Answer:
A) The probability is 0.95 that the percent of adults living in the United States who are satisfied with their health care plans is between 63.6% and 68.4%.
Step-by-step explanation:
A polling agency reported that 66 percent of adults living in the United States were satisfied with their health care plans. The estimate was taken from a random sample of 1,542 adults living in the United States, and the 95 percent confidence interval for the population proportion was calculated as (0.636, 0.684).
This means that we are 95% sure that the true proportion of adults living in the United States who were satisfied with their health care plans is between 0.636 and 0.684.
So the correct answer is:
A) The probability is 0.95 that the percent of adults living in the United States who are satisfied with their health care plans is between 63.6% and 68.4%.
Answer:
Step-by-step explanation:
Let the exponential function be
We substitute (0,9) to get:
The equation now becomes:
We substitute (3,72) to get:
The equation is therefore
Answer:
Null Hypothesis: H_0: \mu_A =\mu _B or \mu_A -\mu _B=0
Alternate Hypothesis: H_1: \mu_A >\mu _B or \mu_A -\mu _B>0
Here to test Fertilizer A height is greater than Fertilizer B
Two Sample T Test:
t=\frac{X_1-X_2}{\sqrt{S_p^2(1/n_1+1/n_2)}}
Where S_p^2=\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2}
S_p^2=\frac{(14)0.25^2+(12)0.2^2}{15+13-2}= 0.0521154
t=\frac{12.92-12.63}{\sqrt{0.0521154(1/15+1/13)}}= 3.3524
P value for Test Statistic of P(3.3524,26) = 0.0012
df = n1+n2-2 = 26
Critical value of P : t_{0.025,26}=2.05553
We can conclude that Test statistic is significant. Sufficient evidence to prove that we can Reject Null hypothesis and can say Fertilizer A is greater than Fertilizer B.
R = 6t.....subbing in (8,48).....t = 8 and r = 48
48 = 6(8)
48 = 48 (correct)
r = 6t...subbing in (13,78)...t = 13 and r = 78
78 = 6(13)
78 = 78 (correct)
so u have 2 sets of points on this line and they are (8,48) and (13,78)
