frito lay case in business statistics

1. FRITO - LAY CASE
By Group 14

2. Question 1 (a)
 p = 0.63 ( 63% of all U.S, Hispanics are Mexican-
American)
 n = 850, þ^ = 0.676
 α = 0.05
 H₀: þ = 0.63
Ha: þ ≠ 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 output
Test and CI for One Proportion
Test of p = 0.63 vs p not = 0.63
Sample X N Sample p 95% CI Z-Value P-Value
1 575 850 0.676471 (0.645021, 0.707921) 2.81 0.005
Using the normal approximation.

4. Question 1 (Part b)
 p = 0.94 ( 94% of all Hispanic grocery shoppers
are women)
 n =689, þ^ = 0.879
 α = 0.05
 H₀: þ = 0.94
Ha: þ < 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 output
Test and CI for One Proportion
Test of p = 0.94 vs p < 0.94
95% Upper
Sample X N Sample p Bound Z-Value P-Value
1 606 689 0.879536 0.899933 -6.68 0.000
Using the normal approximation.

6. Question 1 (Part c)
 H₀: þ = 0.83
H₁: þ ≠ 0.83
 α = 0.05
 n = 438, þ^ =0.792
 p- value = 0.042, which is less than α
 Result: Reject H₀
 With 95% probability, proportion of
Hispanics listen primarily to ads in Spanish
is not equal to 83%

7. Question 2 (Part a)
 μ = 31, n= 24, sample mean = 28.31
 s = 7.0897, α = 0.01
 H₀: μ = 31
 H₁: μ ≠ 31
 t test, df = 23 (Since n small and
population standard deviation unknown)
 t = 2.807 (at 0.005, 23)

8. Minitab output
One-Sample T
Test of mu = 31 vs not = 31
N Mean StDev SE Mean 99% CI T P
24 28.81 7.09 1.45 (24.75, 32.87) -1.51 0.144
Here we see that t= -1.51 which is in range with
calculated t from table.
Also p value is greater than α = 0.01 therefore we
cannot reject the null hypotheses.

9. Question 2 (Part b)
 n= 18, sample mean = $35.667
 s = 19.259, α = 0.05
 H₀: μ = $45
 H₁: μ < $45
Single tail test
 t test, df = 17 (Since n small and
population standard deviation unknown)
 t cal= 1.740 (0.05,17)

10. Minitab output
One-Sample T
Test of mu = 45 vs > 45
95% Lower
N Mean StDev SE Mean Bound T P
18 35.67 19.26 4.54 27.77 -2.06 0.972
Here we see that t= -2.06 which is in range with
calculated t from table.
Also p value is greater than α = 0.05 therefore we
cannot reject the null hypotheses.

11. Thank you