Statistical Test

Dec 6, 2007Download as PPT, PDF2 likes1,160 views

This document provides instructions for conducting two statistical tests: Spearman's Rank Correlation Coefficient and Chi-Square. Spearman's Rank is used to analyze the relationship between two variables like distance and environmental quality. It involves ranking values, calculating differences between ranks, and using a formula to determine if the relationship is statistically significant. Chi-Square analyzes relationships between categorical variables like opinions and demographics. It involves creating a results table, calculating expected values, applying a formula, and determining statistical significance based on degrees of freedom. Both tests are used to evaluate a null hypothesis of no relationship between variables.

Spearman's Rank Correlation Coefficient Chi-Square Both tests look at the relationship between two variables They tell us whether the relationship is significant, i.e. if one is related to the other Statistical Tests

Spearman's Rank Best used to look at either: Distance and Environmental Quality OR Distance and Counts Either look at total environmental quality score for a survey point or just one element. Choose a point on your map to say the distance it is from your survey point. Try and choose a point that will get results! Doesn't matter where your point is as long as you can explain why

1. Create a Null Hypothesis There is no significant relationship between x and y. The Null Hypothesis tells us that there is no significant relationship between two factors. That means we cannot say with any degree of certainty that one affects the other. You state the null hypothesis as shown below before any piece of statistical analysis. The purpose of your statistical analysis is to disprove the null hypothesis. Therefore you can say that there IS a significant relationship between x and y.

2. Create Table Site No. Dist from x Rank Env Q. Rank Diff Diff Sq don't need real distance - use cm on map (1) (2) (1)-(2)  Sum of 1 2 3 4 5 20 15 7 35 25 3 2 1 5 4 2.5 2.5 1 5 4 14 14 12 25 24 0.5 -0.5 0 0 0 0.25 0.25 0 0 0 0.5

Answer is between +1 and -1 +1 - perfect positive relationship 0 - no relationship -1 - perfect negative relationship 3. Apply formula n = your number of sites

4. Can we reject null hypothesis Degrees of Freedom = number of sample sites. If larger than critical value then with 0.05 you are 95% certain that you can reject null hypothesis. 0.01 you are 99% certain you can reject the null

5. Writing it up Introduce: It was thought that litter would be worse closer to the cinema entrance. In order to investigate this relationship a Spearman's Rank test was carried out (figure 2). Null Hypothesis and All your calculations go in figure 2. Then explain what you have found out Environmental survey KQ1 - Counts KQ2

Chi Square Test Best place to use - look at the relationship between people's opinions against their age or gender. For example: Do men feel safer in Cardinal Park than women?

1. Create null hypothesis There is no significant relationship between .... What would it be if we were looking at gender and feelings of safety in Cardinal Park?

2. Create results table If you have a not sure column - you will need to merge these with agree or disagree See worksheet I have given you for an example!

3. Check over 5 Check all cells are over 5. If you have lower you will need to merge categories Again - there is an example of this on the sheet.

4. Calculate Expected You need to know: Total number of questionnaires Total in rows Total in columns Need to do 4 calculations for expected values - see sheet.

5. Apply the Formula Remember to total up

6. Degrees of Freedom - Does it fit the Null Slightly different calculation for degrees of freedom this time. It is the number of rows minus one x number of columns minus one. Again look at the significance level. Can you be 95% or even 99% certain?

7. Write up in same way as you did with Spearman's Introduce: It was thought that women would be more scared in Cardinal Park that men. In order to investigate this relationship a Chi Squared test was carried out (figure 3). Null Hypothesis and All your calculations go in figure 3. Then explain what you have found out

- The chi-square goodness-of-fit test can be used to determine if a frequency distribution fits a specific pattern or theoretical distribution. It compares observed frequencies to expected frequencies. - To perform the test, the chi-square statistic is calculated using the formula (O-E)^2/E, where O is the observed frequency and E is the expected frequency. This value is then compared to a critical value from the chi-square distribution based on the degrees of freedom. - If the chi-square statistic exceeds the critical value, the null hypothesis that the observed and expected frequencies are the same is rejected, indicating a poor fit between the observed and expected distributions.

