Z-test: A psycho-linguist has been studying the frequency of text-speech (e.g.,
ID: 3358081 • Letter: Z
Question
Z-test: A psycho-linguist has been studying the frequency of text-speech (e.g., R U Ok?, LOL ) has been progressing into EMAIL correspondence. He collects a sample of 100 EMAIL messages and computes the mean (average) number of examples of text-speech per message. He also has a huge library of archived EMAIL messages from the year 2001. From these he is able to calculate the population average and standard deviation. Compute the Z-test to evaluate the hypothesis that the use of text-speech has increased in the years since 2001. Use an level of 0.05. These are the obtained values: Population Mean from 2001: 1.7 Population Standard deviation from 2001: 1.26 Sample mean from this year: 1.95 A) Compute the Z-test to evaluate the hypothesis that the use of text-speech has increased in the since 2001. B) What conclusion can the researcher draw from this analysis about the use of text-speech? (Remember to phrase your answer in terms of the independent and dependent variables.)
Explanation / Answer
Given that,
population mean(u)=1.7
standard deviation, =1.26
sample mean, x =1.95
number (n)=100
null, Ho: =1.7
alternate, H1: >1.7
level of significance, = 0.05
from standard normal table,right tailed z /2 =1.645
since our test is right-tailed
reject Ho, if zo > 1.645
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 1.95-1.7/(1.26/sqrt(100)
zo = 1.98
| zo | = 1.98
critical value
the value of |z | at los 5% is 1.645
we got |zo| =1.98 & | z | = 1.645
make decision
hence value of | zo | > | z | and here we reject Ho
p-value : right tail - ha : ( p > 1.98 ) = 0.02
hence value of p0.05 > 0.02, here we reject Ho
ANSWERS
---------------
null, Ho: =1.7
alternate, H1: >1.7
test statistic: 1.98
critical value: 1.645
decision: reject Ho
p-value: 0.02
text-speech has increased in the years since 2001
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