Math scores of 13 year olds.The National Assessment of Ed- ucational Progress tested a simple random sample of 1,000 thirteen year old students in both 2004 and 2008 (two separate simple random samples). The average and standard deviation in 2004 were 257 and 39, respectively. In 2008, the average and standard deviation were 260 and 38, respectively. 2. It is possible that your conclusion in part (1) is incorrect. What type of error is possible for this conclusion? Explain. PART 1D. F=n1 + n2 – 2=1000 + 1000 – 2=1998Determine the critical value for t a using the excel formula, "=T.INV.2T(0.10,1998)".2 1. 6456C.I=(x1 – x2) ± tal2 ×n1n2(39)²=(257 – 260) ± 1. 6456 ×(38)2+10001000=-3.0 ± 1. 6456 * 1. 72192=-3.0 + 2. 8336=(-5. 83, –0. 17)90% confidence interval for the change in average scores from 2004 to 2008 is C.I = (-5.83,-0.17).Since the 90% confidence interval for the change in average scores from 2004 to 2008 includesall the values less than 0, hence there is sufficient evidence to conclude that the average mathscore for 13 year old students has changed from 2004 to 2008.

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Description: Math scores of 13 year olds.The National Assessment of Ed- ucational Progress tested a simple random sample of 1,000 thirteen year old students in both 2004 and 2008 (two separate simple random samples). The average and standard deviation in 2004 wer

Type 1 and 2 error: The type 1 error is rejecting the null hypothesis when the null hypothesis is…

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Math scores of 13 year olds.The National Assessment of Ed- ucational Progress tested a simple random sample of 1,000 thirteen year old students in both 2004 and 2008 (two separate simple random samples). The average and standard deviation in 2004 were 257 and 39, respectively. In 2008, the average and standard deviation were 260 and 38, respectively.

2. It is possible that your conclusion in part (1) is incorrect. What type of error is possible for this conclusion? Explain.

PART 1
D. F=n1 + n2 – 2
=1000 + 1000 – 2
=1998
Determine the critical value for t a using the excel formula, "=T.INV.2T(0.10,1998)".
2 1. 6456
C.I=(x1 – x2) ± tal2 ×
n1
n2
(39)²
=(257 – 260) ± 1. 6456 ×
(38)2
+
1000
1000
=-3.0 ± 1. 6456 * 1. 72192
=-3.0 + 2. 8336
=(-5. 83, –0. 17)
90% confidence interval for the change in average scores from 2004 to 2008 is C.I = (-5.83,
-0.17).
Since the 90% confidence interval for the change in average scores from 2004 to 2008 includes
all the values less than 0, hence there is sufficient evidence to conclude that the average math
score for 13 year old students has changed from 2004 to 2008.

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