This exercise is based on summary statistics rather than raw data. This informat

Question

This exercise is based on summary statistics rather than raw data. This information is typically all that is presented in published reports. You can perform inference procedures by hand from the summaries. Use the conservative Option 2 (degrees of freedom the smaller of n1 1 and n2 1) for two-sample t confidence intervals and P-values. You must trust that the authors understood the conditions for inference and verified that they apply. This isn't always true.

Equip male and female students with a small device that secretly records sound for a random 30 seconds during each 12.5-minute period over two days. Count the words each subject speaks during each recording period, and from this, estimate how many words per day each subject speaks. The published report includes a table summarizing six such studies. Here are two of the six.

a) Find the following two sample t statistic. (Round your answers to three decimal places.)

study 1: t= ???

study 2: t=???

ESTIMATED AVERAGE NUMBER (SD) OF WORDS SAMPLE SIZE SPOKEN PER DAY WOMEN STUDY WOMEN MEN MEN 56 16,177 (7520) 16,569 (9108) 56 2 27 20 16,496 (7914) 12,867 (8343)

Explanation / Answer

Q1.
Given that,
mean(x)=16177
standard deviation , s.d1=7520
number(n1)=56
y(mean)=16569
standard deviation, s.d2 =9108
number(n2)=56
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.004
since our test is two-tailed
reject Ho, if to < -2.004 OR if to > 2.004
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =16177-16569/sqrt((56550400/56)+(82955664/56))
to =-0.248
| to | =0.248
critical value
the value of |t | with min (n1-1, n2-1) i.e 55 d.f is 2.004
we got |to| = 0.24836 & | t | = 2.004
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -0.2484 ) = 0.805
hence value of p0.05 < 0.805,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: -0.248
critical value: -2.004 , 2.004
decision: do not reject Ho
p-value: 0.805

Q2.
Given that,
mean(x)=16496
standard deviation , s.d1=7914
number(n1)=27
y(mean)=12867
standard deviation, s.d2 =8343
number(n2)=20
to =16496-12867/sqrt((62631396/27)+(69605649/20))
to =1.507
| to | =1.507
critical value
the value of |t | with min (n1-1, n2-1) i.e 19 d.f is 2.093
we got |to| = 1.50686 & | t | = 2.093
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 1.5069 ) = 0.148
hence value of p0.05 < 0.148,here we do not reject Ho

ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 1.507
critical value: -2.093 , 2.093
decision: do not reject Ho
p-value: 0.148

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