Respond to the underline part at the bottom of this attachment.
The last time I compared two samples was in an exercise consisting on comparing the temperature of New York City on March on two consecutive years. To do this exercise, I collected the daily average temperatures for the different days of March during 2016 and 2017. Thus, sample 1 consisted on the daily average temperatures recorded during the different days of March 2016, whereas sample 2 consisted on the daily average temperatures recorded during the different days of March 2017.
I then compared the results obtained for the two samples by performing a z test for the comparison of the two-sample means assuming equal variances (Boslaugh, 2012). This test is based on the calculation of:
Â· The mean and standard deviation of the two samples (x 1 , x 2 , s 1 , s 2 )
Â· The calculation of the pooled variance according to the equation:
Â· The calculation of the z statistic according to the equation:
Â· The comparison of the calculated z statistic with the critical z value obtained from the normal distribution table for a target significance level.
I used celsius to make it easier.
The sample means were of 9.48ÂºC and 4.13ÂºC for the daily average temperature of New York City on March 2016 and March 2017, respectively. On the other hand, the standard deviations were of 5.04ÂºC and 5.89ÂºC for the daily average temperatures, respectively. Taking these values into account, the computed z statistic was of 3.84. Since this value is higher than the critical value for a 95% confidence level (1.96), I concluded that the two samples were not comparable, meaning that the daily average temperature on New York City had been different on March 2016 than on March 2017.
I apologize my formulas would not transfer over.
Boslaugh, S. (2012). Statistics in a nutshell. Sebastopol, CA: OReilly Media.
Thanks for sharing your thoughts regarding ANOVA testing and the importance of statistical significance between the means of three or more independent groups. Anova analysis will help determine if the null hypothesis should be rejected or not. I also liked your example of the daily average temperatures. Anova will help determine if thereâ€s any relationship between the temperatures in that given month. As stated by Lind, Marchal & Wathen (2015), Anova will simultaneously determine if there is any relation between the means.Reference:Lind, D.A., Marchal, W.G., & Wathen, S.A. (2015). Statistical Techniques in Business & Economics. McGraw -Hill Education: New York.
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