A study that looked at beverage consumption used sample sizes that were much smaller than previous national surveys. One part of this study compared 20 children who were 7 to 10 years old with 5 who were 11 to 13. The younger children consumed an average of 8.2 oz of sweetened drinks per day while the older ones averaged 14.1 oz. The standard deviations were 10.8 oz and 8.2 oz respectively. Required:
Do you think that it is reasonable to assume that these data are Normally distributed? Explain why or why not?

Step-by-step explanation:

In statistics, about 68 percent of values come in one standard deviation of the mean by using a standard normal model.   Approximately 95% of the data were all within two standard deviations from the mean. Almost all of the data are in the range of three standard deviations of the mean (roughly 99.7%).

The 68-95-99.7 law, also known as the Empirical Rule, is based on this evidence. 68 percent of the data values of a naturally distributed data collection of small children with a mean of 8.2 and a standard deviation of 10.8 would be between -2.2 and 19.0.

Within a mean of 14.1 as well as a standard deviation of 8.2, 68 percent of the data values in a usually distributed data collection of older children would be between 5.9 and 22.3.

However, we cannot conclude that the data is naturally distributed since the real actual data vary from the usual normal curve computed above.

Hence, various measures like either goodness of fit or theory testing, would be used for this.

answer: true

step-by-step explanation:

the results from a pre-test for students for the year 2000 and the year 2010 are illustrated in the box plot. what do these results tell us about how students performed on the 29 question pre-test for the two years?

if we compare only the lowest and highest scores between the two years, we might conclude that the students in 2010 did better than the students in 2010. this conclusion seems to follow since the lowest score of 8 in 2010 is greater in value than the lowest score of 6 in 2000. also, the highest score of 28 in 2010 is greater in value than the highest score of 27 in 2000.

but the box portion of the illustration gives us more detailed information. the middle bar in each box shows us that the median score of 20 in 2000 is greater in value than the median score of 17 in 2010. further, we note that the box and whiskers divide the illustration into four pieces. each of these four pieces represents the same portion of students. so, the upper half of the students in 2000 scored in the same score range as the upper one-fourth of the students in 2010, see the illustration at a score of 20.

by considering the upper one-fourth, upper half, and upper three-fourths instead of just the lowest and highest scores, we would conclude that the students as a whole did much better in 2000 than in 2010. we would conclude that as a whole the students in 2010 are less prepared than the students in 2000.

in this section, we discuss box-and-whisker plots and the five key values used in constructing a box-and-whisker plot. the key values are called a five-number summary, which consists of the minimum, first quartile, median, third quartile, and maximum.

step-by-step explanation: