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Sagot :
The 95% confidence interval for the mean difference of scores [5.573,11.427].
A) Null and Alternative hypothesis are
H₀ : μ₁ - μ₂ =0
Hₐ : μ₁ - μ₂ ≠ 0
B) Margin of Error of 95% confidence interval is 2.927
C) Margin of error increase with increasing the confidence interval. So, the margin of error large at 98% confidence interval.
D) Because zero is not included in the confidence interval, and all values in the confidence interval [5.573, 11.427] are positive.
A sample of scores for 320 students in this new curriculum were compared to scores in a control group (traditional program) of 273 students.
(A) Null and Alternative hypothesis are
H₀ : μ₁ - μ₂ =0 ( There is no difference between the means in performance scores for the two types of math curriculum )
Hₐ : μ₁ - μ₂ ≠ 0 ( There is a difference between the means in performance scores for the two types of math curriculum ) (Claim)
(B)We have given data that ,
Confidence Interval : [5.573, 11.427]
Sample mean (X-bar) is given by:
X-bar =( 5.573 + 11.427)/2 =17/2 = 8.5
Using margin of error formula,
MOE = mean - lower bound of confidence interval
Margin of Error = 8.5 - 5.573= 2.927
=> Margin of Error = 2.927
(C)If a 98% confidence interval would have been created, the margin of error would be larger because the margin of error increases as the confidence level increases from 95% to 98%
(D)Based on the confidence interval the researchers can conclude that there is a difference between the means in performance scores for the two types of math curriculum because 0 is not included in the confidence interval, [5.573, 11.427] and all values in the confidence interval [5.573, 11.427] are positive.
To learn more about Confidence interval, refer:
https://brainly.com/question/17212516
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