Confidence Interval for Age Categorical Variable
I decided to find the confidence interval for the Elementary categorical
variable.
The population proportion for the Elementary variable was 176/654= 0.2691
I chose the confidence level of 95% for my confidence interval for the
sample proportion.
For Sample 1 (Simple Random Sample):
n= 35
x= 10
pHat= 10/35= 0.2857
qHat= 1-0.2857= 0.7143
To find the margin of error for the sample proportion, I used the formula:
E= Za (Square Root(pHatXqHat)/(n))
2
So,
E= 1.96(Square Root (0.2857X0.7143)/(35))= 0.1497
pHat-E<pHat<pHat+E
0.2857 – 0.1497= 0.1360
0.2857 + 0.1497= 0.4354
So the confidence interval for Sample 1 is (0.1360, 0.4354)
For Sample 2 (Systematic Sample):
n= 35
x= 12
pHat= 0.3429
qHat= 0.6571
E= 1.96(Square Root(0.3429X0.6571)/35))= 0.1572
0.3429 – 0.1572= 0.1857
0.3429 + 0.1572= 0.5001
So the confidence interval for Sample 2 is (0.1857, 0.5001)
Confidence Interval for Forced Expiratory Volume
Quantitative Variable
I decided to find the confidence interval for the mean of the Forced
Expiratory Volume in Liters quantitative variable.
The population mean was calculated as 2.637 and the population standard
deviation was calculated as 0.867.
I chose the confidence level of 95% for my confidence interval for the
sample mean.
For Sample 1 (Simple Random Sample):
n= 35
Xbar= 2.930
s= 1.043
I will be calculating the confidence interval for the sample mean by
pretending not to know the population standard deviation. Because of this, I
will use the following formula to find the margin of error:
E= tSub /2 ((s)/(Square Root of n))
E= 2.032 (1.043)/(Square Root of 35))
E= 0.358
Xbar – E
Xbar + E
2.930 – 0.358= 2.572
2.930 + 0.358= 3.288
So the confidence interval for the sample mean is (2.572, 3.288)
For Sample 2 (Systemic Sample):
n= 35
Xbar= 2.476
s= 0.870
E= 2.032 (0.870)/(Square Root of 35))
E= 0.299
2.476 – 0.299= 2.177
2.476 + 0.299= 2.775
So the confidence interval for the sample mean is (2.177, 2.775)
Reflection Paper
The purpose of calculating the confidence intervals for the samples is to provide
an estimate of the true value of the population. By using the confidence levels, one is able
to give a percentage of how accurate the calculations are to the true population value. For
example in the confidence interval calculated for Sample 1’s sample proportion, I
calculated that the true population proportion fell between the values of 0.136 and 0.4354
with 95% confidence. Since I had the advantage of knowing what the population
proportion, I was able to test my calculations. The population proportion was calculated
as 0.2691 and so I made the conclusion that the sample proportion did capture my
population proportion. For both samples of the categorical and quantitative variables, I
was able to capture the true population proportion and population mean.
Hypothesis Test for the Population Proportion
I will be conducting a hypothesis test based on the population proportion of
categorical variable of Elementary age children (ages 6 to 8). The population
proportion of elementary age children is calculated as 26.91%. We need to
test the claim that the population proportion is actually greater than 26.91%
by comparing the proportions of the two samples consisting of 35 trials each.
Sample 1 has a proportion of 28.57% elementary age children and Sample 2
has a proportion of 34.29% elementary age children. The significance level
will be 0.05.
First I need to see whether or not I am able to fulfill the requirements needed
to test this claim.
1. Samples are simple random (the second sample is systemic sampling
which is a version of random sampling)
2. There is a fixed number of trials. Meaning, we will be testing 35 trials.
3. That np>5 (9.4185, yes) and nq>5 (25.5815, yes)
Since all requirements are met, we are able to test this claim. We will be
using the P-value method. We will compare the P-value results to the
significance level of 0.05.
Sample 1
H0 : p= 0.2691
H1: p> 0.2691
q= 0.7309
pHat= 0.2857
X= 10 (0.2857x35)
Since this claim is dealing with proportions, we will use the Z-test formula
to calculate the test statistic, which will in turn help us find the P-value.
