1. For a one-tailed test (lower tail) at 93.7% confidence, Z = 2. For a one-tailed test (upper tail), a sample size of 24 at 90% confidence, t =

3. In a one-tailed hypothesis test (lower tail=left tailed test) and where the population standard deviation is known, the test statistic is determined to be -2.5. What is the p-value for this test?

1) Z = -1.8592

2) t = 1.7139

3) p = 0.0062

Step-by-step explanation:

1) Area: (93.7+100)/2 = 193.7/2 = 96.85

Z = invNorm(0.9685,0,1) = 1.8592, but since the test is one-tailed at the lower tail, Z is actually -1.8592

2) Area: (90+100)/2 = 190/2= 95 where df=n-1=24-1=23

t = invT(0.95,23) = 1.7139

3) Since the population standard deviation is known, our test statistic is z=-2.5, making the p-value for the test p=normalcdf(-1e99,-2.5,0,1) = 0.0062

step-by-step explanation:

for this pair of functions g(x)=x-2 and f(x)=|x| is (g o f)(a)= |a| -2

step-by-step explanation:

g(x)=x-2

f(x)=|x|

(g o f)(a)=?

(g o f)(x)=g(f(x))

(g o f)(x)=g(|x|)

(g o f)(x)=|x|-2

(g o f)(a)=|a|-2