When I use: Stats>Tests>B: 2-sample-2-Prop-Z-Int, I enter X1, N1, N2, X2, CL, Calculate, enter, I get:

ERR: Domain

1: Quit

Help!

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I'm not sure, but it probably gives the error because one of the values you entered isn't valid. It's too big or too small.

For more information, look here:

http://tibasicdev.wikidot.com/2-propzint

If this doesn't solve the problem, please let me know.

Gr Marijn

Posted on Nov 29, 2007

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When I go back to check what's wrong, n1 has been replaced with n1 = 0, I have no idea why the calculator is doing this; I'm on a time crunch and I need help with this as soon as possible. Thank you for reading! BTW, I'm using a TI-84 Plus CE if that is helpful. AmanStino Ultra Slim Lightweight Smart shell Stand Cover with Auto Wake...

Nov 27, 2016 | Office Equipment & Supplies

n1 and n2 are integers. The proportions p1 and p2 must satisfy the conditions of the Hypothesis you are testing. p1 not equal to p2, p1 < p2, p1>p2.

Dec 10, 2013 | Texas Instruments TI-89 Calculator

It's in the STATS menu, not the TEST menu.

Press STAT then left-arrow then up-arrow five times.

Press STAT then left-arrow then up-arrow five times.

Apr 04, 2012 | Texas Instruments TI-84 Plus Silver...

After looking in the manual, I read that

It takes as input the COUNT of successes in each sample (x1 and x2) and the COUNT of observations in each sample (n1 and n2). Thus your values of x1 and x2 must be integers (whole numbers).. Besides I duplicated your example taking x1=21, n1=43, x2=18, and n2=38 and I did not get the domain error.

I was able to draw the graph also.

You might want to check your values.

It takes as input the COUNT of successes in each sample (x1 and x2) and the COUNT of observations in each sample (n1 and n2). Thus your values of x1 and x2 must be integers (whole numbers).. Besides I duplicated your example taking x1=21, n1=43, x2=18, and n2=38 and I did not get the domain error.

I was able to draw the graph also.

You might want to check your values.

Apr 02, 2012 | Texas Instruments TI-84 Plus Silver...

Press the Y= button.

There are several possibilities:

1) Scroll through all equations, Y1 through Y9 and Y0. Make sure they are cleared.

2) The equation you typed in is not entered correctly.

3) Make sure Plot 1, Plot 2, and Plot 3 are NOT highlighted.

There are several possibilities:

1) Scroll through all equations, Y1 through Y9 and Y0. Make sure they are cleared.

2) The equation you typed in is not entered correctly.

3) Make sure Plot 1, Plot 2, and Plot 3 are NOT highlighted.

Feb 01, 2011 | Texas Instruments TI-84 Plus Calculator

The site seems to eat the plus signs I enter, so I will use PLUS to symbolize addition.

To find the equation of the straight line (y = a*x PLUS b) that passes through two points P1(x1,y1) and P(x2,y2) , you need to use

1. the coordinates of the points to calculate the slope a (gradient) as a=(y2-y1)/(x2-x1)

2. Replace the calculated value of a in the equation and write that one of the points ( P1(x1,y1) for example) satisfies the equation. In other words y1=a*x1 PLUS b.

Here y1 and x1 are known values, a has been calculated, and only b is still unknown. You can now use the equation y1=a*x1 PLUS b to calculate b as

b=(y1-a*x1)

Example: Equation of the line through (1,5) and (3,6)

Calculate the slope (gradient) of the line as a=(y2-y1)/(x2-x1) where y2=6, y1=5, x2=3, and x1=1. You should get a=(6-5)/(3-1)=1/2

The equation is y=(1/2)x PLUS b, where b is not known yet.

To find b, substitute the coordinates of one of the points in the equation. Let us do it for (3,6).

The point (3,6) lies on the line, so 6=(1/2)*3 PLUS b.

Solve for b: 6 MINUS 3/2=b, or b=9/2=4.5

Equation is thus y=(x/2) PLUS 9/2 =(x PLUS 9)/2

I trust you can substitute you own values for (x1,y1, x2,y2) to duplicate the calculations above.

To find the equation of the straight line (y = a*x PLUS b) that passes through two points P1(x1,y1) and P(x2,y2) , you need to use

1. the coordinates of the points to calculate the slope a (gradient) as a=(y2-y1)/(x2-x1)

2. Replace the calculated value of a in the equation and write that one of the points ( P1(x1,y1) for example) satisfies the equation. In other words y1=a*x1 PLUS b.

Here y1 and x1 are known values, a has been calculated, and only b is still unknown. You can now use the equation y1=a*x1 PLUS b to calculate b as

b=(y1-a*x1)

Example: Equation of the line through (1,5) and (3,6)

Calculate the slope (gradient) of the line as a=(y2-y1)/(x2-x1) where y2=6, y1=5, x2=3, and x1=1. You should get a=(6-5)/(3-1)=1/2

The equation is y=(1/2)x PLUS b, where b is not known yet.

To find b, substitute the coordinates of one of the points in the equation. Let us do it for (3,6).

The point (3,6) lies on the line, so 6=(1/2)*3 PLUS b.

Solve for b: 6 MINUS 3/2=b, or b=9/2=4.5

Equation is thus y=(x/2) PLUS 9/2 =(x PLUS 9)/2

I trust you can substitute you own values for (x1,y1, x2,y2) to duplicate the calculations above.

Jan 27, 2011 | Texas Instruments TI-84 Plus Calculator

Example: Equation of the line through (1,5) and (3,6)

Calculate the slope (gradient) of the line as a=(y2-y1)/(x2-x1) where y2=6, y1=5, x2=3, and x1=1. You should get a=(6-5)/(3-1)=1/2

The equation is y=(1/2)x PLUS b, where b is not known yet.

To find b, substitute the coordinates of one of the points in the equation. Let us do it for (3,6).

The point (3,6) lies on the line, so 6=3/2 PLUS b.

Solve for b: 6 MINUS 3/2=b, or b=9/2=4.5

Equation is thus y=(x/2) PLUS 9/2 =(x PLUS 9)/2

Calculate the slope (gradient) of the line as a=(y2-y1)/(x2-x1) where y2=6, y1=5, x2=3, and x1=1. You should get a=(6-5)/(3-1)=1/2

The equation is y=(1/2)x PLUS b, where b is not known yet.

To find b, substitute the coordinates of one of the points in the equation. Let us do it for (3,6).

The point (3,6) lies on the line, so 6=3/2 PLUS b.

Solve for b: 6 MINUS 3/2=b, or b=9/2=4.5

Equation is thus y=(x/2) PLUS 9/2 =(x PLUS 9)/2

Jan 10, 2011 | Texas Instruments TI-84 Plus Calculator

Calculate
the slope (gradient) of the line as a=(y2-y1)/(x2-x1) where y2=6, y1=5,
x2=3, and x1=1. You should get a=(6-5)/(3-1)=1/2

The equation is y=(1/2)x PLUS b, where b is not known yet.

To find b, substitute the coordinates of one of the points in the equation. Let us do it for (3,6).

The point (3,6) lies on the line, so 6=3/2 PLUS b.

Solve for b: 6 MINUS 3/2=b, or b=9/2=4.5

Equation is thus y=(x/2) PLUS 9/2 =(x PLUS 9)/2

The equation is y=(1/2)x PLUS b, where b is not known yet.

To find b, substitute the coordinates of one of the points in the equation. Let us do it for (3,6).

The point (3,6) lies on the line, so 6=3/2 PLUS b.

Solve for b: 6 MINUS 3/2=b, or b=9/2=4.5

Equation is thus y=(x/2) PLUS 9/2 =(x PLUS 9)/2

Oct 20, 2010 | Texas Instruments TI-84 Plus Calculator

Having received no response in over a month, I assume this is no longer a problem.

Jun 25, 2010 | Texas Instruments TI-84 Plus Silver...

#include <stdio.h>

int main()

{

int n1, n2, i;

int sum = 0;

n1 = 3;

n2 = 11;

for (i = n1; i <= n2; i++) {

if (i % 2 == 0) {

printf("Adding %d\n", i);

sum = sum + i;

}

}

printf("sum = %d\n", sum);

return 0;

}

int main()

{

int n1, n2, i;

int sum = 0;

n1 = 3;

n2 = 11;

for (i = n1; i <= n2; i++) {

if (i % 2 == 0) {

printf("Adding %d\n", i);

sum = sum + i;

}

}

printf("sum = %d\n", sum);

return 0;

}

Feb 21, 2008 | Computers & Internet

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