A random sample of 100 observations from a quantitative population produced a sa
ID: 3363445 • Letter: A
Question
A random sample of 100 observations from a quantitative population produced a sample mean of 23.9 and a sample standard deviation of 7.9. Use the p-value approach to determine whether the population mean is different from 25. Explain your conclusions. (Use = 0.05. Round your test statistic to two decimal places and your p-value to four decimal places.)
1-2. Null and alternative hypotheses:
H0: = 25 versus Ha: 25
H0: 25 versus Ha: = 25
H0: = 25 versus Ha: < 25
H0: = 25 versus Ha: > 25
H0: < 25 versus Ha: > 25
3. Test statistic: z = 4.
p-value = 5.
Conclusion:
H0 is rejected. There is sufficient evidence to indicate that the mean is different from 25.
H0 is rejected. There is insufficient evidence to indicate that the mean is different from 25.
H0 is not rejected. There is sufficient evidence to indicate that the mean is different from 25.
H0 is not rejected. There is insufficient evidence to indicate that the mean is different from 25.
Explanation / Answer
Given that,
population mean(u)=25
standard deviation, =7.9
sample mean, x =23.9
number (n)=100
null, Ho: =25
alternate, H1: !=25
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 23.9-25/(7.9/sqrt(100)
zo = -1.39
| zo | = 1.39
critical value
the value of |z | at los 5% is 1.96
we got |zo| =1.39 & | z | = 1.96
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != -1.39 ) = 0.16
hence value of p0.05 < 0.16, here we do not reject Ho
ANSWERS
---------------
null, Ho: =25
alternate, H1: !=25
test statistic: -1.39
critical value: -1.96 , 1.96
decision: do not reject Ho
p-value: 0.16
H0 is not rejected. There is insufficient evidence to indicate that the mean is different from 25.
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