In: Statistics and Probability
The pH of 20 randomly selected lakes is measured. Their average pH is 5.7.
Part C. Five of the 20 lakes have a pH less than 4.5.
Part Ci. Build a 95% confidence interval for the proportion of
lakes that have a pH less than 4.5.
Part Cii. Test whether the population proportion of pH’s that are
less than 4.5 differs from 0.5.
Part Ciii. To be accurate to within 0.1 with 95% confidence, how
many lakes should we measure?
ci)
Level of Significance, α =
0.05
Number of Items of Interest, x =
5
Sample Size, n = 20
Sample Proportion , p̂ = x/n =
0.2500
z -value = Zα/2 = 1.960 [excel
formula =NORMSINV(α/2)]
Standard Error , SE = √[p̂(1-p̂)/n] =
0.0968
margin of error , E = Z*SE = 1.960
* 0.0968 = 0.1898
95% Confidence Interval is
Interval Lower Limit = p̂ - E = 0.250
- 0.1898 = 0.060
Interval Upper Limit = p̂ + E = 0.250
+ 0.1898 = 0.440
95% confidence interval is (
0.0602 < p < 0.440
)
----------
cii)
Ho : p = 0.5
H1 : p ╪ 0.5
(Two tail test)
Level of Significance, α =
0.05
Number of Items of Interest, x =
5
Sample Size, n = 20
Sample Proportion , p̂ = x/n =
0.2500
Standard Error , SE = √( p(1-p)/n ) =
0.1118
Z Test Statistic = ( p̂-p)/SE = ( 0.2500
- 0.5 ) / 0.1118
= -2.2361
p-Value = 0.0253 [excel formula
=2*NORMSDIST(z)]
Decision: p-value<α , reject null hypothesis
There is enough evidence to conclude that
population proportion of pH’s that are less than 4.5
differs from 0.5
----------------
ciii)
sample proportion , p̂ =
0.25
sampling error , E = 0.1
Confidence Level , CL= 0.95
alpha = 1-CL = 0.05
Z value = Zα/2 = 1.960 [excel
formula =normsinv(α/2)]
Sample Size,n = (Z / E)² * p̂ * (1-p̂) = (
1.960 / 0.1 ) ² *
0.25 * ( 1 - 0.25 ) =
72.03
so,Sample Size required=
73