Statistics Homework: KU Spring 2019 - Task Solutions

This document presents solutions to a statistics homework assignment, addressing concepts from confidence intervals and hypothesis testing to sample size calculations and margin of error. The assignment includes calculations of proportions, confidence intervals (95% and 99%), and conservative margin of error for different scenarios. It also covers hypothesis testing, including null and alternative hypotheses, p-values, and levels of significance, with examples involving salary, birth weight, and other variables. The solutions provide step-by-step calculations and interpretations, making it a comprehensive resource for students studying elementary statistics. The solutions cover tasks 7.3, 7.4 and 9.1 from the assignment brief.

STATISTICS
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Task - 7.3
Question 1
Sample = 180 bulbs
Number of defective bulbs = 27
Proportion of defective bulbs in factory =?
95% confidence interval =?
Margin of error e =?
Left end point (LEP)=?
Right end point (REP)=?
Conservative margin of error (E)=?
Now,
Proportion of defective bulbs in factory = 27/180 = 0.15
The z value for 95% confidence interval = 1.96
Margin of error = ± z value √ p ( 1−p )
n =±1.96 √ 0.15∗1−0.15
180 =± 0.0522
Left end point (LEP) = p – Margin of error = 0.15 – 0.0522 = 0.0978
Right end point (REP)= p + Margin of error = 0.15 +0.0522=0.2022
Conservative margin of error (E) = C- MOE = (REP-LEP)-MOE = (0.2022-0.0978)-0.0522
=0.0522
Question 2
Sample = 176 people
Number of people found to be infected with HIV = 44
Proportion of infected people =?
99% confidence interval =?
Margin of error e =?
Left end point (LEP)=?
Right end point (REP)=?
Conservative margin of error (E)=?
Now,

Proportion of infected people = 44/176 = 0.25
The z value for 99% confidence interval = 2.58
Margin of error = ± z value √ p ( 1−p )
n =±2.58 √ 0.25∗1−0.25
176 =± 0.0842
Left end point (LEP) = p – Margin of error = 0.25 – 0.0842 = 0.1658
Right end point (REP)= p + Margin of error = 0.25 +0.0842=0.3342
Conservative margin of error (E)= C- MOE = (REP-LEP)-MOE = (0.3342-0.1568)- 0.0842 =
0.0932
Question 3
Sample size =?
Proportion within 0.01 from sample proportion i.e. E = 0.01
95% level of confidence
Proportion = 0.15
n = {(z value)^2 *p(1-p)}/(Margin of error)^2 = (1.96*1.96*0.15*(1-0.15)/(0.01*0.01)
Required sample size n= 4898
Question 4
Sample size =?
Proportion within 0.02 from sample proportion i.e. E = 0.02
99% level of confidence
Proportion = 0.5
n = {(z value)^2 *p(1-p)}/(Margin of error)^2 = (2.58*2.58*0.5*(1-0.5)/(0.02*0.02)
Required sample size n =4147
Task 7.4
Question 1
95% confidence interval for variance of the salary =?
Left end point (LEP)=?
Right end point (REP)=?
Now,

It can be concluded with 95% confidence that the mean salary variance of the mathematics
faculty member falls between 94.77 and 280.91thousands dollars.
Now,
Left end point (LEP) of 95% CI of variance
Right end point (REP)of 95% CI of variance
Question 2
Sample size = 16
Sample variance = s^2 = $313

99% confidence interval for variance of tuition
It can be said with 99% confidence that the mean amount paid per semester by students
would fall between $143.13 and $1020.45.
Finally,
Left end point (LEP)
Right end point (REP)

Question 3

Finally,
Left end point (LEP)
Right end point (REP)

Task 9.1
Question 1
(a) Null and alternative hypotheses
Null hypothesis H0 :μ=51 KMean salary of new KU graduates is 51K.
Alternative hypothesis Ha : μ>51 K Mean salary of new KU graduates is more than 51K.
Now,
Sample mean salary x=54 K
Standard deviation σ=5 K
Sample size n = 15
z stat = x−μ
s . e . = 54−51
5
√ 15
=2.324
As per the sign of the alternative hypothesis, the test would be right tailed test.
The right tailed p value for the calculated z stat = 0.010063
(b) Level of significance = 0.05
Here, p value < level of significance which implies that null hypothesis would be rejected.
Answer: YES
(c) The lowest level of significance at which we can conclude that graduate salary has
increased is 2%.
Question 2

(a) Null and alternative hypotheses
Null hypothesis H0 :μ=112 ouncesMean birth weight of babies in US is 12 ounces.
Alternative hypothesis Ha : μ>1 12ounces Mean birth weight of babies in US is higher than
12 ounces.
Now,
Sample mean salary x=115 ounces
Standard deviation σ=17 ounces
Sample size n = 96
z stat = x−μ
s . e . = 115−112
17
√ 96
=1.729
As per the sign of the alternative hypothesis, the test would be right tailed test.
The right tailed p value for the calculated z stat = 0.041905
(b) Level of significance = 0.05
Here, p value < level of significance which implies that null hypothesis would be rejected.
Answer: YES
(c) The lowest level of significance would be 5%.
Question 3
(a) Null and alternative hypotheses
Null hypothesis H0 :μ=25 sMeantime of athleteis 25 sec .
Alternative hypothesis Ha : μ<25 s Mean time of athlete is lower than 25 sec .
Now,
Sample mean salary x=23.9 s
Standard deviation σ=3 s
Sample size n = 29
z stat = x−μ
s . e . = 23.9−25
3
√ 29
=−1.975

As per the sign of the alternative hypothesis, the test would be left tailed test.
The right tailed p value for the calculated z stat = 0.024134
(b) Level of significance = 0.05
Here, p value < level of significance which implies that null hypothesis would be rejected.
Answer: YES
(c) The lowest level of significance to conclude improvement is 3%.