Statistical Problems: One Sample t-test, Two Sample t-test, Two Sample Proportion z-test, One-way ANOVA

Statistical Problems
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1.
A. The Population Mean rental for one-bedroom apartments in Los Angeles = μ≤1500 dollars
Sample size = n=100 one bedroom apartment
Sample Mean = x
¿
=1523 dollars
Sample standard deviation = s=120dollars
The claim of the Department of Housing about the average rental for one bedroom apartment in
Los Angeles was tested at 5% level of significance by one sample t-test.
Null Hypothesis: H0: ( μ≤1500 ) was tested against the one tail alternate hypothesis
HA : ( μ>1500 ) at 5% level of significance.
The test statistic was calculated as
tcal= x
¿
−μ
s
√ n
=1523−1500
120
√ 100
=23
12 =1 . 917
at 99 degrees of
freedom.
At 5% level the tcrit =1. 66 (right tail) for 99 degrees of freedom.
Hence, tcal>tcrit at 5% level of significance
Again, the p-value was evaluated as P ( t>1 . 917 ) =0 .029 < α = 0.05
The null hypothesis was rejected at 5% level of significance, since p-value was less than the
level of significance.
B. The confidence interval was calculated as
x
¿
±tcrit
s
√ n =[ x
¿
−t0. 025
s
√ n , x
¿
+t0 . 025
s
√ n ]=[1523−1 . 98∗12, 1523+1 . 98∗12 ]=[ 1499. 19 , 1546 . 81]
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With 95% confidence it was possible to estimate that the population mean (average rental for one
bedroom apartment in Los Angeles) was somewhere between $ 1499.19 and $ 1546.81.
2. For males, mean = x
¿
1=10 . 9 , sample standard deviation = s= √ 8 . 4=2 . 898 , and sample size =
n1=41
For females, mean = x
¿
2=10. 1 ,sample standard deviation = s= √ 8 . 8=2. 966 , and sample size =
n2=41
The difference in the average daily commute distance in Long Beach Population between a
males and females was tested by two-sample independent t-test at 5% level of significance.
The variances of the two populations were not known, and therefore independent t-test (two-tail)
for unequal variances was conducted at 5% level of significance (Dean, Sullivan, & Soe, 2014).
Null Hypothesis: H0: ( μ1=μ2 ) : There is no difference in the average daily commute distance.
Alternate Hypothesis: HA: ( μ1≠μ2 ) : There is significant difference in the average daily
commute distance.
The test statistic was
t= x1
¿
−x
¿
2
√ s1
2
n1
+ s2
2
n2
=10 . 9−10 . 1
√ 8. 4
41 + 8 . 8
41
= 0 . 8
√ 0 . 419 =1. 236 at 5 % level
Where DF =
=
( s1
2
n1
+ s2
2
n2 )
( s1
2
n1 ) 2
n1−1 +
( s2
2
n2 )
n2−1
2
2
= ( 0 . 419 ) 2
( 0. 205 ) 2
40 + ( 0 .215 ) 2
40
= 0. 1755
0 . 00105+0 . 00115 =79 . 77≈80
The p-value was evaluated as P (|t|>1. 236 ) =0 . 22
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