The annual salaries (in dollars) or 14 randomly chosen te fi ters are istod. At

Question

The annual salaries (in dollars) or 14 randomly chosen te fi ters are istod. At ?-O 05 is there enough evidence to support te clam rat tw standard d Assume the population is normally distributed. Complete parts (a) through (e) below ation of ho an Maes s dn -bom SSS 07 50.751 40.939 52.409 46538 41.758 40.196 51.102 52.089 43,869 34,946 35.007 28,236 32,776 37,845 Click the icon to view the Chi-Square Distribution Table EEB (a) Write the claim mathematically and identity Ho and Ha OA. Ho?-5350, Ha : ?#5350 (Clam) O B. Ho os 5350 (cami, Ha: ? > 5350 ? ?. ?? o2 5350. Ha

Explanation / Answer

Define X : annual salaries (in $)

X ~ Normal(mu,sigma2)

To test the null hypothesis

H0 : sigma=sigma0 ( a specified value)

Against the alternative hypothesis

H1 : sigma not equal to sigma0

The test statistic is given by

T= {(n-1)*s2}/ (sigma02)

Where s is the sample standard deviation

n = sample size

Under H0 , T follows chi square distribution with df (n-1)

We reject H0 against H1 if observed T > c1 or T<c2

Where c1 and c2 are such that

P[T>c1 or T< c2]=alpha

Assigning equal error probability to both the tails

P[T>c1]=alpha/2= P[T<c2]

Which implies c1 = X2alpha/2;(n-1)

And c2 = X21-(alpha/2);(n-1)

Where X2alpha/2;(n-1) denotes the upper alpha/2 point of chi square distribution with df (n-1)

i.e. we rejet H0 at alpha % level of significance if

T> X2alpha/2;(n-1) or T< X21-(alpha/2);(n-1)

Here n=14

alpha=0.05

sigma0 =5350

sigma02 = 28622500

s2 = {1/(n-1)}*[sum{(x-x bar )2}] =60559188.4

x bar = sample mean =(1/n)*sum(x)

alpha/2= 0.025

observed value of the test statistic T ={(14-1)*60559188.4}/ 28622500 = 27.505

from the statistical table

X2alpha/2;(n-1) = X20.025;13 = 24.736

X21-(alpha/2);(n-1) =X20.975;13 =5.009

hence the critical values are (5.009,24.736?)

(a) option (A) is correct

the critical values are (5.009,24.736?)

(b) option (A) is correct

(c) observed value of T is the answer, i.e. 27.505

(d) since observed value of T (27.505) > X20.025;13 (=24.736)

we reject H0 against H1 at 5 % level of significance.

option (C) is correct.

(e) since H0 is rejected against H1 at 5% level of significance we can conclude that there is enough evidence that the annual salaries differs from $5350 significantly.

option (C ) is correct

Dear student hope I am able to help you to give you clear idea.

if you like my answer please rate it .

thank you !

happy learning !

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.