Download Running Head: INFERENTIAL STATISTICS Student Name

Running Head: INFERENTIAL STATISTICS
Student Name:
Institution:
Professor:
Course Title:
Date of Submission:
1
Running Head: INFERENTIAL STATISTICS
Question 1:
(a). There are two independent samples. Let the sample from West Virginia be n1 and the
sample from Alabama be n2.
n1=20
δ 1=1.9
µ1=5.6
Does this support Fred’s claim?
N2=26
δ 2=1.3
µ2=4.8
We proceed to test the following set of hypothesis.
H0: µ1-µ2 ≥0
H1: µ1- µ2 <0
(b)
Mean
Dependent
Assume δ1=δ2
Proportion
Variance
One-sided
Difference of means
Independent
Assume δ1 ≠δ2
Difference of proportion
Difference of variances
Two-sided
2
Running Head: INFERENTIAL STATISTICS
(c ): The correct standardized test-statistic is the z test-statistic
Z=(mean for sample 1-mean for sample 2)/√(n1/s12+n2/s22)
Z=(5.6-4.8)/√(3.61/20+1.69/26)=1.6146
The Critical z: =1.644852 at the 0.05 level of significance. (Read from statistical tables).
Since the critical z value for the two independent samples is greater than the z statistic, we fail to
reject the null hypothesis at the 0.05 level of significance and conclude that Fred’s claim is
supported by the available evidence at the 0.05 level of significance.
Question 2:
(a)95% Confidence interval for δ
Sample mean=579
Sample standard deviation=24
Sample size=61
Sample standard deviation±Z(alfa/2)*s/√(n-1)
24±1.96*(24/√60)=(17.92716211<δ<30.07283789
Hence, 95% Confidence Interval for the St. Dev.:
17.92716211 < δ < 30.07283789
3
Running Head: INFERENTIAL STATISTICS
(b) The 95% Confidence interval for µ
Sample mean=579
Sample standard deviation=24
Sample size=61
Sample mean±Z(alfar/2)*s/√n-1=579±1.96*(24/√60)
572.9271621<µ<585.0728379
95% Confident the population mean is within the range:
572.8533 < mean <585.1467
(c ):
H0: µ=550
H1:µ≠550
Level of significance=0.05
Sample mean=579
Sample standard deviation=24
Sample size=61
Z=(579-550)/(24/√61)=29
4
Running Head: INFERENTIAL STATISTICS
From statistical table, the critical z value is 1.97, Since the z statistic is greater than the critical
z value, we reject the null hypothesis at the 0.05 level of significance and conclude that the average
Big Mac has no 550 calories.
(d): Test the claim that the standard deviation of the calories in a Big Mac is at most 20 at the 0.05
level of significance.
H0: δ≤20
H1: δ>20
Sample mean=579
Sample standard deviation=24
Sample size=61
Z=(24-20)/(20/√61=1.56
Z test Statistic=1.56
Critical z value = 1.96 (Read from statistical table at the 0.05 level of significance).
Since the z statistic is less than the critical z value we fail reject the null hypothesis at the 0.05
level of significance and conclude that the standard deviation of calories in a Big Ma is at most 20.
Question 3:
5
Running Head: INFERENTIAL STATISTICS
(a): Test the gerrymanderer’s claim at the 98% Confidence level.
Neighborhood A
n1=45
No.of successes=(45-17)=28
Neighborhood B
n2=62
No. of successes=(62-28)=34
H0: p1≤p2
H1: p1>p2
Level of significance=0.02
Pooled proportion: 0.5794393
Z=(p^-p)/√(pq)/n=(0.5794393-0.451612903)/√(0.5794393*0.451612903)/45=1.67
Critical z value is=2.06 (Read from z statistical table)
since the z test statistic is less than the critical z value, we fail to reject the null hypothesis at the
0.02 level of significance and conclude that there is no evidence to support the gerrymanderer’s
belief.
(b):
H0: p≤0.6
H1: p>0.6
Level of significance=0.02
6
Running Head: INFERENTIAL STATISTICS
Sample size=62
No. of successes=34
Sample proportion: (34/62)= 0.548387
Z statistic=(0.548387-0.6)/√(0.4*0.6)/62=/-13.3333/=13.3333
The critical z value at the 0.02 level of significance is 2.06
Since the z test statistic is greater than the critical z value, we reject the null hypothesis at the 0.02
level of significance and conclude that the claim is supported at the 0.02 level of significance.
(c ).
H0: p=0.377777777
H1: p≠0.377777777
Sample size n=45, No. of successes=28
Sample proportion: 0.6222222
Z=(0.377777777-0.6222222)/(√(0.6222222*0.3777778)/45)=/-3.382/=3.382
Critical Z value(Read from the table)=2.58
7
Running Head: INFERENTIAL STATISTICS
Since the Z statistic is greater than the critical z value, we reject the null hypothesis at the 0.01
level of significance and conclude that Neighborhood A is not really a republocratic dominated
area.
Question 4:
(a): H0: Heart rate of people who have had a heart attack are not higher than for those who have
had a heart attack.
H1: Heart rates of people who have had a heart attack are higher than for those who have never
had a heart attack.
Heart Attack(Sample 1)
No Heart Attack(Sample 2)
Sample size=5
Sample size=5
Sample mean=123.6
Sample mean=108.4952
Sample standard deviation=11.08152
Sample standard deviation=5.07937
Not eq. vars: No Pool (and df calculated with Formula 9-1
Z statistic=(Mean of sample 1-mean of sample2)/(√(s12/n1+s22/n2)
Z=(123.6-108.4952)/(√(122.8000855/5)+(25.7999996/5)=2.770706046
The critical Z value at the 0.1 level of significance is =1.28
Since the z statistic is greater than the critical z value, we reject the null hypothesis at the 0.1 level
of significance and conclude that heart rates for those who had heart attack are higher than those
who never had heart attack.
(b):
H0: D=0
8
Running Head: INFERENTIAL STATISTICS
H1: D≠0 , where D is the difference between the two groups “Before and After”
Before
After
Sample size=5
Sample size=5
Sample mean=123.6
Sample mean=120
Sample standard deviation=11.08152
Sample standard deviation=13.58308
Not eq. vars: No Pool (and df calculated with Formula 9-1)
Before
115
125
111
128
139
Mean difference=(18/5)=3.6
After
112
130
105
115
138
d
3
-5
6
13
1
T=(Mean difference-D0)/(s/√n) where n=5
S=√(3.6-3)2+(3.6+5)2+(3.6-6)2+(3.6-13)2+(3.6-1)2=169.44
T=3.6/(169.44/√5)=0.047508526
T critical=t(0.05, 4)=2.776
Since the T statistic is less than the t critical value, we fail to reject the null hypothesis at the 0.05
level of significance and conclude that exercise does not reduce heart rates for patients who have
had heart attack.
9
Running Head: INFERENTIAL STATISTICS
10