A federal funding program is available to low-income neighborhoods. To qualify for
funding, a neighborhood must have a mean household income of less than $15,000
per year. Neighborhoods with mean annual household income of $15,000 or more
do not qualify. Funding decisions are based on a sample of residents in the
neighborhood. A hypothesis test with a .02 level of significance is conducted. If the
funding guidelines call for a maximum probability of .05 of not funding a
neighborhood with a mean annual household income of $14,000, what sample size
should be used in the funding decision study? Use σ = $4000 as a planning value.
Case(i)
Consider the hypothesis test first.
The null hypothesis is H0: The mean house hold income μ < 15000
Alternative hypothesis is Ha: The mean household income μ ≥ 15000
The hypothesis will be rejected at 2% level if (Xbar - 15000)√n/σ ≥ 2.0537
That is Xbar ≥ 15000 + 2.0537 σ/√n
P(Xbar ≥ 15000 + 2.0537 σ/√n /μ=14000) ≤ 0.05
P( Z ≥ (15000 + 2.0537 σ/√n -14000)*√n/σ) ≤ 0.05
P( Z ≥ (1000 + 2.0537 σ/√n ) *√n/σ) ≤ 0.05
P( Z ≥ 1000 √n/σ + 2.0537 ) ≤ 0.05
1000 √n/σ + 2.0537 ≥ 1.644854 from table of areas under the standard normal
curve.
1000 √n/4000 + 2.0537 ≥ 1.644854 using the value 4000 for σ
√n/4 + 2.0537 ≥ 1.644854
√n/4 ≥ 1.644854 - 2.0537 = -0.4089
This is true for all vaues of n.
Case(ii)
Consider the following hypothesis test first.
The null hypothesis is H0: The mean house hold income μ ≥ 15000
Alternative hypothesis is Ha: The mean household income μ < 15000
The hypothesis will be rejected at 2% level if (Xbar - 15000)√n/σ ≤- 2.0537
That is Xbar ≤ 15000 - 2.0537 σ/√n
The neighborhood will be disqualified for funding if is Xbar > 15000 - 2.0537 σ/√n
P(Xbar > 15000 - 2.0537 σ/√n /μ=14000) ≤ 0.05
P( Z ≥ (15000 - 2.0537 σ/√n -14000)*√n/σ) ≤ 0.05
P( Z ≥ (1000 - 2.0537 σ/√n ) *√n/σ) ≤ 0.05
P( Z ≥ 1000 √n/σ - 2.0537 ) ≤ 0.05
1000 √n/σ - 2.0537 ≥ 1.644854 from table of areas under the standard normal
curve.
1000 √n/4000 - 2.0537 ≥ 1.644854 using the value 4000 for σ
√n/4 - 2.0537 ≥ 1.644854
√n/4 ≥ 1.644854 + 2.0537 =3.6986
𝑛 ≥ (3.6986 × 4)2 = 218.8
The minimum sample size is 219