In: Statistics and Probability
You want to test whether salaries have gone up compared to the year 2000. The 2000 Census shows that the average annual salary of all Americans was $35,152. You take an SRS of 10 Americans, ask for their annual salary for the year, and get the following results in thousands of dollars:
Annual Salary |
45 |
16 |
24 |
83 |
103 |
31 |
69 |
94 |
72 |
20 |
A) Determine the population(s) and parameter(s) being discussed.
B) Determine which tool will help us find what we need (one sample z test, one sample t test, two sample t test, one sample z interval, one sample t interval, two sample t interval).
C) Check if the conditions for this tool hold.
D) Whether or not the conditions hold, use the tool you choose in part B. Use C=95% for all confidence intervals and alpha=5% for all significance tests.
* Be sure that all methods end with a sentence describing the results *
A.
Given that,
population mean(u)=35152
sample mean, x =55700
standard deviation, s =32441.5714
number (n)=10
null, Ho: μ=35152
alternate, H1: μ!=35152
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.262
since our test is two-tailed
reject Ho, if to < -2.262 OR if to > 2.262
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =55700-35152/(32441.5714/sqrt(10))
to =2.0029
| to | =2.0029
critical value
the value of |t α| with n-1 = 9 d.f is 2.262
we got |to| =2.0029 & | t α | =2.262
make decision
hence value of |to | < | t α | and here we do not reject
Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != 2.0029 )
= 0.0762
hence value of p0.05 < 0.0762,here we do not reject Ho
ANSWERS
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B.
t test for single mean because given sample standard
deviation
C.
null, Ho: μ=35152
alternate, H1: μ!=35152
test statistic: 2.0029
critical value: -2.262 , 2.262
D.
decision: do not reject Ho
p-value: 0.0762
we do not have enough evidence to support the claim that the
average annual salary of all Americans was $35,152.