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Sagot :
P-value is equal to 0.0206 and the significance levels at which there is sufficient evidence to reject H0 are at = 0.05 and [tex]\alpha[/tex]=0.1 and value of 99% CI = 7.132, 10.218.
Here we have,
n= 36
e= 9.06
S.E. = 4.256
Hypothesis
H0: μD = 0
Ha: μD > 0
This a right-tailed test
Test statistics,
t = 2.12
P-value at t= 2.12 and degree of freedom(df) = n-1 =36-1 = 35
P-value = t-dist(t, df, tails)
= t-dist(2.12, 35, L)
= 0.020585
= 0.0206 Round to 4 decimals
If pvalue>[tex]\alpha[/tex], then we reject the null hypothesis.
So, [tex]\alpha[/tex]= 0.05 and [tex]\alpha[/tex]=0.1 we reject the H0.
For sample of n=45 df will be= 45 -1 =44
t0.025 = 2.015
Mean weight = (7.520 + 9.830)/2 = 8.675
Margin of error = 9.830 - 8.675 = 1.155
SE = 1.155/2.015 = 0.573
99%tc = 2.692
Margin of error = 2.692 x 0.573 = 1.543
99% CI = 8.675 ± 1.543
= 7.132, 10.218
99% CI = 7.132, 10.218.
To learn more about the null hypothesis visit: https://brainly.com/question/28920252
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