Berry waked the dog for 20 minutes.
2 answers:
You might be interested in
Answer:
The calculated |t| = |-26.517|=26.517 > 2.326 at 0.01 level of significance
Null hypothesis is rejected at 0.01 level of significance
There is difference between mean savings account balance in 2013 is different from the mean savings account balance in 2012
Step-by-step explanation:
<u><em>Step(i)</em></u>:-
Mean of the Population = 1900
Given random sample size 'n' = 45
Mean of the sample x⁻ = 1000
Standard deviation of the sample = 225
<u><em>Null Hypothesis</em></u>:-
There is no difference between mean savings account balance in 2013 is different from the mean savings account balance in 2012.
<u><em>Alternative Hypothesis</em></u> :-
There is difference between mean savings account balance in 2013 is different from the mean savings account balance in 2012
<u><em>Step(ii</em></u>):-
Test statistic
t = -900 /33.54 = -26.517
|t| = |-26.517|=26.517
Degrees of freedom
γ = n-1 = 45-1 =44
t₍₀.₀₁ , ₄₄₎ = 2.326
The calculated |t| = |-26.517|=26.517 > 2.326 at 0.01 level of significance
Null hypothesis is rejected at 0.01 level of significance
There is difference between mean savings account balance in 2013 is different from the mean savings account balance in 2012