Question

A Lansing company is planning to introduce a new detergent powder in the market, but before that they would like to do a survey and get an idea about the proportion of Michigan families who may buy this new product. A random sample of 93 Michigan families was given free sample of the new detergent powder to use and their comments were collected after a week. 48 of the families said that they were satisfied with the product (i.e. they would buy the product if it is introduced to the market).

1. Here the population of interest is:
a) 93 families, who were given free sample
b) all Michigan families
c) 48 satisfied families
d) proportion of satisfied customers

2. Here the sample is:
a) 93 families, who were given free sample
b) all Michigan families
c) 48 satisfied families
d) proportion of satisfied customers

3.Here the variable "satisfaction" is:
a) quantitative
b) qualitative
c) qualitative for some families, and quantitative for other

4.What is the target parameter in this problem?
a)93 sampled families
b)proportion of families willing to buy the new detergent
c) proportion of satisfied families in the sample
d) no parameter is involved in this problem

5. Here, the standard error of sample proportion takes the value: [Answer to 4 decimal places]

A: 0.0518

B: 0.0996

C: 0.1008

D: 0.1128

E: 0.1205

F: 0.1819

6. Here, the margin of error for the 98% confidence interval of sample proportion is [Answer to 4 decimal places]

A: 0.0518

B: 0.0720

C: 0.1205

D: 0.1945

E: 0.2124

F: 0.2402

7. The 98% confidence interval of proportion of Michigan residents willing to buy the new detergent powder:
(0.3914, 0.6408)
(0.4097, 0.6226)
(0.4067, 0.6255)
(0.3956, 0.6367)

8. Is the sample size sufficiently large to compute the above 98% confidence interval?
a) Yes, because n ≥ 30.
b) It is not possible to answer this question without knowing what is the number of failure here.
c) No, the sample size is actually not sufficiently large.
d) Yes, because n[^(p)] ≥ 10 and n(1−[^(p)]) ≥ 10.

Answer 1

1) b) all Michigan families

2) a) 93 families, who were given free sample

3) b) qualitative

4) b)proportion of families willing to buy the new detergent

5)

sample success x =		48
sample size n=		93.0
pt estiamte p̂ =x/n=		0.51613
standard error =se= √(p*(1-p)/n) =		0.0518

6)

for 98 % CI value of z=		2.326
margin of error E=z*std error   =		0.1205

7)

lower bound=p̂ -E                       =		0.3956
Upper bound=p̂ +E                     =		0.6367
from above 98% confidence interval for population proportion =(0.3956,0.6367)

8)

d) Yes, because n[^(p)] ≥ 10 and n(1−[^(p)]) ≥ 10.