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This document discusses hypothesis testing using a single sample. It explains that a hypothesis test involves a null hypothesis (H0) which is initially assumed to be true, and an alternative hypothesis (Ha) which is the competing claim. The test aims to reject the null hypothesis in favor of the alternative. A test statistic is calculated from sample data and compared to a significance level (α) to determine whether to reject H0. Examples are provided to illustrate hypotheses about population means, proportions, and their tests.

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1) The chi-square test of independence is used to determine if there is a relationship between two categorical variables. It compares observed frequencies to expected frequencies if the null hypothesis of independence is true. 2) A contingency table is constructed with the observed frequencies. Expected frequencies are calculated for each cell based on row and column totals and the grand total. 3) The chi-square statistic is calculated by summing the squared differences between observed and expected frequencies divided by the expected frequency for each cell. This value is then compared to a critical value from the chi-square distribution to determine if the null hypothesis should be rejected.

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This document discusses statistical hypothesis testing, including: - The process involves formulating a null and alternative hypothesis, determining a significance level, identifying an appropriate test, making conclusions by comparing results to a critical value, and interpreting findings. - Common tests include those concerning population means, comparing two population means, and comparing two population proportions. - Interpretation of significance levels ranges from very strong to little evidence against the null hypothesis depending on the level of alpha. - Examples demonstrate hypothesis tests for a population mean, comparing two population means, and interpreting results to either reject or fail to reject the null hypothesis.

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TEST #1 Perform the following two-tailed hypothesis test, using a .05 significance level: · Intrinsic by Gender · State the null and an alternate statement for the test · Use Microsoft Excel (Data Analysis Tools) to process your data and run the appropriate test. Copy and paste the results of the output to your report in Microsoft Word. · Identify the significance level, the test statistic, and the critical value. · State whether you are rejecting or failing to reject the null hypothesis statement. · Explain how the results could be used by the manager of the company. TEST #2 Perform the following two-tailed hypothesis test, using a .05 significance level: · Extrinsic variable by Position Type · State the null and an alternate statement for the test · Use Microsoft Excel (Data Analysis Tools) to process your data and run the appropriate test. · Copy and paste the results of the output to your report in Microsoft Word. · Identify the significance level, the test statistic, and the critical value. · State whether you are rejecting or failing to reject the null hypothesis statement. · Explain how the results could be used by the manager of the company. GENERAL ANALYSIS (Research Required) Using your textbook or other appropriate college-level resources: · Explain when to use a t-test and when to use a z-test. Explore the differences. · Discuss why samples are used instead of populations. The report should be well written and should flow well with no grammatical errors. It should include proper citation in APA formatting in both the in-text and reference pages and include a title page, be double-spaced, and in Times New Roman, 12-point font. APA formatting is necessary to ensure academic honesty. Be sure to provide references in APA format for any resource you may use to support your answers. Making Inferences When data are collected, various summary statistics and graphs can be used for describing data; however, learning about what the data mean is where the power of statistics starts. For example, is there really a difference between two leading cola products? Hypothesis testing is an example of making these types of inferences on data sets. Hypothesis Tests Claims are made all the time, such as a particular light bulb will last a certain number of hours. Claims like this are tested with hypothesis testing. It is a straight forward procedure that consists of the following steps: 1. A claim is made. 2. A value for probability of significance is chosen. 3. Data are collected. 4. The test is performed. 5. The results are analyzed. Hypothesis tests are performed on the mean of the population. µ It is not possible to test the full population. For example, it would be impossible to test every light bulb. Instead, the hypothesis test is performed on a sample of the population. Setting up a Hypothesis Test When performing hypothesis testing, the test is setup with a null hypothesis (or claim) and the alternative hypothesis. ...

