Statistical computation (Coursework Sample)

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Math Problem, Mathematics
Introduction
Question one
Standard deviation=578 pounds,α =0.05
Solution
Ho: µ=3000
H1: µ>3000 this will be right tailed test
N=60 mean=3120 S=578 α =0.05 Test Value= 1.61
C.V= 1-0.05=0.95the shaded region
From the table Zcv= 1.65 C.V=1.65
Test value: Z= mean-µS/‚n
= (3120-3000)/ 578/‚60
=1.61
Reason= the test value doesn’t fall within the shaded region hence no evidence that the average production of peanuts has increased. (C.V=1.65)
Question 8
Solution
Average=$59.93
H0= $59,593 H1=$59,593 (this will be the claim) -1.65 0
Critical value= α 0.0500
-1.65
Test value= Z= mean-µS/‚n
Z= (58,800-59593) / 1500/‚30
Z=-2.90
Analysis= the test value is less than critical value
We reject the H0
Conclusion= the state employees earn less than the federal employees
Question 11
H0: 500 σ=42. σ=0.01
H1: 476Z= mean-µS/‚n
Z= (500-476) /42/ ‚50
=4.04
Analysis = the test value is more than the critical value
We accept the H0,
The mean differs from 500.
Question 16
Solution
H0: µ= 52
H1:µ ≠ 52 Z= mean-µS/‚n
Z= (56.3-52) / 3.5 / ‚50
Z=8.69 -1.96 0 1.96
Reason= NO, the Z value is into the rejected region. Therefore we reject his claim.
Section 8-3
Question 8
Solution
H0 : µ
H1: µ<25.4 α=0.1, df=n-1 t=(22.1-25.4)/ 5.3/‚25 =-3.11 CV: t0.1= -1.38.
Comparing: -3.11< -1.38 (outside CV).
Conclusion: reject H0 suggesting that the average commute is indeed shorter.
Question 14
Solution
Claim µ<= 110 calories, if false µ> 110 calories.
H0: µ= 110 calories. H1 :µ> 110 calories. µ0=110 µ>µ0
p- Value= 0.000309 hence reject H0. There is sufficient evidence to reject the claim that the average content of calories is not more than 110.
Question 17
Solution
At α= 0.05, n=20, mean = 3.85 and σ=2.519. d.f= 19.
H0: µ= 5.8 H1: µ≠ 5.8(claim) t= mean-µS/‚n= (3.85-5.8)/ 2.519/‚20= -3.46.
p- Value<0.01.
if p-value <=α, 0.01<=0.05 hence true, we reject the null hypothesis. There is no enough evidence to support the claim that the mean is not 5.8
Section 8.4
Question 10
Ho: µ=500
H1: µ=420
σ=0.05
α=0.86Z= mean-µS/‚n
Z= (500-420) /0.86/0.05‚380
1.13
The proportion does not differ from the national percentage.
Question 14
Solution
Ho: p=0.517
Ha: p≠0.517 n=200, x=115, p^=115/200=0.575
Z = (0.575-0.517)/ ((0.517)(0.483)/200)=1.64
P-value = 2P (Z>1.64) =0.101
Conclusion: since the p-value (0.101) >0.05
Don´t reject H0
The evidence supports the claim there is no enough evidence to say that the percentage has changed.