Anewspaper conducted a statewide survey concerning the 1998 race for state senator. the newspaper took a srs of n=1100 registered voters and found that 570 would vote for the republican candidate. let p represent the proportion of registered voters in the state who would vote for the republican candidate. we test h0: p=.50 ha: p> .50 (a) what is the z-statistic for this test? (b) what is the p-value of the test?

Part A

p = 0.5 is the hypothesized population proportion

x = 570 is the number of successes

n = 1100 is the sample size

phat = x/n = 570/1100 = 0.518182 is the approximate sample proportion

SE = standard error

SE = sqrt(p*(1-p)/n)

SE = sqrt(0.5*(1-0.5)/1100)

SE = 0.015076 which is approximate

----

We have enough to compute the z test statistic

z = (phat - p)/(SE)

z = (0.518182 - 0.5)/(0.015076)

z = 1.206023

z = 1.21

==============================

Part B

Compute the probability of getting a z score larger than 1.21

P(Z > 1.21) = 1 - P(Z < 1.21)

P(Z > 1.21) = 1 - 0.8869 .... see note below

P(Z > 1.21) = 0.1131

note: the value 0.8869 comes from the table that I've attached as an image below. I have marked in red the proper row and column used to look up the value. The z table is commonly found in statistics textbooks toward the back sections. There are also many free online z tables you can search out as well. I used a z table found online. Alternatively, you can use your TI83 or TI84 calculator to compute the area under the normal curve. You would use the normalCDF function.

its a little blurry tbh

step-by-step explanation:

c.

step-by-step explanation:

she should of added 9^2 and 12^2. that is because it is a^2+b^2=c^2. c^2 is the hypotenuse and should always be bigger than the other two numbers. which is why she should of added 9 squared and 12 squared together instead of 15 squared.