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Statistics T test 2 populations, Exercises of Statistics

This document contains questions about statistics and how to do them using the T-test and 2 populations formula step by step

Typology: Exercises

2020/2021

Available from 06/12/2023

silvia-maulani
silvia-maulani 🇮🇩

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At the 5% significance level, do the data provide sufficient evidence to conclude that the mean
time served for fraud is less than that for firearms offenses? (Note:
)10.12 , .90 , x8.78 , s.64x1= s1= 4 2= 1 2= 4
We know that :
, 0.12x1= 1 ,.90s1= 4 0n1= 1
,8.78 x2= 1 .64s2= 4 0n2= 1
.05 df 8 α = 0 = n1+n2 2 = 1
H0: μ1= μ2
H1: μ1< μ2
.77177 sp=n+n−2
1 2
(n−1)s+(n−1)s
1 122 22=10+10−2
(10−1)4.90 +(10−1)4.64
2 2 4
.05811 t=xx
1 2
sp+
1
n1
1
n2
=10.12−18.78
4.77177+
1
10 1
10
4
ritical value T.10092 c= α,df =T0.05,18 = 2
Since the , so we reject the null hypothesis.Tcalculated <Tα,df
Therefore, the data provide sufficient evidence to conclude that the mean time served for fraud
is less than that for firearms offenses.
10.39 Doing Time. The Federal Bureau of Prisons publishes data in Prison Statistics on the
times served by prisoners released from federal institutions for the first time. Independent
random samples of released prisoners in the fraud and firearms offense categories yielded the
following information on time served, in months.
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At the 5% significance level, do the data provide sufficient evidence to conclude that the mean time served for fraud is less than that for firearms offenses? (Note: x 1 = 10.12 , s 1 = 4 .90 , x (^) 2 = 1 8.78 , s (^) 2 = 4 .64 ) We know that : x 1 = 1 0.12 , s 1 = 4 .90 , n 1 = 10 x 2 = 1 8.78 , s 2 = 4 .64 n 2 = 10 α = 0 .05 df = n 1 + n 2 − 2 = 18 H 0 : μ 1 = μ 2 H 1 : μ 1 < μ 2

sp =.

√ n^^1 +^ n^^2 −

( n (^) 1 −1) s (^) 12 +( n (^) 2 −1) s (^22) =

(10−1)4.90 +(10−1)4.64^2 2 ≈ 4

t = (^) spx^^1 − x^2 −.

n^1 1 n^1 2

101 101 ≈^4

c ritical value = T (^) α, df = T (^) 0.05,18 = 2.

Since the T (^) calculated < T (^) α, df , so we reject the null hypothesis. Therefore, the data provide sufficient evidence to conclude that the mean time served for fraud is less than that for firearms offenses.

10.39 Doing Time. The Federal Bureau of Prisons publishes data in Prison Statistics on the times served by prisoners released from federal institutions for the first time. Independent random samples of released prisoners in the fraud and firearms offense categories yielded the following information on time served, in months.

10.40 Gender and Direction. In the paper “The Relation of Sex and Sense of Direction to Spatial Orientation in an Unfamiliar Environment” (Journal of Environmental Psychology, Vol. 20, pp. 17–28), J. Sholl et al. published the results of examining the sense of direction of 30 male and 30 female students. After being taken to an unfamiliar wooded park, the students were given some spatial orientation tests, including pointing to the south, which tested their absolute frame of reference. The students pointed by moving a pointer attached to a 360◦ protractor. Following are the absolute pointing errors, in degrees, of the participants.

At the 1% significance level, do the data provide sufficient evidence to conclude that, on average, males have a better sense of direction and, in particular, a better frame of reference than females? (Note: x 1 = 37.6, s 1 = 38.5, x 2 = 55.8, and s 2 = 48.3.) We know that : x 1 = 37.6, s 1 = 38.5, n 1 = 30 x 2 = 55.8, s 2 = 48.3 n 2 = 30

α = 0 , 01 df = n 1 + n 2 − 2 = 68

c ritical value = T α, df = T 0.01,68 = ± 2.

H 0 : μ 1 = μ 2 H 1 : μ 1 > μ 2

sp = 43,

√ n^^1 +^ n^^2 −

( n (^) 1 −1) s (^) 12 +( n (^) 2 −1) s (^22) =

(30−1)38.5 +(30−1)48.3^2 2 ≈

T (^) calculated = (^) sx^^1 −^ x^2 −.

p √ n^11 + n^12

c ritical value = T (^) α, df = T (^) 0.01,68 = 2.

Since the T (^) calculated < T (^) α, df , so we fail to reject the null hypothesis.

Therefore, the data does not provide sufficient evidence to conclude that on average, males have a better sense of direction and, in particular, a better frame of reference than females.