Na gallop poll of 1012 randomly selected adults, 9% said that cloning of humans should be allowed. we are going to use a .05 significance level to test the claim that less than 10% of all adults say that cloning of humans should be allowed. the test statistic for this test is -1.06. what is the p-value for this test statistic?

0.1446

Step-by-step explanation:

Let p be the population proportion of adults say that cloning of humans should be allowed.

As per given , the appropriate set of hypothesis would be :-

Since is left-tailed and sample size is large(n=1012) , so we perform left-tailed z-test.

Given : The test statistic for this test is -1.06.

P-value for left tailed test,

P(z<-1.06)=1-P(z<1.06)        [∵P(Z<-z)=1-P(Z<z]

=          (Using z-value table.)

Hence, the p-value for this test statistic= 0.1446

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.