Comparison Between Group D And Group C

From the above output, we can see that the p-value is 0.000186, which is smaller than 0.05 (if we select a 0.05 significance level).

This information is true because the data have a p-value of 0.055793. This also allows us to support the null hypothesis, and reject the previously stated hypothesis. Shifting next to hypothesis 2, using the chi squared test to find a p-value of 0.00000000605505 to determine that acidity of a region had a significant relationship between the amount of isopods and the amount of acid or base present. This means that we should reject the null hypothesis, and support our original hypothesis. Now, hypothesis 3 had a p-value of 0.001823 showing us that the relationship between the isopod count and the amount of light present is significant. This information disproves the null hypothesis, as well as the previously stated hypothesis because the isopods were in fact drawn to a darker environment. Lastly, hypothesis 4 gives a p-value of 0.0000000153118 expressing that the relationship between temperature and amount of isopods are significant. This allows us to determine that we should reject the null hypothesis as well as the original hypothesis due to the fact that there was a significant amount of isopods in the cold

All the p-values are greater than 0.05, therefore there is a statistical difference between each transect.

However, we must perform some analysis on this data to confirm that the results for our class really are significantly different from the average. To do this, we performed a chi-square analysis on our data. Our chi-square value is 14.733, and the degrees of freedom are 6. The resulting p-value falls between 0.05 and 0.02. Therefore, we conclude that the data from our class indeed does not fit the average.

Data that is statistically significant helps us to understand whether there is a relationship within the null hypothesis. T statistical significant results showed that the participants who consumed caffeine performed worse than participants who did not was less than 5% (p < .05). This means for the author to determine whether a test is significant it must be less than .05, which is the significance level. The P value (probability value) is defined as the probability of obtaining a result equal to or "more extreme" than what was observed (Wikipedia). A p-value helps determine the significance

To first understand the importance of the clinical reaction time test, the physiological pathways involved in reaction time must first be

The reaction timer from Maths Is Fun (2014) is specific to three decimal places. The accuracy of the chosen source reduces the risk of statistical measures being slightly greater or less than they would be if only one or two decimal places were provided. Unlike other reaction timers that are available online, this particular source requires the subject to complete five trials before the mean is calculated. Undertaking multiple trials will be vital to this investigation because if the subject was to anticipate the event or have one delayed response, the trials that follow would reveal such errors, therefore increasing the reliability of the results.

Note: the program we used on this worksheet said the results were not significant, but then in statistical notion it had “p < 0.05”. That is confusing because I believe it should be “p >

The hypothesis was, If any of the products (orange juice, coca cola, taco sauce) cleans pennies, then the coca cola will clean the pennies the best. This was rejected. None of the products actually took of dirt because when the pennies were weighed it never decreased it just stayed at 2 grams each time. Before any of the trials the pennies weighed 2 grams. After the first trial (30 minutes) the weight was still 2 grams. For the last trial after the 30 minutes it still stayed the same which was 2 grams. So all the data and the average was at zero on the graphs because there was no dirt taken off any of the pennies. In this case the graphs showed no trend, they were all at zero. The products were suppose to clean the pennies but it did not work out as planned so they did not clean the

The standard Deviation comparing both sets of data only shows a slight difference on the numbers (a difference of only 1.38187054), but in the histograms both show a huge difference on the frequencies of each data set, but shows a similarity on the cumulative percentage. Data set #2 seem to be more stable even though the grades of the student were lower than data set #1. Data set#2 shows a relative stability in regards the frequency on the histograms, when comparing the frequency histograms data set #2 has more frequency on number two’s than data set #1 on frequency on the number three’s. In this case Data set #2 shows a normal distribution behavior.

Name two ethnic or cultural groups other than African Americans and Jews that have made stellar contributions to the development of the united states. List at least ten achievers from each group, and try to form intergroup pairs as Cornel West does in his essay. Also compare and contrast the groups as a whole and share your overall perspective on their achievements.

Both data sets contain outliers, as clearly shown in the box and whisker plots. The control group contains two outliers, both of which are 70. The experimental group contains four outliers: 0, 10, 40 and 70. In all calculations, the outliers have been omitted in order to ensure more accurate calculations. The median test score/center of the data for both the control group and experimental group is around 90 percent. The mean or average test score for the experimental group is 93.5 and for the control group, the average test score is an 89.3. The standard deviation of the group that chewed gum is 6.895 (about 7) and the group that did not chew gum had a standard deviation of 7.164 (about 7.2.) This shows that the experimental group has less variance in comparison to the control group. Without doing an actual hypothesis test, just based off of the graphs and raw data, we can guess that our original hypothesis (those who chew gum will have better test scores than those who do not) might be correct and the conclusion will be that we fail to reject the null hypothesis. Based on the dotplot and histogram, the assumption of normality is plausible for the control group but not for the experimental

To examine responder type differences in comparison processes, we included the NDE slopes of the children as the dependent measure in a one-way ANCOVA with grade as a covariate. After controlling for the effect of grade, the responder type group difference was nonsignificant (F < 1.0). This suggests that responding in linear versus exponential fashion on the estimation task is unrelated to the speeded comparisons of number magnitude, a somewhat surprising outcome given that both tasks are assumed to rely on access to a mental number line, that is, a mental representation of

Line 184: a p-value of p=0.09 does not indicate a trend. If you want to talk about a trend just fit a simple regression line to each individual's data across the testing days and see if the average slope is different from zero.

The p value is .001, which shows that the data is not normal. However, this test is more reliable with larger sample sizes. Therefore, if there was a larger sample size here, the results of this test may differ from above.