Question

9. The Wechsler IQ test is designed so that the mean is 100 and the standard deviation is 15 for the population of normal adults. Listed below are IQ scores of randomly selected professional pilots. It is claimed that because professional pilots are a more homogeneous group than the general population. Use a 0.05 significance level to test the claim that pilots have IQ scores with a standard deviation less than 15. Do not use the p-value. 121 116 115 121 116 107 127 98 116 101 130 114

Answer 1

We frame the Hypothesis

   Pilots have IQ Scores with a Standard Deviation = 15

Но

Pilots have IQ Scores with a Standard Deviation less than15.

H1

(Or)

$H_{0}: \sigma =15$

H:0 < 15

Since ; So, it's a One Tailed ( Left Tailed ) Hypothesisi Test. So, we don't use modulous in the numerator of the formula.

H:0 < 15

To test the ; we use Z - Statistic

он

$Z = \frac{s - \sigma }{S.E(s)}$

Where = Sample Standard Deviation

S

$\sigma$ = Population Standard Deviaton

$S.E(s) = \sqrt{\frac{s^{2}}{2n}}$

Where = Sample Variance

2 S

n = Size of the Sample = 12

	X
	121
	116
	115
	121
	116
	107
	127
	98
	116
	101
	130
	114
S.D	9.0998
VARIANCE$(s^{2})$	82.8056

Therefore

$S.E(s) = \sqrt{\frac{s^{2}}{2n}} = \sqrt{\frac{82.8056}{2*12}} = \sqrt{\frac{82.8056}{24}} = \sqrt{3.4502}= 1.8575$

Therefore

$Z = \frac{s - \sigma }{S.E(s)}$

$\Rightarrow Z = \frac{9.0998 - 15 }{1.8575}$

$\Rightarrow Z = \frac{-5.9002 }{1.8575}$

$\therefore \, \, \, Z_{cal} = -3.1759$

Given Level of Sifnificance = 5% = 0.05

$\Rightarrow Z_{cri} = Z_{\alpha } = Z_{5\%} = Z_{0.05} = -1.6449$

Since So, We Reject
он
at 5% Level of Significant.

Therefore we conclude that " Pilots have IQ Scores with a Standard Deviation less than15".

i.e $" \sigma < 15"$

NOTE: Since it's the test base on Left Tailed Hypothesis; So, We Rejected
он
though .

EXCEL COMMAND for $Z_{cri}$ is

=NORM.S.INV(0.05)