1. The mean gross annual incomes of certified welders are normally distributed w
ID: 3359547 • Letter: 1
Question
1. The mean gross annual incomes of certified welders are normally distributed with the mean of $37,682 and a standard deviation of $3,768. The ship building association wishes to find out whether their welders earn more or less than $37,682 annually. The alternate hypothesis is that the mean is not $37,682. Which of the following is the alternate hypothesis?
H1: $37,682
H1: µ < $37,682
H1: µ $37,682
H1: µ = $37,682
2.
A sample of 35 observations is selected from a normal population. The sample mean is 30, and the population standard deviation is 4. Conduct the following test of hypothesis using the .05 significance level.
3.
As a condition of employment, Fashion Industries applicants must pass a drug test. Of the last 210 applicants, 25 failed the test.
Develop a 95% confidence interval for the proportion of applicants that fail the test. (Round your answers to 3 decimal places.)
No
A sample of 35 observations is selected from a normal population. The sample mean is 30, and the population standard deviation is 4. Conduct the following test of hypothesis using the .05 significance level.
H0 : 29 H1 : > 29Explanation / Answer
Given that,
population mean(u)=29
standard deviation, =4
sample mean, x =30
number (n)=35
null, Ho: <=29
alternate, H1: >29
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 = 30-29/(4/sqrt(35)
zo = 1.479
| zo | = 1.479
critical value
the value of |z | at los 5% is 1.645
we got |zo| =1.479 & | z | = 1.645
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value : right tail - ha : ( p > 1.479 ) = 0.0696
hence value of p0.05 < 0.0696, here we do not reject Ho
ANSWERS
---------------
a. "One-tailed"-the alternate hypothesis is greater than direction.
b. reject Ho, if zo > 1.645
c. test statistic: 1.479
d. decision: do not reject Ho
e. p-value: 0.0696
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