Morestatistics22 091208004743-phpapp01mandrewmartin

1) The document discusses hypothesis testing and statistical inference using examples related to coin tossing. It explains the concepts of type I and type II errors and how hypothesis tests are conducted. 2) An example is provided to test the hypothesis that the average American ideology is somewhat conservative (H0: μ = 5) using data from the National Election Study. The alternative hypothesis is that the average is less than 5 (HA: μ < 5). 3) The results of the hypothesis test show the observed test statistic is lower than the critical value, so the null hypothesis that the average is 5 is rejected in favor of the alternative that the average is less than 5.

8 Statistical SignificanceOK, measures of association are one .docxevonnehoggarth79783

8 Statistical Significance OK, measures of association are one important thing to examine in data. There is another important thing to consider. If you find associations, are they a result of associations that exist in the population or are those associations simply a result of sampling error? Tests of statistical significance estimate the chances that your associations are a result of the population, and not simply sampling error. Chi-Square Tests Chi-Square tests are appropriate for nominal and ordinal variables. When you calculate Chi-Square, you determine the probability that your answer is a result of your sample and not the population. So, a probability of .05 (p = .05) means that the association that you found in your analysis would occur only 5 out of 100 times if there actually was no association in the population. If you had a p = .001, this means that out of 1,000 samples, you would find the association simply as a result of your sampling error 1 time. Convention suggests that probabilities of .05, 01, and .001 support a differing levels of statistical significance of your conclusions. Let’s go back to our variables SEX and HAPPY. You actually do the same thing that you did when you calculated lambda, except you also check chi-square in the statistics box. Here are the exact steps: · Analyze > Descriptive Statistics > Crosstabs · Dependent variable as the Row variable (Mneumonic suggestion: remember DR, dependent belongs on the row) · Independent variable as the column variable · Statistics: Lambda or Gamma AND Chi-Square · Continue > OK Most of your output will look be the same as when we calculated lambda. There is one new box: Look at the Asymp. Sig (2 sided). This means that out of 1,000 chances, 883 times you would get the lambda of 0 totally by accident of your sample! Is that statistically significant? Well, not for scientific research. People play the lottery with far worse odds than this, but remember, for a result to be statistically significant in social science research, the probability must be .05 or less. Let’s think about our query about the relationship between gender and happiness. We have discovered that there is no association between the two and there is no statistical significance. What does that mean? Well, I guess that is a good thing for men and women. It is not a good finding however if you were expecting to find an association between the variables! T Tests T tests are used to determine statistical significance of scale (ratio/interval) variables. If you want to examine the associations of nominal/scale variables, you may be quickly overwhelmed with data. If you do discover that you would like to discover the statistical significance of scale variables, I would suggest that you use an Independent-sample t test. You can use your output for the Pearson’s r and it will tell you your level of significance. Here is the output for our analysis of the association between AGE and SIBS: SPSS has cal.

data handling class 8HimakshiKava

1) The document discusses different types of data (raw, discrete, continuous) and frequency distributions (grouped, ungrouped, cumulative, relative). 2) It explains the concept of probability and key terms like random experiment, sample space, events. Probability is calculated as the number of desirable events divided by the total number of outcomes. 3) The document also covers the arithmetic mean, which is the average value of a data set calculated by summing all values and dividing by the number of values.

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This document provides an overview of how to perform and interpret a chi-square test of goodness of fit. The chi-square test determines whether observed data fits the expected results based on a theoretical model or hypothesis. An example calculation is shown for offspring ratios from a genetic cross. The chi-square value is compared to a critical value from tables to determine if the null hypothesis should be rejected or failed to reject.

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What So Funny About Proportion Testv3ChrisConnors

The document discusses hypothesis testing and proportion tests. It provides an overview of hypothesis testing terminology and steps. It also gives examples of using one-proportion and two-proportion tests to analyze business data on regulatory compliance documentation and workload balance between regions. The null hypothesis is tested in each example to determine if there are statistically significant differences between the proportions.

