29. To test whether the mean math score is higher for males students than for fe

Question

29. To test whether the mean math score is higher for males students than for females students, a math test was given to a group of 8th graders. The results of the test are summarized in the following table. Assume that the distribution of math scores for both groups is normally distributed with equal standard deviations. Sample size Mean Standard deviation Males n1 = 14 X1=55.9 S1=12.96 Females n2 =25 X2=48.4 S2=12.65 Given that the test statistic is 1.526, is there enough evidence to conclude that males perform better in math than females? Test at the 5% level of significance. (a) yes, because the test statistic is in the rejection region; (b) no, because the test statistic is in the rejection region; (c) yes, because the test statistic is not in the rejection region; (d) no, because the test statistic is not in the rejection region; (e) aconclusion is not possible 30. The mean age of residents in a large city is 35 with a standard deviation 13. If 100o residents are randomly selected from this city, the probability that their average age is less than 32 is about: (a) 1711 (b) .5910 (c) 4090 (d) .9896 (e) 0104

Explanation / Answer

29.
Given that,
mean(x)=55.9
standard deviation , s.d1=12.96
number(n1)=14
y(mean)=48.4
standard deviation, s.d2 =12.65
number(n2)=25
null, Ho: u1 = u2
alternate, H1: u1 > u2
level of significance, = 0.05
from standard normal table,right tailed t /2 =1.687
since our test is right-tailed
reject Ho, if to > 1.687
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (13*167.962 + 24*160.023) / (39- 2 )
s^2 = 162.812
we use test statistic (t) = (x-y)/sqrt(s^2(1/n1+1/n2))
to=55.9-48.4/sqrt((162.812( 1 /14+ 1/25 ))
to=7.5/4.259
to=1.761
| to | =1.761
critical value
the value of |t | with (n1+n2-2) i.e 37 d.f is 1.687
we got |to| = 1.761 & | t | = 1.687
make decision
hence value of | to | > | t | and here we reject Ho
p-value: right tail -ha : ( p > 1.7608 ) = 0.04326
hence value of p0.05 > 0.04326,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 > u2
test statistic: 1.761
critical value: 1.687
decision: reject Ho
p-value: 0.04326
we have enough evidence to support the claim that males perform better than the females
option:C
yes, because the test statistic is not in the rejection region

30.
NORMAL DISTRIBUTION
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
mean ( u ) = 35
standard Deviation ( sd )= 13/ Sqrt ( 100 ) =1.3
sample size (n) = 100
probability that the average age is less than 32 is
P(X < 32) = (32-35)/13/ Sqrt ( 100 )
= -3/1.3= -2.3077
= P ( Z <-2.3077) From Standard NOrmal Table
= 0.0105
= 0.0104
option:e

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