Ateacher wants to estimate the mean time (in minutes) that students take to go from one classroom to the next. his research assistant uses the sample time of 42 students to report the confidence interval as [7.40, 8.60]. [you may find it useful to reference the t table.] a. find the sample mean time used to compute the confidence interval. (round intermediate calculations to 4 decimal places and final answer to the nearest whole number.) b. determine the confidence level if the sample standard deviation used for the interval is 1.606. (round intermediate calculations to at least 4 decimal places. round "t" value to 3 decimal places and final answer to the nearest whole number.)

a:  We can't determine this from the given information

b: 98%

Step-by-step explanation:

For a:

n = 42

The confidence interval has equal time on each side of µ, so we can add the two end points and divide them by 2 to find the middle of the interaval:

7.4 + 8.6 = 16

16/2 = 8

Now subtract 7.4 from 8 to find the distance from the mean to the end of the interval

8 - 7.4 = 0.6

So the sample mean, plus the calculated error was 0.6 minutes.

We don't have a way of calculating the sample mean with the given information.  We could only find the sample standard deviation and the variance.

For b:  

We have:

E = 0.6

s = 1.606

n = 42

See attached photo for the calculation of this value

The value is 2.421.

Using a sample size of 42, our degrees of freedom are 41.  Use the t-distribution chart to see which level of confidence has 2.421 under it.

The level of confidence is: 98%

We need 41 degrees of freedom, but the chart has only 40, then 45.  We can see that 40 has 2.423, and the values go down as the degrees of freedom go up, so 41 will correlate to 2.421

the answer is (-8,-5)

step-by-step explanation:

the endpoint to midpoint is half the length. the length is (-2,-1) so when you are at the midpoint just add it to it to find your answer -6+(-2)= -8 and -4+(-1)=-5 so (-8,-5)

what is the question?

step-by-step explanation: