Answer:
Test statistic:
z = (0.67 - 0.63) / 0.033 = 1.21
p - Value = P(z > 1.21) = 0.106
My conclusion: Fail to reject null hypothesis
No, there is no evidence of a difference
The common difference of the sequence is -3 and the fifth term is 5
<h3>How to determine the common difference?</h3>
The sequence is given as:
17, 14, 11, 8....
The common difference is
d = T2 - T1
So, we have
d = 14 - 17
Evaluate
d = -3
Hence, the common difference of the sequence is -3
<h3>How to determine the
fifth term?</h3>
The fifth term is calculated as:
T5 = a + 4d
Where
a = T1 = 17
d = -3
So, we have:
T5 = 17 - 4 * 3
Evaluate
T5 = 5
Hence, the fifth term is 5
<h3>How to determine the
nth term?</h3>
The nth term is calculated as:
Tn = a + (n - 1)d
Hence, the nth term is Tn = a + (n - 1)d
Read more about arithmetic sequence at:
brainly.com/question/6561461
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Answer:
this is the answer and the work btw I just took a screenshot this I have way better handwriting
Answer:
not statistically significant at ∝ = 0.05
Step-by-step explanation:
Sample size( n ) = 61
Average for student leader graduates to finish degree ( x') = 4.97 years
std = 1.23
Average for student body = 4.56 years
<u>Determine if the difference between the student leaders and the entire student population is statistically significant at alpha</u>
H0( null hypothesis ) : u = 4.56
Ha : u ≠ 4.56
using test statistic
test statistic ; t = ( x' - u ) / std√ n
= ( 4.97 - 4.56 ) / 1.23 √ 61
= 2.60
let ∝ = 0.05 , critical value = -2.60 + 2.60
Hence we wont fail to accept H0
This shows that the difference between the student leaders and the entire student population is not statistically significant at ∝ = 0.05
Answer:
1.49 Grams
Step-by-step explanation:
