SURVEY DATA FOR ICE CREAM SOLD VERSUS THE TEMPERATURE, statistics homework help

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Locate the results of a recent survey that shows at least two variables in a newspaper, magazine, or internet article. Outline the survey data so that your peers can understand the variables and results, and then identify at least one key formula from this module that you could use to evaluate the data. Provide a brief explanation of why you selected the formula you did and why it matters.

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1. Analysis of Variance We have discussed how to develop interval estimates and how to conduct hypothesis tests for situations involving a single population mean and a single population proportion. So what if we wanted to identify means between different populations? For example, what if we wanted to learn if there was a difference in post-graduate earnings between men and women? To identify any differences, we would take two independent random samples – one for each population of men and of women – and calculate the mean. We would assume that the standard deviation has been provided. We would then subtract the two means to obtain the point estimator of the difference, and determine the standard error for them using the formula: As before, we would determine the significance level and find the range of error. If, however, the standard deviations are not known for given populations, we can use the standard deviations from the samples and the formula: We can then calculate the degrees of freedom for t based on the formula: and determine the range for the p-value. In an actual study, we would create the hypotheses, determine the samples, and then run the statistical analysis. For the example above, let: μ1= the population mean post-graduate earnings of men μ2= the population mean post-graduate earnings of women The hypotheses would then be written as follows: H0: μ1−μ2 = 0 (There is no difference between post-graduate earnings between men and women.) Ha: μ1−μ1 ≠0 (There is a difference between post-graduate earnings between men and women.) We could then use a simple random selection procedure to select data from men and women. For example, we could take a list of men and a list of women and select data from every odd numbered man and woman. We would then determine the mean post-graduate earnings between men and between women, calculate the sample standard deviation, then calculate the degrees of freedom for 95% confidence level, and see if the p-value is between .10 and .05 (two-tailed test). If it does not lie between the p-values, then you can say that H0 is not rejected. In general, the hypotheses for the difference of two means are written as Two-Tailed: H0: μ1= μ2 Ha: μ1≠ μ2 or μ1−μ2≠ 0 Lower Tail: H0: μ1≥ μ2 Ha: μ1< μ2 or μ1−μ2< 0 Upper Tail: H0: μ1≤ μ2 Ha: μ1> μ2 or μ1−μ2> 0 The data from studies conducted to generate data can be used in the statistical procedure called analysis of variance or ANOVA. ANOVA can be used to test data for three of more populations in much the same way that we have already done with two populations. To use ANOVA analysis, you must assume the following criteria: Critaria1 Each population has a normal distribution. Criteria2 The variance for each population is the same for all populations. Criteria3 The observations are independent. ANOVA is used when the data are divided into groups according to only one factor, such as the mean of post-graduate pay. The analysis is usually: a. Is there a significant difference between the groups? b. If there is a difference, which groups are significantly different from which others? In other words, to determine whether or not the population means are equal, we have to see whether or not there is variability among the sample means and variability of the data within each sample. Statistical tests are then provided to compare group means, group medians, and group standard deviations. For example, if we have three sets of data regarding post-graduate pay increases between an M.S. in Leadership, an M.S. in Management, and an M.S. in Organizational Leadership, and we wanted to know if there was a difference in pay between them, we would use ANOVA. For this example, we would let: μ1= the population mean post-graduate pay increase of graduates with an M.S. in Leadership μ2 = the population mean post-graduate pay increase of graduates with an M.S. in Management μ3 = the population mean post-graduate pay increase of graduates with an M.S. in Organizational Leadership The hypotheses would then be written as follows: H0: μ1=μ2=μ3 ( The population means are equal.) Ha: Not all the population means are equal to each other. After calculating the means, we would calculate the sample standard deviations and then the Estimate of Standard Deviation = (Number of Data)(Sample Standard Deviation squared). We would need to use the same formula for the sample variance data, and then compare the information to see if the ratio is 1. If it is not 1, we can reject the null hypothesis. 2. Analysis of Correlation The test of independence seeks to identify whether or not variables in a given sample have a relationship. For example, does being a man or a woman impact the amount of post-graduate pay increase? The following is a contingency table with data collected for graduates: Male Female $0-$3,000 456 709 $3,001-$6,000 302 958 $6,001-$9,000 909 387 $9,001-$12,000 536 282 More than $12,000 736 815 The hypotheses are: H0: Gender and post-graduate pay are independent of each other Ha: Gender and post-graduate pay are dependent In general, if one makes a contingency table (as above), the hypotheses for tests of independence are written as: H0: The row and column are independent of each other Ha: The row and column are dependent The test statistic used is the same as the chi-square goodness-of-fit test. The principle behind the test for independence is the same as the principle behind the goodness-of-fit test. In fact, you can think of the test for independence as a goodness-of-fit test where the data is arranged into table form called a contingency table. For a chi-square goodness-of-fit test, we determine whether or not the population proportions are equal to given values b1, b2, …,bk (where all the bi’s add up to 1) : H0: p1=b1, p2=b2, p3 = b3,..., pk=bk Ha: H0 is not true or if all the population proportions are equal to each other: H0: p1= p2=p3= ... = pk Ha: H0 is not true The test statistic for goodness of fit compares the sample of observed results with the expected results under the assumption that the null hypothesis is true. The formula is fi = observed frequency for category i ei = expected frequency for category i k = the number of categories Note: The test statistic has a chi-square distribution with k – 1 degrees of freedom provided that the expected frequencies are 5 or more for all categories. The following are properties of the test for independence:  The data are the observed frequencies.  The data is arranged into a contingency table that lists all possible combinations of possible relationships.  The degrees of freedom are the degrees of freedom for the row variable times the degrees of freedom for the column variable. It is the product of the two degrees of freedom.  It is always a right tail test.  It has a chi-square distribution.  The expected value is computed by taking the row total times the column total and dividing by the grand total.  The value of the test statistic doesn't change if the order of the rows or columns is switched.  The value of the test statistic doesn't change if the rows and columns are interchanged. where fij = observed frequency for contingency table category in row i and column j eij = expected frequency for contingency table category in row i and column j based on the assumption of independence Note: With n rows and m columns in the contingency table, the test statistic has a chi-square distribution with (n - 1)(m -1) degrees of freedom provided that the expected frequencies are five or more for all categories. So, to test for goodness of fit and test of independence, we would follow these steps: Step 1 State the null and alternative hypotheses. Step 2 Select a random sample and record the observed frequencies. Step 3 Determine the expected frequency in each category by multiplying the category probability by the sample size. Step 4 Compute the test statistic. Step 5 Accept or reject the H0 and the Ha with consideration to the level of significance. Statistical data often has relationships between variables. Your ability to test for such relationships will allow you to draw conclusions from the information that can be helpful in understanding the characteristics of your studied population.
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Running head: SURVEY DATA FOR ICE CREAM SOLD VERSUS THE TEMPERATURE 1

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SURVEY DATA FOR ICE CREAM SOLD VERSUS THE TEMPERATURE

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Survey: Number of ice cream sol...


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I was struggling with this subject, and this helped me a ton!

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