The Administrator data contains a sample of school administrator salaries in Ohi

Question

The Administrator data contains a sample of school administrator salaries in Ohio. Using this data, answer the 8 questions below in the space provided. This question has 3 pages.

1Formulate hypotheses that can be used to determine whether the population mean annual administrator salary in Ohio is smaller than the national mean of $90,000. Use the standard mathematical notation for stating the null and alternative hypotheses as well as an English description of each hypothesis. Compute the test t, the critical t and the p-value. Test your hypotheses at alpha = 0.05 and state your conclusion using both the p-values and the test values.

Ho:____________ Description:_______________

Ha:____________ Description:_______________

Critical t:____________

Test t :____________

P-value :____________

Conclusion:________________________________________________________

2) Assuming a normal distribution, compute the following probabilities.Use standard mathematical notation.

a) Probability that a single person drawn at random has a salary less than $90,000

b) Probability that a single person drawn at random has a salary greater than $100,000

c) Probability that a single person drawn at random has a salary between $80,000 and $110,000

d) Probability that 36 peopole drawn at random have a mean salary less than $90,000

Ohio Salary 77600 76000 90700 97200 90700 101800 78700 81300 84200 97600 77500 75700 89400 84300 78700 84600 87700 103400 83800 101300 94700 69200 95400 61500 68800 78700 84600 87700 103400 83800

Explanation / Answer

1. let X denotes the annual administrator salary in Ohio.

assumption is X~N(u,sigma2) with both the parameters unknown.

we have to Formulate hypotheses that can be used to determine whether the population mean annual administrator salary in Ohio is smaller than the national mean of $90,000.

so H0:u=90000 description: the mean annual administrator salary in Ohio is equal to $90000

Ha: u<90000 description: the mean annual administrator salary in Ohio is less than $90000

to test this we have a sample of size n=30 with sample mean=Xbar=$85667 and standard deviation=s=$10668

since the population standard deviation sigma is unknown it is estimated by s.

level of significance=alpha=0.05

so the test t is given by T=(Xbar-90000)*sqrt(n)/s which under H0 follows a t distribution with df n-1

and since the alternative hypothesis is a left tailed one hence

critical t is -talpha;n-1  the lower alpha point of a t distribution with df n-1

and H0 is rejected iff t<-talpha;n-1   where t is test t

and p value is P[T<test t] where T follows a t distribution with df n-1

critical t=-t0.05;29= -1.69913   [using MINITAB]

test t=t=(85667-90000)*sqrt(30)/10668=-2.22467

p value = P[T<-2.22467]=0.0170228

so test t<critical t and p value<level of significance=alpha=0.05

hence conclusion: at 5% level of significance H0 is rejected and the conclusion is that the population mean annual administrator salary in Ohio is smaller than the national mean of $90,000

2. now assumption is X follows a normal distribution with mean=85667 and standard deviation=10668

so Z=(X-85667)/10668~N(0,1)

a) we need to find P[X<90000]=P[(X-85667)/10668<(90000-85667)/10668]=P[Z<0.406]=0.6576287 [answer] [using R}

b) P[X>100000]=P[(X-85667)/10668>(100000-85667)/10668]=P[Z>1.34355]=0.08954697 [answer] [using R]

c) P[80000<X<110000]=P[(80000-85667)/10668<(X-85667)/10668<(110000-85667)/10668]=P[-0.5312<Z<2.2809]

=P[Z<2.2809]-P[Z<-0.5312]=0.9887228-0.2976401=0.6910827 [answer] [using R]

d) now if we take a sample of n=36 people and sample mean be Xbar then Xbar would follow a normal distribution with E[Xbar]=85667 and standard deviation=10668/sqrt(36)=10668/6=1778

so P[Xbar<90000]=P[(Xbar-85667)/1778<(90000-85667)/1778]=P[Z<2.437]= 0.9925952 [answer] [using R]

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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