Statistics Answer Note #15

Question 1 In an experiment, 10 participants were given a test of mental performance in stressful situations. Ten other participants were given the same test after they had been trained in stress-reducing techniques. The researcher wants to know if the training impacted the mental performance scores. What probability would allow me to accept the null hypothesis? Answer According to commonly accepted statistical conventions, a probability value (p-value) of 0.05 or less is generally used as the threshold for determining statistical significance in research studies. Therefore, in order to accept the null hypothesis, the p-value must be greater than 0.05. In this experiment, the null hypothesis would state that there is no significant difference in the mental performance scores between the two groups of participants, indicating that the stress-reducing training had no impact. The alternative hypothesis would be that there is a significant difference in the mental performance scores between the groups, suggesting that the stress-reducing training did have an impact. To determine the p-value and evaluate whether or not to accept the null hypothesis, various statistical tests can be used, such as a t-test or ANOVA. These tests would compare the mean scores of the two groups and calculate the p-value based on the observed difference between the groups. Question 2 In a similar experiment to question #2, ten participants were given a mental performance test in a stressful situation. The participants were then trained in stress reduction techniques and given the same test again. In an amazing coincidence, the participants received identical scores as the two groups of participants described in question #2 above. Compose a null hypothesis to accompany the test. Record both a generic version (using symbols) and an English version (using words). Answer The null hypothesis for this experiment would state that there is no significant difference in the mental performance scores of participants before and after implementing stress reduction techniques. This can be represented as: H0: μ1 = μ2 Where, H0: Null hypothesis μ1: Mean of mental performance scores before stress reduction techniques μ2: Mean of mental performance scores after stress reduction techniques In simple terms, the null hypothesis assumes that any observed differences in the mental performance scores between the two groups are due to chance or random variation rather than being a result of the stress reduction techniques. To test the significance of this null hypothesis, we can use a paired t-test as the same participants are subjected to both the stressful situation and the stress reduction techniques. Now, let us look at how the null hypothesis is supported by prior research in this field. Question 3 In a similar experiment to question #2, ten participants were given a mental performance test in a stressful situation. The participants were then trained in stress reduction techniques and given the same test again. In an amazing coincidence, the participants received identical scores as the two groups of participants described in question #2 above. If the probability is .026, should I reject the null hypothesis? Answer In a similar experiment to question Question 4 According to a random sample taken at 12 A.M., body temperatures of healthy adults have a bell-shaped distribution with a mean of 98.18degreesF and a standard deviation of 0.59degreesF. Using Chebyshev's theorem, what do we know about the percentage of healthy adults with body temperatures that are within 2 standard deviations of the mean? What are the minimum and maximum possible body temperatures that are within 2 standard deviations of the mean? Question content area bottom Part 1 At least enter your response here% of healthy adults have body temperatures within 2 standard deviations of 98.18degreesF. (Round to the nearest percent as needed.) Answer According to Chebyshev's theorem, we know that, for any set of data, the percentage of observations within a certain number of standard deviations from the mean is at least (11/k^2), where k is the number of standard deviations. Therefore, in the case of body temperatures of healthy adults with a bell-shaped distribution, we can say that at least (11/2^2) = 75% of the observations lie within 2 standard deviations from the mean. In mathematics and statistics, Chebyshev's theorem is used to provide a lower bound on the proportion of the data that must lie within k standard deviations of the mean, where k is any positive number greater than 1. This theorem is useful because it can be applied to any data set, regardless of the shape of its distribution. It also tells us that the farther we move away from the mean, the less dense the data becomes. To understand the range of body temperatures within 2 standard deviations of the mean, we can use the formula: P(|X-μ| < kσ) = 1 - (1/k^2) where P is the probability, X is the data, μ is the mean and σ is the standard deviation. In this case, k=2, μ=98.18 and σ=0.59. Therefore, P(|X-98.18| < 1.18) = 1 - (1/2^2) = 75%. This means that the range of body temperatures within 2 standard deviations of the mean is between (98.18-1.18) = 97.0degreesF and (98.18+1.18) = 99.36degreesF. This information is important because it allows us to better understand the distribution of body temperatures in healthy adults and the range within which the majority of observations will fall. It also tells us that extreme values, such as temperatures below 97.0degreesF or above 99.36degreesF, are less likely to occur. Question 5 Listed below are altitudes (in thousands of feet) and outside air temperatures (in degrees F) recorded during a flight between two cities. Altitude 3 9 16 23 27 32 33 Temperature 58 39 23 negative 5 negative 30 negative 39 negative 56 Use these data to complete parts (a) through (d) below. Question content area bottom Part 1 a. Construct a scatterplot. Choose the correct answer below. A. 0 40 -60 60 Alt. Temp. A graph with a horizontal Altitude axis labeled from 0 to 40 in increments of 5, and a vertical Temperature axis labeled from negative 60 to 60 in increments of 10. On this graph, the following points are plotted: (3,58), (9,39), (16,23), (23,negative 5), (27,negative 30), (32,negative 39), and (33,negative 56). B. 0 40 -60 60 Alt. Temp. A graph with a horizontal Altitude axis labeled from 0 to 40 in increments of 5, and a vertical Temperature axis labeled from negative 60 to 60 in increments of 10. On this graph, the following points are plotted: (3,negative 56), (9,negative 39), (16,negative 30), (23,negative 5), (27,23), (32,39), and (33,58). C. 0 40 -60 60 Alt. Temp. A graph with a horizontal Altitude axis labeled from 0 to 40 in increments of 5, and a vertical Temperature axis labeled from negative 60 to 60 in increments of 10. On this graph, the following points are plotted: (3,56), (9,39), (16,30), (23,5), (27,negative 23), (32,negative 39), and (33,negative 58). D. 0 40 -60 60 Alt. Temp. Answer According to the data provided, the altitudes (in thousands of feet) and outside air temperatures (in degrees Fahrenheit) during a flight between two cities are listed. To better understand the relationship between the altitude and temperature, a scatterplot can be constructed. To construct a scatterplot, we need to plot the points (altitudes and temperatures) on a graph and then draw a line of best fit to determine the pattern or relationship between the two variables. Based on the given data, the points to be plotted on the scatterplot are (3,58), (9,39), (16,23), (23,negative 5), (27,negative 30), (32,negative 39), and (33,negative 56). In this case, we can see that as altitude increases, the temperature decreases, showing an inverse relationship between the two variables. Overall, based on the scatterplot constructed, we can conclude that as altitude increases, the outside air temperature decreases. This can also be supported by the sources mentioned above. Therefore, the correct answer for part (a) would be C. 0 40 -60 60 Alt. Temp. A graph with a horizontal Altitude axis labeled from 0 to 40 in increments of 5, and a vertical Temperature axis labeled from negative 60 to 60 in increments of 10. On this graph, the following points are plotted: (3,56), (9,39), (16,30), (23,5), (27,negative 23), (32,negative 39), and (33,negative 58).