Question

5. A pilot training manager has hypothesized that 30 percent of the pilot want more flying hours during night compare to daylight flying.

Sample Size = 235

% in Favor (p) = 0.47

% not in Favor (q) = __________

In the random interview it seem only 45 percent preferred night flying.

Develop Hypothesis:

Null Hypothesis (H_o)                       : ___________________________________________

Alternate Hypothesis (H₁)             : ___________________________________________

Test the hypothesis using 9 % significance level

Decision Rule

Reject Ho

Fail to reject Ho

Interpret

PLEASE USE THIS FORMULA ONLY. AND INTERPRET THE ANSWER IN 2 SENTENSES

  π - Z √((p .  q)/n) <-------->.   π + Z √((p .  q)/n)

Answer 1

Answer)

Null hypothesis Ho : P = 0.3 (30%)

Alternate hypothesis Ha : P not equal to 0.3 (30%)

N = 235

First we need to check the conditions of normality that is if n*p and n*(1-p) both are greater than 5 or not

N*p = 70.5

N*(1-p) = 164.5

Both the conditions are met so we can use standard normal z table to estimate the P-Value

Test statistics z = (oberved p - claimed p)/standard error

Standard error = √{claimed p*(1-claimed p)/√n

Observed P = 0.47

Claimed P = 0.3

N = 235

After substitution

Test statistics Z = 5.69

From z table, P(z>5.69) = 0

As our test is two tailed

So, P-value is = 2*0 = 0

As the obtained P-value is less than 0.09 (given significance)

Reject Ho

We do not have enough evidence to conclude that 30 percent of the pilot want more flying hours during night compare to daylight flying.