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16. In a study of memory recall, 5 people were given 10 minutes to memorize | Complete Solution

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16. In a study of memory recall, 5 people were given 10 minutes to memorize a list of 20 words .Each was asked to list as many of the words as he or she could remember both 1 hour and 24hours later. The result is shown in the following table.
              Number of Words Recalled
Subject   1 hour later        24 hours later
1                    13                        13
2                    18                        14
3                   12                          11
4                   15                          14
5                  11                           11

Is there evidence to suggest that the mean number of words recalled after 1 hour exceeds the mean recall after 24 hours? Assume we want to use a 0.05 significance level to test the claim.
(a) Identify the null hypothesis and the alternative hypothesis.
(b) Determine the test statistic. Show all work; writing the correct test statistic, without supporting work, will receive no credit.
(c) Determine the P-value. Show all work; writing the correct P-value, without supporting work, will receive no credit.
(d) Is there sufficient evidence to support the claim that the mean number of words recalled after 1 hour exceeds the mean recall after 24 hours? Justify your conclusion.


18. The UMUC Daily News reported that the color distribution for plain M&M’s was: 35%
brown, 30% yellow, 15% orange, 10% green, and 10% tan. Each piece of candy in a random
sample of 200 plain M&M’s was classified according to color, and the results are listed below.
Use a 0.05 significance level to test the claim that the published color distribution is correct.
Show all work and justify your answer.

Color     Brown     Yellow    Orange    Green    Tan
Number    80    45    30    30    15

(a) Identify the null hypothesis and the alternative hypothesis.
(b) Determine the test statistic. Show all work; writing the correct test statistic, without supporting
work, will receive no credit.
(c) Determine the P-value. Show all work; writing the correct P- value, without supporting work,
will receive no credit.
(d) Is there sufficient evidence to support the claim that the published color distribution is correct?
Justify your answer.


19. A STAT 200 instructor believes that the average quiz score is a good predictor of final exam
score. A random sample of 10 students produced the following data where x is the average quiz
score and y is the final exam score.
x 72 96 55 65 100 50 85 70 74 85
y 72 98 50 70 96 60 83 75 77 87
(a) Find an equation of the least squares regression line. Show all work; writing the correct
equation, without supporting work, will receive no credit.
(b) Based on the equation from part (a), what is the predicted final exam score if the average quiz
score is 80? Show all work and justify your answer

 

20. A study of 10 different weight loss programs involved 200 subjects. Each of the 10 programs
had 20 subjects in it. The subjects were followed for 12 months. Weight change for each subject
was recorded. We want to test the claim that the mean weight loss is the same for the 10
programs.
(a) Complete the following ANOVA table with sum of squares, degrees of freedom, and mean
square (Show all work):

Source of Variation
    Sum of Squares
(SS)    Degrees of Freedom
(df)    Mean Square
(MS)
Factor
(Between)    53.6        
Error
(Within)            
Total     553.05    199    N/A

(b) Determine the test statistic. Show all work; writing the correct test statistic, without supporting work, will receive no credit.
(c) Determine the P-value. Show all work; writing the correct P-value, without supporting work, will receive no credit.
(d) Is there sufficient evidence to support the claim that the mean weight loss is the same for the 10 programs at the significance level of 0.05? Explain.

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[Solved] 16. In a study of memory recall, 5 people were given 10 minutes to memorize | Complete Solution

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19) Answer: a) a= =(768*58976-752*59890)/(10*58976-752^2) 1...
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16. In a study of memory recall, 5 people were given 10 minutes to memorize | Complete Solution

19) Answer: a) a= =(768*58976-752*59890)/(10*58976-752^2) 10.56596 b= =(10*59890-752*768)/(10*58976-752^2) 0.880772 b) when avg score is 80, predicted y= 10.56596+0.0880772*80 17.612136 ...

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