u4) A company gihs dding the amount of time (in hours ach week tereste by girls.

Question

u4) A company gihs dding the amount of time (in hours ach week tereste by girls. Independent weekly time on their phones w n ensimple random samples of 81 boys and 82 girls were selected, and each child's as monitored. The summary statistics are provided in the table Boys r Girls x1-29.5 hr | x2 = 35.9 hr 51-8.5 hr | s2 = 5.3 hr n1-31n2 32 Use a 0.05 significance level to test the mean am significantly different. Perform a complete hypothesis test and answer th the claim that the mean amount of time on their phones for boys and girls is e questions below. Read the

Explanation / Answer

Q14.

Given that,
mean(x)=29.5
standard deviation , s.d1=8.5
number(n1)=31
y(mean)=35.9
standard deviation, s.d2 =5.3
number(n2)=32
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.042
since our test is two-tailed
reject Ho, if to < -2.042 OR if to > 2.042
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =29.5-35.9/sqrt((72.25/31)+(28.09/32))
to =-3.573
| to | =3.573
critical value
the value of |t | with min (n1-1, n2-1) i.e 30 d.f is 2.042
we got |to| = 3.57299 & | t | = 2.042
make decision
hence value of | to | > | t | and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -3.573 ) = 0.001
hence value of p0.05 > 0.001,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: -3.573
critical value: -2.042 , 2.042
decision: reject Ho
p-value: 0.001

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