In 2004, registered nurses earned an average annual salary of $52,330. A survey

Question

In 2004, registered nurses earned an average annual salary of $52,330. A survey was conducted of 41 California nurses to determine if the annual salary is higher than $52,330 for California nurses. The sample average was $61,121 with a sample standard deviation of $7,849. (a) Conduct a hypothesis test at the 5% significance level. Clearly state the null and alternate hypotheses, the type of test you will use (right- left- or two-tailed test), the test statistic, and the p-value. (b) Clearly state whether you will or will not) reject the null hypothesis (and why). (c) What is the Type l error? (d) What is the Type ll error? Note: If you use a student-t distribution for this problem, you may assume that the underlying population is normally distributed. (However, in general, a statistician would first need to verify that this assumption is reasonable before applying a t-test.)

Explanation / Answer

a.
we use t test for single mean because they given sample standard deviation and sample mean so that we go for t test
Given that,
population mean(u)=52330
sample mean, x =61121
standard deviation, s =7849
number (n)=41
null, Ho: =52330
alternate, H1: >52330
level of significance, = 0.05
from standard normal table,right tailed t /2 =1.684
since our test is right-tailed
reject Ho, if to > 1.684
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =61121-52330/(7849/sqrt(41))
to =7.172
| to | =7.172
critical value
the value of |t | with n-1 = 40 d.f is 1.684
we got |to| =7.172 & | t | =1.684
make decision
hence value of | to | > | t | and here we reject Ho
p-value :right tail - Ha : ( p > 7.1716 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: =52330
alternate, H1: >52330
test statistic: 7.172
critical value: 1.684
decision: reject Ho
p-value: 0

b.
we have enough evidence to support the claim that annual salary is higher than 52330$ for california nurses

c.
Given that,
Standard deviation, =7849
Sample Mean, X =61121
Null, H0: =52330
Alternate, H1: >52330
Level of significance, = 0.05
From Standard normal table, Z /2 =1.645
Since our test is right-tailed
Reject Ho, if Zo < -1.645 OR if Zo > 1.645
Reject Ho if (x-52330)/7849/(n) < -1.645 OR if (x-52330)/7849/(n) > 1.645
Reject Ho if x < 52330-12911.605/(n) OR if x > 52330-12911.605/(n)
-----------------------------------------------------------------------------------------------------
Suppose the size of the sample is n = 41 then the critical region
becomes,
Reject Ho if x < 52330-12911.605/(41) OR if x > 52330+12911.605/(41)
Reject Ho if x < 50313.546 OR if x > 54346.454
Suppose the true mean is 52330
Probability of Type I error,
P(Type I error) = P(Reject Ho | Ho is true )
= P(50313.546 < x OR x >54346.454 | 1 = 52330)
= P(50313.546-52330/7849/(41) < x - / /n OR x - / /n >54346.454-52330/7849/(41)
= P(-1.645 < Z OR Z >1.645 )
= P( Z <-1.645) + P( Z > 1.645)
= 0.05 + 0.05 [ Using Z Table ]
= 0.1

d.
Given that,
Standard deviation, =7849
Sample Mean, X =61121
Null, H0: =52330
Alternate, H1: >52330
Level of significance, = 0.05
From Standard normal table, Z /2 =1.6449
Since our test is right-tailed
Reject Ho, if Zo < -1.6449 OR if Zo > 1.6449
Reject Ho if (x-52330)/7849/(n) < -1.6449 OR if (x-52330)/7849/(n) > 1.6449
Reject Ho if x < 52330-12910.8201/(n) OR if x > 52330-12910.8201/(n)
-----------------------------------------------------------------------------------------------------
Suppose the size of the sample is n = 41 then the critical region
becomes,
Reject Ho if x < 52330-12910.8201/(41) OR if x > 52330+12910.8201/(41)
Reject Ho if x < 50313.6687 OR if x > 54346.3313
Implies, don't reject Ho if 50313.6687 x 54346.3313
Suppose the true mean is 52330
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(50313.6687 x 54346.3313 | 1 = 52330)
= P(50313.6687-52330/7849/(41) x - / /n 54346.3313-52330/7849/(41)
= P(-1.6449 Z 1.6449 )
= P( Z 1.6449) - P( Z -1.6449)
= 0.95 - 0.05 [ Using Z Table ]
= 0.9
For n =41 the probability of Type II error is 0.9

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