided. The function assumes a two tailed test.

Question 3: Define the meaning of each output value.

Answer: From the above given input, he function processes the information and outputs the following attributes which can be accessed by assigning a name to the model then reference each attribute by its name

Statistic - it is the test statistic of the test (a chi-square statistic)

Parameter - it is the degrees of freedom of the model.

P value - it is obtained from chi-square cdf curve. It gives the probability of our test statistic or a value more extreme on the event of true null hypothesis.

Estimate- it gives the estimates observed probability distribution, they are the values x/n explained on the input section.

Conf.int - gives the confidence interval of the population proportion.

Null. Value - it is the value of the null hypothesis if supplied.

Alternative - specifies the type of the test performed as declared on the previous section.

From the provided problem

From the problem the chi-square statistic is 0.16, the parameter (df) = 1 i.e the number of degrees of freedom.the p value = .6892, the confidence interval is (0.3846,0.5768), the hypothesized value(null value) is 0.5,the alternative hypothesis is two tail while the default is not to correct for the bias using yates correction factor.

Interpreting the output: The test performed here is whether the probability of a head is equal to the probability of a tail or the case is otherwise. As discussed on the introduction section we only have one proportion here, if the two proportions are equal we expect the proportion equal 0.5 to be tails and the other 0.5 to be heads so the hypothesis is.

H0 : p = 0.5 p is the proportion of heads on the population

H1 : p ≠0.5

The chi-square statistic is 0.16 with one degree of freedom, the p value is .6892 which is greater than 0.05, it indicates that at 5% level of significance, there is insufficient evidence to support the claim that the proportion of heads is different from the proportion of tails.

The 95% confidence is (0.3846, 0.5768). This means that we can be 95% confident that the population proportion of heads is a member of the interval above. Our hypothesized value is .5 which is a member of the interval too. Indicating that it's usual to have p as .5 hence the null hypothesis cannot be rejected. There is no enough evidence to warrant that.

GET READYMADE PROPORTION TEST ASSIGNMENT SOLUTIONS - 100% PLAGIARISM FREE WORK DOCUMENT AT NOMINAL CHARGES!

Tag This :- EM1911SON1306STAT_E Proportion Test Assignment Help

### Assignment Samples

 MGMT6012 Management Perspectives Assignment Help mgmt6012 management perspectives assignment help - the report is a reflection on organizational culture and scenarios in the management context. Influenza And Epidemiology Assignment Help influenza and epidemiology assignment help - Describe the chosen communicable disease, including causes, symptoms, mode of transmission, complications. Leadership Values Assignment Help leadership values assignment help - Decision based on the Leadership Values and Ethics Reflection Paper Assignment Help reflection paper assignment help - based on the ability to capture the value of the seminar series, critique its contribution to their professional development ISY1003 Foundations of Programming Assignment Help isy1003 foundations of programming assignment help - Australian Institute of Higher Education, Australia. Discuss the elements of effective programming style Digital Revolution Assignment Help digital revolution assignment help - Tuxedo Zone, Federation University - calculates the projected annual outgoing costs of running Tuxedo Zone Ethical Responses Assignment Help ethical responses assignment help - This document refers to the ethical responses by different students. Each of them has presented their responses in different

Get Academic Excellence with Best Skilled Tutor! Order Assignment Now!