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# Critical Value and Decision

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In an experiment to determine the effect of nutrition on the attention spans of elementary school students, a group of 15 students were randomly assignment to each of three meal plans: no breakfast, light breakfast, and full breakfast. Their attentions spans (in minutes) were recorded during a morning reading period and are shown below. Is there a significant difference in attention span based on the three meal plans? Identify the appropriate test and use the 0.05 significance level in determining your answer. Assume the populations are normal with equal standard deviations.

No Breakfast Light Breakfast Full Breakfast
8 14 10
7 16 12
9 12 16
13 17 15
10 11 12

Sum X 47 70 65
Sum X^2 463 1006 869
Means: 9.4 14 13
Variances: 5.3 6.5 6

1. Identify the claim
2. Critical Value(s)
3. Test Value

Plants emit gases that trigger the ripening of fruit, attract pollinators, and cue other physiological responses. The hydrocarbons emitted by the potato plant were measured and compared to the plant weight. Weight (x) was measured in grams and hydrocarbon emissions (y) were measured in hundreds of nanograms for 11 plants. Let alpha = 0.05 and use the traditional method to test for a relationship between plant weight and hydrocarbon emissions. Write the hypotheses in sentence form, not with symbols.

X 57 85 57 65 52 67 62 80 77 53 68
Y 8.0 22.0 10.5 22.5 12.0 11.5 7.5 13.0 16.5 21.0 12.0

1. Critical Value(s)
2. Decision

##### Solution Summary

In an experiment to determine the effect of nutrition on the attention spans of elementary school students, a group of 15 students were randomly assignment to each of three meal plans: no breakfast, light breakfast, and full breakfast. Their attentions spans (in minutes) were recorded during a morning reading period and are shown below. Is there a significant difference in attention span based on the three meal plans? Identify the appropriate test and use the 0.05 significance level in determining your answer. Assume the populations are normal with equal standard deviations.

No Breakfast Light Breakfast Full Breakfast
8 14 10
7 16 12
9 12 16
13 17 15
10 11 12

Sum X 47 70 65
Sum X^2 463 1006 869
Means: 9.4 14 13
Variances: 5.3 6.5 6

1. Identify the claim
2. Critical Value(s)
3. Test Value

Plants emit gases that trigger the ripening of fruit, attract pollinators, and cue other physiological responses. The hydrocarbons emitted by the potato plant were measured and compared to the plant weight. Weight (x) was measured in grams and hydrocarbon emissions (y) were measured in hundreds of nanograms for 11 plants. Let alpha = 0.05 and use the traditional method to test for a relationship between plant weight and hydrocarbon emissions. Write the hypotheses in sentence form, not with symbols.

X 57 85 57 65 52 67 62 80 77 53 68
Y 8.0 22.0 10.5 22.5 12.0 11.5 7.5 13.0 16.5 21.0 12.0

1. Critical Value(s)
2. Decision

##### Solution Preview

In an experiment to determine the effect of nutrition on the attention spans of elementary school students, a group of 15 students were randomly assignment to each of three meal plans: no breakfast, light breakfast, and full breakfast. Their attentions spans (in minutes) were recorded during a morning reading period and are shown below. Is there a significant difference in attention span based on the three meal plans? Identify the appropriate test and use the 0.05 significance level in determining your answer. Assume the populations are normal with equal standard deviations.

No Breakfast Light Breakfast Full Breakfast
8 14 10
7 16 12
9 12 16
13 17 15
10 11 12

Sum X 47 70 65
Sum X^2 463 1006 869
Means: 9.4 14 13
Variances: 5.3 6.5 6

1. Identify the claim
Solution. Assume that their mean attentions spans are , respectively. Then we can set up the null hypothesis as follows.
:
2. Critical Value(s)
3. Test Value ...

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###### Education
• BSc , Wuhan Univ. China
• MA, Shandong Univ.
###### Recent Feedback
• "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
• "excellent work"
• "Thank you so much for all of your help!!! I will be posting another assignment. Please let me know (once posted), if the credits I'm offering is enough or you ! Thanks again!"
• "Thank you"
• "Thank you very much for your valuable time and assistance!"

##### Measures of Central Tendency

Tests knowledge of the three main measures of central tendency, including some simple calculation questions.