One of your peers claims that boys do better in math classes than girls. Together you run two independent simple random samples
and calculate the given summary statistics of the boys and the girls for comparable math classes. In Calculus, 15 boys had a mean percentage of 82.3 with standard deviation of 5.6 while 12 girls had a mean percentage of 81.2 with standard deviation of 6.7. What assumptions need to be made in order to determine the 90% confidence interval for the difference in the mean percentage scores for the boys in calculus and the girls in calculus? Supposing the assumption is true, calculate the interval.
To test if boys are better in math classes than girls two random samples were taken:
X₁: score of a boy in calculus
X₂: Score in the calculus of a girl
To estimate per CI the difference between the mean percentage that boys obtained in calculus and the mean percentage that girls obtained in calculus, you need that both variables of interest come from normal populations.
To be able to use a pooled variance t-test you have to also assume that the population variances, although unknown, are equal.
Then you can calculate the interval as:
[(X[bar]_1-X[bar_2) ± * ]
[(82.3-81.2) ± 1.708* (6.11*]
Using a 90% confidence level you'd expect the interval [-2.94; 5.14] to contain the true value of the difference between the average percentage obtained in calculus by boys and the average percentage obtained in calculus by girls.
- A change of variable is a technique employed in solving many differential equations that are of the form: y ' = f ( t , y ).
- Considering a differential equation of the form y' = f ( αt + βy + γ ), where α, β, and γ are constants. A substitution of an arbitrary variable z = αt + βy + γ is made and the given differential equation is converted into a form: z ' = g ( z ).
- This substitution basically allow us to solve in-separable differential equations by converting them into a form that can be separated, followed by the set procedure.
- We are to solve the initial value problem for the following differential equation:
First Step: Make the appropriate substitution
- We will use a arbitrary variable ( z ) and define the our substitution by finding a multi-variable function f ( t , y ) that is a part of the given ODE.
- We see that the term ( t + y ) is a multi-variable function and also the culprit that doesn't allow us to separate our variables.
- Usually, the change of variable substitution is made for such " culprits ".
- So our substitution would be:
Second Step: Implicit differential of the substitution variable ( z ) with respect to the independent variable
- In the given ODE we see that the variable ( t ) is our independent variable. So we will derivate the supposed substitution as follows:
Remember: z is a multivariable function of "t" and "y". So we perform implicit differential for the variable " z ".
Third Step: Plug in the differential form in step 2 and change of variable substitution of ( z ) in the given ODE.
- The given ODE can be expressed as:
... Separable ODE
Fourth Step: Separate the variables and solve the ODE.
- We see that the substitution left us with a simple separable ODE.
Note: If we do not arrive at a separable ODE, then we must go back and re-choose our change of variable substitution for ( z ).
- We will progress by solving our ODE:
c: The constant of integration
Fifth Step: Back-substitution of variable ( z )
- We will now back-substitute the substitution made in the first step and arrive back at our original variables ( y and t ) as follows:
Sixth Step: Apply the initial value problem and solve for the constant of integration ( c )
- We will use the given initial value statement i.e y ( 3 ) = 6 and evaluate the constant of integration ( c ) as follows:
Seventh Step: Express the solution of the ODE in an explicit form ( if possible ):
3 assembly lines can fill 24 orders in 10 hours , therefore in 1 hour 3 assembly lines can fill 2.4 orders (24/10) , consequently, 1 assembly line can fill 0.8 orders (2.4/3) in 1 hour. Therefore, 7 assembly lines can fill 5.6 (7×0.8)orders in an hour and thus the 7 assembly lines in 5 hours would fill (5.6 ×5)= 28 orders. Therefore, the answer is 28 orders