Z= ((phat – p)/(Square Root(pq/n)))
Z= ((0.2857 – 0.2691)/(Square Root(0.2691x0.7309/35)))
Z= 0.0166/0.0750
Z= 0.22
On the calculator we come up with a P-value of 0.4123. Since this number is
greater than the significance level of 0.05, we failed to reject the null
hypothesis. This means that there is not sufficient evidence to support the
claim that the actual population proportion is greater than 0.2691.
Sample 2
H0: p= 0.2691
H1: p> 0.2691
q= 0.7309
pHat= 0.3429
X= 12
Z= ((0.3429 – 0.2691)/Square Root(0.2691x0.7309/35)))
Z= 0.0738/0.750
Z= 0.98
On the calculator we come up with a P-value of 0.1626. Since this number is
greater than the significance level of 0.05, we failed to reject the null
hypothesis again with Sample 2. Again, we do not have sufficient evidence
to support the claim that the actual population proportion is greater than
0.2691.
Hypothesis Testing for the Population Mean
I will be conducting a hypothesis test based on the population mean of
quantitative variable of Forced Expiratory Volume in Liters. There is a claim
that the population mean of the forced expiratory volume is calculated as
2.637. After gathering two samples, each with the sample size of 35, we find
that Sample 1 has a mean of 2.930, standard deviation of 1.043 and Sample
2 has a mean of 2.476, standard deviation of 0.870. We need to test the claim
that the population mean is actually 2.637. The significance level will be
0.05.
First I need to see whether or not I am able to fulfill the requirements needed
to test this claim.
1. Samples are simple random (the second sample is systemic sampling
which is a version of random sampling)
2. The distribution is normal
3. That the sample size is at least 30 or greater if the distribution is not
normal.
Since all requirements are met, we are able to test this claim. We will be
using the P-value method. We will compare the P-value results to the
significance level of 0.05.
Sample 1
H0: µ = 2.637
H1: µ ≠ 2.637
Sample Mean: 2.930
Sample Standard Deviation: 1.043
n= 35
For the purpose of this test, we will act as if we do not know the population
standard deviation, since in most real life cases we do not know the
population standard deviation. We will use the T test statistic to find the Pvalue.
T= ((xbar - µ)/((s)/Square Root (n)))
T= ((2.930 – 2.637)/((1.043)/Square Root(35)))
T= 0.293/0.176
T= 1.665
On the calculator we come up with a P-value of 0.106. Since this number is
greater than the significance level of 0.05, we failed to reject the null
hypothesis. Therefore we do not have sufficient evidence to warrant the
rejection of the claim that the actual population mean is equal to 2.637.
Sample 2
H0: µ = 2.637
H1: µ ≠ 2.637
Sample Mean: 2.476
Sample Standard Deviation: 0.870
n= 35
Again, we will use the T test statistic to find the P-value
T= ((2.476 – 2.637)/((0.870)/Square Root(35)))
T= -0.161/0.147
T= -1.095
On the calculator we come up with a P-value of 0.281. Since this number is
greater than the significance level of 0.05, we failed to reject the null
hypothesis again with Sample 2. Therefore we do not have sufficient
evidence to warrant the rejection of the claim that the actual population
mean is equal to 2.637.
Reflection Paper
For both the population proportion and population mean hypothesis testing, I was
able to fulfill the conditions to perform the tests. In the population proportion testing,
both samples were simple random testing (mainly the first sample, but the second sample
was also a form of randomized testing). Also, the samples were able to meet the
condition of there being a set number of trials, which in this case was 35. Lastly, the
population proportion multiplied by the sample size was greater than 5. For the
population mean testing, the first requirement was the same as the population proportion,
but the difference in requirements was connected to the distribution. Since I did not know
whether or not the sample distribution was normal, I had to rely on the fact that the
sample size was larger than 30, which is was. Therefore, I was able to continue to
perform the tests based on meeting all of the conditions.
In both samples of the population proportion, I was unable to reject the null
hypothesis of the claim that the population proportion was actually greater than 26.91%.
This meant that I was unable to prove that the population proportion was greater than
26.91%. If there was a Type I error to this result that would mean that the population was
actually greater than 26.91%, but I have wrongfully proved that it was not greater.
Again, in both samples of the population mean, I failed to reject the null
hypothesis that the population mean was not equal to 2.637. This meant that I was not
able to prove that the claim was false. If there was a Type I error to this result, it would
mean that I was able to prove the claim was false, however, I failed to demonstrate this
and the claim will still be perceived as true.