Download Project and Reflection

Lee 1
Jinhee Lee
Math1040-012
Tiffany Hilton
Dec 05, 2103
Project 5-6
Objection; 1. Compute a hypothesis test for the population proportion
2. Complete a hypothesis test for the population mean.
3. In the reflection, discuss how your sample meets the conditions for performing these tests,
explain the meaning of the conclusions as it relates your population, and discuss Type I
error as it relates to your hypothesis test.
1. For one value of your categorical variable,
 From Simple random, n=31, smoker-2 (6.45%), nonsmoker-29 (93.55%)
*In a random sample of adults 31, adult 29 said that they do non-smoker. Use a 0.05 significance
level to test the claim that more than 75% of adults are non-smokers.
STEPS:
 1st step; Null Hypothesis H0: p=.75, H1: p>.75claim
 2nd step; right tailed; Zαfind the Critical Value-Z.05=1.645
 3rd step; compute the Test Statistic
n=31,
p̂ (sample proportion) = 29/31=93.55%-->.9355
p (proportion in claim) =.75
q = .25
.9355-.75/


=2.387
th
4 step; determine the p-value, Right-tailed: P (Z>2.387) .0087
5th step: make a decision; reject H0 or Fail to reject H0
Critical Value Method: compare T.S. and R.R.: T.S.=2.387>C.V.1.645so reject H0
P-Value Method: compare p-value and α: reject H0 if p-value (.0087) _< α (.015) So reject H0
 6th step:
Conclusion-There is sufficient evidence to support the claim that more than 75% of adults are non-smokers.
2. For quantitative samples, -population:
Column
n
Mean
Variance Std. dev. Median Range Min Max Q1 Q3 IQR
Height - inches
654 61.143578 32.530058 5.7035128
61.5
28
46
74 57 65.5
8.5
Simple Random from data3 quantitative (height)
Column
n
Mean
Variance
Std.
Dev.
Height(inches) 31 60.822582 34.942474 5.911216
Median Range Min Max Q1
61.5
24
48
Q3
Std. Err. IQR
72 56.5 64.5 1.0616857
* A simple random sample of 31 height is obtained and the sample mean is 60.823 inches.
The population standard deviation is 5.704. Use a 0.05 significance level to test the claim that the mean
height of the population is equal to 61.144, as is commonly believed.
STEPS:
8
Lee 2
 1st step: state the Hypothesis and identify the claim
ClaimH0:
=61.144,
H1 :
≠61.144
 2nd step: Find Critical Value
Two tailed: tα-two tails, n=31, n-1=30, df=30, α=.05 C.V. +_tα-two tails= +1.697, -1.697
 3rd step: compute the Test statistic: -.313
= 60.823-61.144/5.704/
3 1 = -.313

