2. Question 1 (a) p = 0.63 ( 63% of all U.S, Hispanics are
Mexican- American) n = 850, ^ = 0.676 = 0.05 H: = 0.63 Ha: 0.63
(double tail) Z from table = 1.96 Z observed value = 2.81 Therefore
we reject the null hypotheses. Percentage of U.S. Hispanics that
are Mexican- American is not 63%.
3. Minitab outputTest and CI for One ProportionTest of p = 0.63
vs p not = 0.63Sample X N Sample p 95% CI Z-Value P-Value1 575 850
0.676471 (0.645021, 0.707921) 2.81 0.005Using the normal
approximation.
4. Question 1 (Part b) p = 0.94 ( 94% of all Hispanic grocery
shoppers are women) n =689, ^ = 0.879 = 0.05 H: = 0.94 Ha: <
0.94 Z from table = -1.645 Z observed value is -6.68 Therefore
reject null hypotheses. Percentage of Hispanic grocery shoppers
that are women has fallen.
5. Minitab outputTest and CI for One ProportionTest of p = 0.94
vs p < 0.94 95% UpperSample X N Sample p Bound Z-Value P-Value1
606 689 0.879536 0.899933 -6.68 0.000Using the normal
approximation.
6. Question 1 (Part c) H: = 0.83 H: 0.83 = 0.05 n = 438, ^
=0.792 p- value = 0.042, which is less than Result: Reject H With
95% probability, proportion of Hispanics listen primarily to ads in
Spanish is not equal to 83%
7. Question 2 (Part a) = 31, n= 24, sample mean = 28.31 s =
7.0897, = 0.01 H: = 31 H: 31 t test, df = 23 (Since n small and
population standard deviation unknown) t = 2.807 (at 0.005,
23)
8. Minitab outputOne-Sample TTest of mu = 31 vs not = 31 N Mean
StDev SE Mean 99% CI T P24 28.81 7.09 1.45 (24.75, 32.87) -1.51
0.144Here we see that t= -1.51 which is in range with calculated t
from table.Also p value is greater than = 0.01 therefore we cannot
reject the null hypotheses.
9. Question 2 (Part b) n= 18, sample mean = $35.667 s = 19.259,
= 0.05 H: = $45 H: < $45Single tail test t test, df = 17 (Since
n small and population standard deviation unknown) t cal= 1.740
(0.05,17)
10. Minitab outputOne-Sample TTest of mu = 45 vs > 45 95%
Lower N Mean StDev SE Mean Bound T P18 35.67 19.26 4.54 27.77 -2.06
0.972Here we see that t= -2.06 which is in range with calculated t
from table.Also p value is greater than = 0.05 therefore we cannot
reject the null hypotheses.