More Statisticsmandrewmartin

The document discusses various statistical concepts related to hypothesis testing, including: - Types I and II errors that can occur when testing hypotheses - How the probability of committing errors depends on factors like the sample size and how far the population parameter is from the hypothesized value - The concept of critical regions and how they are used to determine if a null hypothesis can be rejected - The difference between discrete and continuous probability distributions and examples of each - How an observed test statistic is calculated and compared to a critical value to determine whether to reject or not reject the null hypothesis

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This document discusses statistical analysis and its application to genetics problems. It introduces the chi-square test, which allows scientists to determine if experimental results differ significantly from expected values based on chance alone. The key steps are to form a null hypothesis with expected results, collect data from an experiment, calculate the chi-square value and compare it to critical values to determine if the null hypothesis can be rejected or not. Examples show how to apply the chi-square test to genetic crosses and determine if observed offspring ratios match expected Mendelian ratios.

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PPT.4.pptx..............................NarutoooUzumakiii

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Statistical Test

1. Spearman's Rank Correlation Coefficient Chi-Square Both tests look at the relationship between two variables They tell us whether the relationship is significant, i.e. if one is related to the other Statistical Tests

2. Spearman's Rank Best used to look at either: Distance and Environmental Quality OR Distance and Counts Either look at total environmental quality score for a survey point or just one element. Choose a point on your map to say the distance it is from your survey point. Try and choose a point that will get results! Doesn't matter where your point is as long as you can explain why

3. 1. Create a Null Hypothesis There is no significant relationship between x and y. The Null Hypothesis tells us that there is no significant relationship between two factors. That means we cannot say with any degree of certainty that one affects the other. You state the null hypothesis as shown below before any piece of statistical analysis. The purpose of your statistical analysis is to disprove the null hypothesis. Therefore you can say that there IS a significant relationship between x and y.

4. 2. Create Table Site No. Dist from x Rank Env Q. Rank Diff Diff Sq don't need real distance - use cm on map (1) (2) (1)-(2)  Sum of 1 2 3 4 5 20 15 7 35 25 3 2 1 5 4 2.5 2.5 1 5 4 14 14 12 25 24 0.5 -0.5 0 0 0 0.25 0.25 0 0 0 0.5

5. Answer is between +1 and -1 +1 - perfect positive relationship 0 - no relationship -1 - perfect negative relationship 3. Apply formula n = your number of sites

6. 4. Can we reject null hypothesis Degrees of Freedom = number of sample sites. If larger than critical value then with 0.05 you are 95% certain that you can reject null hypothesis. 0.01 you are 99% certain you can reject the null

7. 5. Writing it up Introduce: It was thought that litter would be worse closer to the cinema entrance. In order to investigate this relationship a Spearman's Rank test was carried out (figure 2). Null Hypothesis and All your calculations go in figure 2. Then explain what you have found out Environmental survey KQ1 - Counts KQ2

8. Chi Square Test Best place to use - look at the relationship between people's opinions against their age or gender. For example: Do men feel safer in Cardinal Park than women?

9. 1. Create null hypothesis There is no significant relationship between .... What would it be if we were looking at gender and feelings of safety in Cardinal Park?

10. 2. Create results table If you have a not sure column - you will need to merge these with agree or disagree See worksheet I have given you for an example!

11. 3. Check over 5 Check all cells are over 5. If you have lower you will need to merge categories Again - there is an example of this on the sheet.

12. 4. Calculate Expected You need to know: Total number of questionnaires Total in rows Total in columns Need to do 4 calculations for expected values - see sheet.

13. 5. Apply the Formula Remember to total up

14. 6. Degrees of Freedom - Does it fit the Null Slightly different calculation for degrees of freedom this time. It is the number of rows minus one x number of columns minus one. Again look at the significance level. Can you be 95% or even 99% certain?

15. 7. Write up in same way as you did with Spearman's Introduce: It was thought that women would be more scared in Cardinal Park that men. In order to investigate this relationship a Chi Squared test was carried out (figure 3). Null Hypothesis and All your calculations go in figure 3. Then explain what you have found out

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