determine the p-value=Use Technology
TI83/84: STATTEST2T-Tests
0= 61.144
=60.823, s=5.704, n=31,
≠
hit and calculate enter
t= -.313
p=.756
 5th Make a decision-Reject H0 or Fail to reject H0
Critical Value Method: +_2.042, T.S= -.313 so fail to reject H0
P-value method: exact p-value: p=.756, α =.05  p> α so fail to reject H0
Approximately p-value:
1. Determine df=30,
2. Find test statistic lies relative to the values on the row
3. Look at the top of the two columns on either side of where the test statistic lies to find a range of
p-values for the hypothesis test.t= .313 p-value>.20
4. Choose the p-values according to whether the test is one-or two –tailed.
 6th conclusion
There is insufficient evidence to reject the claim that mean height of the population is 61.144
4th
Reflection:
For the categorical data, in a random sample of adults 31, adult 29 said that they do non-smoker.
Use a 0.05 significance level to test the claim that more than 75% of adults are non-smokers.
STEPS:
The sample meets the conditions for performing these tests,
1st step; Null Hypothesis H0: p=.75, H1: p>.75claim
2nd step; right tailed; Zαfind the Critical Value-Z.05=1.645
3rd step; compute the Test Statistic
n=31,
p̂ (sample proportion) = 29/31=93.55%-->.9355
p (proportion in claim) =.75
q = .25
.9355-.75/
=2.387
4th step; determine the p-value, Right-tailed: P (Z>2.387) .0087
5th step: make a decision; reject H0 or Fail to reject H0
 Critical Value Method: compare T.S. and R.R (reject region). T.S.=2.387>C.V.1.645so reject H0
 P-Value Method: compare p-value and α: reject H0 if p-value (.0087) _< α (.015)
So reject H0
6th The conclusion is “There is sufficient evidence to support the claim that more than 75% of adults
are non-smokers.”
Type I error as it relates to your hypothesis test.
Lee 3
Type I error would be the case to reject the H0 when there is actually 75% of non-smokers.
For quantitative data, a simple random sample of 31 height is obtained and the sample mean is 60.823
inches. The population standard deviation is 5.704. Use a 0.05 significance level to test the claim that
the mean height of the population is equal to 61.144, as is commonly believed.
The sample meets the conditions for performing these tests,
1st step: state the Hypothesis and identify the claim
ClaimH0:
=61.144,
H1 :
≠61.144
2nd step: Find Critical Value
Two tailed: tα-two tails, n=31, n-1=30, df=31, α=.05 C.V. +_tα-two tails=+_2.042
3rd step: compute the Test statistic: -.313
= 60.823-61.144/5.704/
31= -.313
4th determine the p-value=Use Technology
TI83/84: STATTEST2T-Tests
0= 61.144
=60.823, s=5.704, n=31,
≠
hit and calculate enter
t= -.313
p=.756
5th Make a decision-Reject H0 or Fail to reject H0
 Critical Value Method: +_2.042, T.S= -.313 so fail to reject H0
 P-value method: exact p-value: p=.756, α =.05  p> α so fail to reject H0
 Approximately p-value:
1. Determine df=30,
2. Find test statistic lies relative to the values on the row
3. Look at the top of the two columns on either side of where the test statistic lies to find a range of
p-values for the hypothesis test.t= .313 p-value >.20
4. Choose the p-values according to whether the test is one-or two –tailed.
6th conclusion
There is not sufficient evidence to reject the claim that mean height of the population is 61.144
Type I error as it relates to your hypothesis test.
Type I error would be the case to reject the H0 when the true mean height of the population is equal to
61.144.
Lee 4
Project 6
During the semester, part1-part3 project included group workings and individual works, part4 -part6
project was individual working. These concepts can make possible to understand many statistic materials.
Part1and part2 project was to understand categorical data, and we can distinguish simple random sample,
stratified sample, cluster sample, and systematic sample. Also, we should be able to compute Bar graph,
Pareto chart and Pie chart for each sample. Our group misunderstood some search the k-value and simple
random sample. It was okay because human learn through the many mistakes, but our group members’
reflection was very stranger to me because she said, “I don’t want to lose 25 point in group working.” Part3
project was to understand quantitative date’s meaning and it is to understand and compute the population
mean and population standard deviation. It was also to understand kinds of samples and to create frequency
histogram and box ploy for each sample. Reflection paper required compare and contrast the result of two
samples’ population. Personally, I don’t like group working because I have suggestion about systematic
sample method, but my group members just ignored that. I knew to search systematic sample method
because I asked it to my professor at prat1 and part2 project’s submission day. To search systematic sample,
k-value must be covered all population and to satisfy the condition n-value, 30<n<35, so k-value is
654/31which is about 21. Because this k-value must be 21 and I used part4 project. Part4 was to describe
categorical data’s confidence interval and to describe quantitative sample’s population parameter. I
successfully performed it. Part5 was hypothesis test for the population and the population mean. It also
distinguishes Type I error in my hypothesis tests.
Reflection: I learned and understood a lot of statistic methods and concepts through the projects. I will apply
some medical science section base on this statistics methods. Before I take the statistic class, I
have absolutely belief about statistic data, but I realize the data is sample data so there can exist
some differences compare with population. Because of this, the confidence level is very
important in statistics data. The project make to understand confidence level which is closer 100
percent is higher confidence in some data.
For example, I am very interesting about the relationship between some symptoms or diseases
and patience’s life-style which excise times and frequencies, eating pattern or eating habits,
stress management, and life of philosophy. I will study in medical section, and I want to know
reducing some diseases just only change their life-style instead of using treatments or medicine.
How do I applicant this concept and statistic data? I will collect the sample data and I will show
the relationship with some diseases and their life-style.