A sample of 64 observations is selected from a normal population. The sample mean is 215, and the population standard deviation is 15. Conduct the following test of hypothesis using the .03 significance level. What is the p-value?H0 : μ ≥ 220H1 : μ < 220

If we compare the p value and the significance level given we see that then we have enough evidence to conclude that the true mean is significantly lower than 220 at 3% of significance.

Step-by-step explanation:

Data given

represent the sample mean

represent the population standard deviation

sample size      

represent the value that we want to test    

represent the significance level for the hypothesis test.    

z would represent the statistic (variable of interest)      

represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.      

We need to conduct a hypothesis in order to check if the true mean is lower than 220, the system of hypothesis would be:      

Null hypothesis:      

Alternative hypothesis:      

The statistic is given by:

(1)      

Calculate the statistic      

We can replace in formula (1) the info given like this:      

P-value      

Since is a one-side lower test the p value would be:      

Conclusion      

If we compare the p value and the significance level given we see that then we have enough evidence to conclude that the true mean is significantly lower than 220 at 3% of significance.

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.