Question

1. Suppose that we assume the population standard deviation is o = 5 and we are testing: Ho: = 50 vs. Ha:y> 50 and that we want the following Powers (probability of detection); PowerMar = 50.25) = .85. PowerMar = 50.40) - .98. a) How much data is needed to satisfy the first requirement? Answer b) How much data is needed to satisfy the second requirement? Answer c) How much data is needed to satisfy both requirements? Answer

Answer 1

Here

H₀ : $\mu$ = 50

H_a : $\mu$ > 50

$\sigma$ = 0.5

let required sample size = n

so

standard error (SE)= 0.5/$\sqrt n$

now,

we will reject the null hypothesis when $\bar x$ < 50 +  0.5/$\sqrt n$  

a.)

Power ($\mu_{alt}$ = 50.25) = 0.85

so here

Power = Pr($\bar x$ > 50 +  0.5/$\sqrt n$ ; 50.25 ; 0.5/$\sqrt n$) = 0.85

or

Pr($\bar x$ < 50 +  0.5/$\sqrt n$ ; 50.25 ; 0.5/$\sqrt n$) = 1 - 0.85 = 0.15

Here Z value for the given p -value is

Z = NORMSINV(0.15) = -1.036433

so here

(50 + 0.5/$\sqrt n$ - 50.25)/0.5/$\sqrt n$ = -1.036433

(0.5/$\sqrt n$ - 0.25) = -0.51822/$\sqrt n$

1.01822/$\sqrt n$ = 0.25

n = 16.5883

hence,

n=17

b.)

Power ($\mu_{alt}$ = 50.40) = 0.98

Power = Pr($\bar x$ > 50 +  0.5/$\sqrt n$ ; 50.40 ; 0.5/$\sqrt n$) = 0.98

or

Pr($\bar x$ < 50 +  0.5/$\sqrt n$ ; 50.40 ; 0.5/$\sqrt n$) = 1 - 0.98 = 0.02

Here Z value for the given p -value is

Z = NORMSINV(0.02) = -2.05375

so here

(50 + 0.5/$\sqrt n$ - 50.40/0.5/$\sqrt n$ = -2.05375

(0.5/$\sqrt n$ - 0.40) = -1.02687/$\sqrt n$

1.52687/$\sqrt n$ = 0.40

n = 14.57

Hence

n=15

a.)ans:17

b.)ans:15

c.)ans:data needed to satisfy both the conditions=17