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2 (00 polnt) Reaoareh ProposalL You wore given a data set with the folowing varlables based on a survey of 850 adults: Verlab
Distance to the nearest grocery Numerical store Distance to health care provider Numerical Number of vehicles Military servic
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Answer #1

This is a problem related ot identifying the correct statistical procedure /test for the data set given above.

We are required to provide details of 3 statistical procedures to illucidate our points.

But before doing so ,let us try to firstly understand as to what are the dataset variables ,thier types and nature

We can see that

1. The variables are of two types =qunantitative and qualitative (numerical )

2. The variables are either dependent or independent .

So based on the nature of the variables,we should perform the following statistical procedures to test the sanctity of the variable dataset.

1. regression analysis (qualitative variable ,quantitative variable)

1. Chi square test involving two qualitative groups

3. T test involving two groups which are independet of each other and which are qualitative.

So lets start the process

1. Regress analysis

Suppose as a researcher i need to know if the annual income of a perosn depends on

1. age

2. gender,

3. race

4. education

5. Experience

siginifcently or not

Then we need to conduct regression analysis in which we take a sample of say 60 subjects and ask them about there6 details.Based on thier output we construct a regression equation in excel or any onlnie software system .

This will give us the coefficent of all variables and we will thus be able to draw the regression equation.

The analysis also gives the p vlaues of all independent variables . if p value < 0.05 ,it means the variable is significant.in the regression equation.

Thus we can estimate if the regression (relationship among the indepedent and dependent variables is ) significant or not.

2. chi square test

This is usually done for variable which are qualitative in nature.

for example as a part of research we can research if the marital status of a person impacts his ability to get house on rent of not. So here we are doing statistical procedure on two given variables marital statusand rent. We can easily do it on race /age/ college education /annual income etc.

To start with the statistical procedure, we shall assume that there is no impact on rent given the marital status of the person.

The Process

defining the hypothesis

Ho: Variable A and Variable B are independent.

Ha: Variable A and Variable B are not independent.

The move to analysis plan .

The analysis plan describes how to use sample data to accept or reject the null hypothesis. The plan should specify the following elements.

  • Significance level. Often, researchers choose significance levels equal to 0.01, 0.05, or 0.10; but any value between 0 and 1 can be used.
  • Test method. Use the chi-square test for independence to determine whether there is a significant relationship between two categorical variables.

Using sample data, we find the degrees of freedom, expected frequencies, test statistic, and the P-value associated with the test statistic as per the given formula

  • Degrees of freedom. The degrees of freedom (DF) is equal to:

    DF = (r - 1) * (c - 1)

    where r is the number of levels for one catagorical variable, and c is the number of levels for the other categorical variable.
  • Expected frequencies. The expected frequency counts are computed separately for each level of one categorical variable at each level of the other categorical variable. Compute r * c expected frequencies, according to the following formula.

    Er,c = (nr * nc) / n

    where Er,c is the expected frequency count for level r of Variable A and level c of Variable B, nr is the total number of sample observations at level r of Variable A, nc is the total number of sample observations at level c of Variable B, and n is the total sample size.
  • Test statistic. The test statistic is a chi-square random variable (Χ2) defined by the following equation.

    Χ2 = Σ [ (Or,c - Er,c)2 / Er,c ]

    where Or,c is the observed frequency count at level r of Variable A and level c of Variable B, and Er,c is the expected frequency count at level r of Variable A and level c of Variable B.
  • P-value. The P-value is the probability of observing a sample statistic as extreme as the test statistic. Since the test statistic is a chi-square, we use the Chi-Square Distribution Calculator to assess the probability associated with the test statistic. Use the degrees of freedom computed above.

Interpret Results

If the sample findings are unlikely, given the null hypothesis, the researcher rejects the null hypothesis. Typically, this involves comparing the P-value to the significance level, and rejecting the null hypothesis when the P-value is less than the significance level.

3). t test

This statistical procedure is conducted when we know the given variables quantitatively and are to compare the degree of difference between the two given samples.

Research problem :-

suppose we want to know if free medical check up facilitates number of medical visits to medical centres , then we can conduct the t test wherein we can take two sample of 30-40 subjects and tell one gropu to go to free medical center and other group to paid medical centeres . we count thier vists and compare .

The level of different with siginificance is given by the t test .

We assume that there will be no differnece and check if our assumption is right or not.

The solution to the problem takes place in four statisitcal steps:

(1) state the hypotheses,

(2) formulate an analysis plan,

(3) analyze sample data, and

(4) interpret results.

We work through those steps below:

  • State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

    Null hypothesis: μ1 - μ2 = 0

    Alternative hypothesis: μ1 - μ2 ≠ 0

    Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.
  • Formulate an analysis plan. For this analysis, the significance level is 0.10. ( but we can choose any other level of siginificance like 0.05, 0.01 etc also)b
  • Using sample data, we will conduct a two-sample t-test of the null hypothesis.
  • <Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

    SE = sqrt[(s12/n1) + (s22/n2)]

    S

    DF = (s12/n1 + s22/n2)2 / { [ (s12 / n1)2 / (n1 - 1) ] + [ (s22 / n2)2 / (n2 - 1) ] }

    t = [ (x1 - x2) - d ] / SE

    where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is the size of sample 1, n2 is the size of sample 2, x1 is the mean of sample 1, x2 is the mean of sample 2, d is the hypothesized difference between the population means, and SE is the standard error.

    We use the t Distribution Calculator to find P(t < calculated t) and P(t > calculated t) = 0.027.

  • Thus, the P-value =sum of both p values

  • Interpret results. Check of p value >0.05 or not. if not then we aceept the null hypothesis and vice versa.

Thus we have discussed the three statisitcal prcedures that can be used to analyse any combination of datset variables .

We have discussed all the details of the processes as required as well along with all the assumptions we have made in the due course of explaination.

Now we need ot know of the relevance of p value

Usually

If the sample findings are unlikely, given the null hypothesis, the researcher rejects the null hypothesis.

Typically, this involves comparing the P-value to the given significance level, and rejecting the null hypothesis when the P-value is less than the significance level.

Now what does this mean ?

It means that if my p-value is lower than significance level (0.05,0.10,0.01, ...) then we will reject our assumption.

if the pvalue is 0.037 .

Then pvalue <0.05 and hence we reject the assumption made in the research question and accept the alternate hypothesis .

However please not that this is a rejection at 0.05 level only.

If we take the siginifcance elvel to be 1 % then we will accept this result since pvalue in this case >0.01

simliarly we reject 0.004 p vlaue and accept 0.385 pvalue.

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