Tmgt 361-001 Assignment 3

To find the coefficient of skewness (a measure of the degree of skewness), the mean, mode and standard deviation was needed. Due to the large data size, a computer program was used to obtain the necessary information. The data set was inserted into the program (One Variable Analysis by Haese and Harris Publications) which then analysed it and produce the required result. The information collected is displayed below with the result for the mean rounded to 87 from 86.964 and the standard deviation to 8.2 from 8.2375. This was done for convenience however it did reduce the precision of the

15 In testing the hypotheses: H0 β1 ’ 0: vs. H1: β 1 ≠ 0 , the following statistics are available: n = 10, b0 = 1.8, b1 = 2.45, and Sb1= 1.20. The value of the test statistic is:

1. Are any of the lab values in Table 1 out of normal range? Do you see some that are too high or too

If 9 t tests were conducted and the set alpha for this study is 0.05, then the alpha level that should be used to determine the differences between the two groups is 0.05/9=0.0056 and the resulting alpha will be used to determine significant differences.

A researcher would like to know if there is a significant difference in clothing purchases between full-time working women, part-time working women, and women who are homemakers. ANOVA

The null hypothesis was that the female and male shoe sizes have an equal mean while the alternative hypothesis was that female and male shoe sizes do not have an equal mean. With the degrees of freedom being 33, the t-statistic is -8.27. The probability that -8.27 is ≤-1.69 is 7.5×10-10 for the one-tailed test. Also, the probability that -8.27 is ≤ ±2.03. is 1.5×10-9 for the two-tailed test. Due to both probabilities being under the alpha value of 0.05, the null hypothesis is rejected, and the alternative hypothesis is accepted at the 95% confidence level.

The degree of freedom in this experiment was 3. The chi-square value that was calculated was 1.4481. The p-value was 0.6943.

According to the above results from MINITAB, the p-value of 0.038 is smaller than the significance level of 0.05; consequently, the null hypothesis will be rejected. There is sufficient evidence to support the

The second research question addressing peer relationships, the variables that will be used in the analysis include friendships and the rate of smoking. Lastly, the third research question, the variables that will be used in the analysis include sex and rate of beginning smoking. The first research question will be analyzed using a one-way ANOVA, the second research question will be analyzed using a chi-square test and the last research question will be analyzed using a chi-square test as well. The data being categorical is what determines the test above. By converting the SAS data to excel, the test will run on Graph Pad Prism 7 to show the association between smoking and Latino youth. It will also, reveal if the participants are more at risk or a higher risk for smoking depending upon the second

1. What are the two groups whose results are reflected by the t ratios in Tables 2 and 3? The two groups are pre-test and 3-month measures and pre-test and 6-month measures.

2In this exercise I am going to test whether there is association between the variable drug and with two level topiramate and placebo and the variable quit smoking coded (yes) or (no). After running the analysis for computing the chi-squared I obtain three different statistical tables as follows: Case processing Summary, Drug Quit Smoking Crosstabulation, and Chi Square Test which fills the statistics.

Insert a complete data table, including appropriate significant figures and units, in the space below. Also include any observations that you made over the course of Part II.

I rejected the null hypothesis and found out the p-value is smaller than the significance level. The p-value is greater than the significance level 0.05.

With a P-value of 0.00, we have a strong level of significance. No additional information is needed to ensure that the data given is accurate.

In this case, level of significance, α was not provided. Therefore, the analysis will be evaluated based on two α values which are: