I really need help with this i have tried everything! Please provide instructions so I could learn how to do it! You are interested in finding a 98% confidence interval for the average number of days of class that college students miss each year. The data below show the number of missed days for 11 randomly selected college students. Round answers to 3 decimal places where possible. 9, 1, 2, 2, 9, 10, 1, 3, 7, 4, 2 a. To compute the confidence interval use a ? distribution. b. With 98% confidence the population mean number of days of class that college students miss is between ? and ? days. c. If many groups of 11 randomly selected non-residential college students are surveyed, then a different confidence interval would be produced from each group. About ? percent of these confidence intervals will contain the true population mean number of missed class days and about ? percent will not contain the true population mean number of missed class days.

Step-by-step explanation:

a. When the sample size is below 30 or the population standard deviation is not known, use a t distribution rather than a normal distribution.

b. The confidence interval is:

CI = x ± CV × SE

where x is the sample mean,

CV is the critical value (in this case, t).

and SE is the standard error (the sample standard deviation).

For the data given, the mean is x = 4.545, and the standard deviation is SE = 3.503.

There are 11−1 = 10 degrees of freedom, and the confidence level is 98%.  For a t-table, the critical value is CV = 2.764.

So the confidence interval is:

CI = 4.545 ± 2.764 × 3.503

CI = (-5.1, 14.2)

c.  The confidence level is 98%, so 98% of these confidence intervals will contain the true population mean, and 2% will not.

step-by-step explanation:

the distance (d) between two points (x1,y1) and (x2,y2) is given by the formula

d = √   ((x2-x1)2+(y2-y1)2)

d = √ (-11-6)2+(4-3)2

d = √ )2+(1)2)

d = √ (289+1)

d = √ 290

the distance between the points is 17.0293863659264

the midpoint of two points is given by the formula

midpoint= ((x1+x2)/2,(y1+y2)/2)

find the x value of the midpoint

xm=(x1+x2)/2

xm=(6+-11)/2=-2.5

find the y value of the midpoint

ym=(y1+y2)/2

ym=(3+4)/2=3.5

the midpoint is: (-2.5,3.5)

graphing the two points, midpoint and distance

p1 (6,3)

p2 (-11,4)

midpoint (-2.5,3.5)

the length of the black line is the distance between the points (17.0293863659264)

find the slope of the line connecting the two points

slope =   (y2-y1)   =   (4-3)   =   (1)   = -0.0588235294117647

(x2-x1) (-11-6) (-17)

find the equation of the line passing through the two points

the general equation for a straight line is

y = mx + b

where m represents the slope of the line which we found in the previous step to be -0.0588235294117647

y = -0.06x + b

we substitute x and y for the values from one of our points (6,3)

3 = -.06×6 + b

3 = -0.35 + b

3--0.35 = b

3.35 = b

knowing both b and m, we can contruct the equation of the line

y= -0.06x+ 3.35

x and y intercepts

the x-intercept is a point on the graph where y is zero

using the equation we found in the previous step and substituting zero for y

y= -0.06x+ 3.35

0= -0.06x+ 3.35

0.06x= 3.35

x= 3.35/0.06 = 57.00

the x intercept for this straight line is 57.00

the y-intercept is a point on the graph where x is zero

using the equation we found in the previous step and substituting zero for x

y= -0.06×0+ 3.35

y= 3.35

the y intercept for this straight line is 57.00

the distance (d) between two points (x1,y1) and (x2,y2) is given by the formula

d = √   ((x2-x1)2+(y2-y1)2)

d = √ (-11-6)2+(4-3)2

d = √ )2+(1)2)

d = √ (289+1)

d = √ 290

the distance between the points is 17.0293863659264

the midpoint of two points is given by the formula

midpoint= ((x1+x2)/2,(y1+y2)/2)

find the x value of the midpoint

xm=(x1+x2)/2

xm=(6+-11)/2=-2.5

find the y value of the midpoint

ym=(y1+y2)/2

ym=(3+4)/2=3.5

the midpoint is: (-2.5,3.5)

graphing the two points, midpoint and distance

p1 (6,3)

p2 (-11,4)

midpoint (-2.5,3.5)

the length of the black line is the distance between the points (17.0293863659264)

find the slope of the line connecting the two points

slope =   (y2-y1)   =   (4-3)   =   (1)   = -0.0588235294117647

(x2-x1) (-11-6) (-17)

find the equation of the line passing through the two points

the general equation for a straight line is

y = mx + b

where m represents the slope of the line which we found in the previous step to be -0.0588235294117647

y = -0.06x + b

we substitute x and y for the values from one of our points (6,3)

3 = -.06×6 + b

3 = -0.35 + b

3--0.35 = b

3.35 = b

knowing both b and m, we can contruct the equation of the line

y= -0.06x+ 3.35

x and y intercepts

the x-intercept is a point on the graph where y is zero

using the equation we found in the previous step and substituting zero for y

y= -0.06x+ 3.35

0= -0.06x+ 3.35

0.06x= 3.35

x= 3.35/0.06 = 57.00

the x intercept for this straight line is 57.00

the y-intercept is a point on the graph where x is zero

using the equation we found in the previous step and substituting zero for x

y= -0.06×0+ 3.35

y= 3.35

the y intercept for this straight line is 57.00 and therefore the awnser is c.