Arandom sample of 64 freshmen spent an average of 14 hours per week watching television. the sample standard deviation was 32 hours per week. what is the critical value for a 95% confidence interval for the population mean?

+/- 1.9983

Step-by-step explanation:

The critical values for a confidence interval for the population mean, with population standard deviation not known, are obtained from the student's t distribution.

The sample size is given as 64. The degrees of freedom will thus be;

sample size - 1 = 64 - 1 = 63

The confidence level is given as 95%. The level of significance will thus be;

100 - 95 = 5%.

The area in the right-tail will thus be 2.5%. Therefore, we looking for a t-value such that the area to its right is 2.5% and the degrees of freedom being 63. From the t-tables we have;

+/- 1.9983

idk

step-by-step explanation:

answer: c, the green line

step-by-step explanation:

there is no black line, eliminate b.

simple interest implies periodic payments, but the red line shows no payments, eliminate d.

simple interest has equal payments. if paid say every day, the graph is about a straight line (actually a stairstep).

the yellow line is not straight, eliminate a.

check: the green line is straight, answer is c.

simple interest with positive interest rate gives a straight line sloping up to the right. green line:

g(t) = $100 + ($200-$100) ÷ (20 years) × (t years)

= $100 + (4%/year) × ($100)(t years)

incidentally, the yellow line curving upward represents continuously compounded interest

y = $100 × 2^(t years / (14 assuming the yellow line hits $200 at 14 years. 2^0 is 1, so y(0) = $100; 2^(14/14) is 2, so y(14) = $200.