Statistical Analysis

Statistical Analysis
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2
Problem 1
Solution.
Given that
Ho: Null Hypothesis P≥0.05 (There are more female drivers than male
drivers)
Ha: Alternative Hypothesis (P<0.05 (There are fewer female drivers than
male drivers)
n=sample size is 900
p= sample of proportion is (468/900) = 0.52 which is the probability of
female drivers.
q=1-p=(1-0.52)=0.48
But SE= √ ( pq ) /¿ n ¿
¿ √(0.52∗0.48)/900=0.0167
Now determining the Test statistic:
Z= (0.52-0.5)/0.0167=1.200
From the table P(Z<1.200) = 0.8849
Therefore, P-Value= 1- 0.8849=0.1151
Since the P-value = 0.1151is greater than = 0.05, We fail to reject the
null hypothesis.
Conclusion
There is no enough evidence to reject the claim that there are more
female drivers in the USA than male drivers.
Problem 2.
Solution
Ho: Null Hypothesis: P=98.6°F
Ha: Alternative Hypothesis: >98.6 °F
Given
n=100 x=98.6 °F , σ =0.6 °F , μ =98.6°F
Z=( x−μ
s
√ n
)
Z=( 98.8−98.6
0.6
√ 100
)
Z=3.333
Now from the table
P(z=3.3333)=0.9996
Therefore, P-value =0.9996
Since the P-Value=0.9996 is greater than = 0.05, we fail to reject the
claim that the mean body temperature of the population is equal to 98.6
°F
Conclusion
There is no sufficient evidence to conclude that the common belief is
wrong.

3
Problem 3
Solution
Ho: Null Hypothesis, Ho: μ<20
Ha: Alternative Hypothesis, Ha: μ>20
Given n=25, x=15, μ<20 σ ¿ 4, μ=20
t=( x−μ
σ
√ n
)
t=(15−20
4
√ 25
)
t=¿-6.25
The critical value for left tail is -1.711
Computing Z statstics
Z=(x−μ)/σ
Z=15−20
4
Z=-1.25
The computed critical value (-1.711) is less than -1.25.
Conclusion
There is enough eveidence to support the claim that the mean tar content
of filtered 100 mm cigarettes is less than 20 mg, which is the mean for
unfiltered king size cigarettes
Problem 4
Solution
Given that
N’=2500
X’=15
Therefore, sample proportion of N’
=15/2500
=0.006
N=7500
X=15