Download σ < = 2.355 = 4.492 - Emily Miller`s ePortfolio

Emily Miller
Math 1040
Project Part 4
Part A:
A confidence interval is an estimate of a population parameter, something you observe,
or from a sample of the population parameter. In other words the purpose of a confidence
interval will show a close or true value of the parameter.
For a 99% confidence interval estimate for the true population, we gather the appropriate
data and compute the interval.
All class yellows = n = 1867
All class yellow candies = x= 387
𝑥
=387/1867=0.207=𝑝̂
𝑛
99% confidence interval = 1-0.99=0.01
Zα/2=0.01/2=0.005 look up in the body of z score table = 2.575
√𝑝̂ (1-𝑝̂ )/n = √. 207 ∗ .793/1867 = 0.009
E=Z𝛼/2√𝑝̂ 𝑞̂/n-2.575*.009 = 0.23
Confidence interval = 𝑝̂ ±E = (.184, .23)
It is a true population parameter, because the 99% confidence level is between .184, and
.23.
My single bag had 23 yellow candies out of 63 total candies
23
= 0.365 My bag would not be a good estimate or unusual because my proportion was
not within the confidence interval.
63
To construct 95% confidence interval, gather required data:
n= 31
df= 30
𝑥̅ = 60.2
s= 3.127
E= 1.147
𝑠
3.127
tα/2*√𝑛= 2.042* √31 = E =1.147
60.2 ± 1.147 = (59.053, 61.347)
I can say with 95% confidence that the population mean of candies per a bag of skittles
will be 59.053 to 61.347 of candies per bag, which is a good representation of the sample
population.
My single bag contained 63 candies that that was not in the confidence interval because it
was not between 59.053, and 61.347
98% confidence interval for the estimate of the standard deviation is computed with the
following information
S=3.127
N=31
√
√
√
(𝑛−1)𝑠 2
<σ<√
2
𝑥𝑅
30∗3.1272
50.892
30∗3.1272
14.954
(𝑛−1)𝑠 2
𝑥𝐿2
= 2.355
= 4.492
2.355 < σ < 4.492
We can say with a 98% confidence that the mean of candies per bag are within 2.355 and
4.492 of the population standard deviation.
Conditions for true proportion:
1- Must be SRS
2- Conditions for binomial distribution must be satisfied: fixed # of trials,
independent trials, 2 possible outcomes, probability of success is constant for each
trial.
3- Sample size large enough, at least 5 successes and 5 failures
Conditions for population mean
1- Must be SRS
2- Population is normally distributed or n>30, with our class
Conditions for estimating a population standard deviation:
1- Must be SRS
2- Population must have normally distributed values (regardless of sample sizes)
The entire class used a convenience sample to select the bags of skittles for this project
so the requeriment for that was not met. The other conditions howeverwere met, we had
all conditions required to get the ture proportion. The conditions for estimating a
population standard deviation I don’t think were meet because the population is not
normally distributed according to the graphs from the part 3 b of the project the
histogram is skewed to the left.
Part B:
The meaning and purpose of a hypothesis test uses data from a study to show you if
something is likely or unlikely to occur. In other words it shows the outcome of a study.
For a .05 level to test the claim that 20% of all skittles candies are red we need the
following information to compute.
p ̂=.184
p=.2
q=.8
n=1867
a=.05
H0: p=.20
H1: p≠.20
𝑝̂−𝑝 .184−.2
Z= 𝑝𝑞 = 0.2∗0.8 = -1.73= test statistic
√
𝑛
√
1867
P-value = 0.0418
Significance level (a) = .05/2=.025
Critical values are -1.96 and 1.96
-1.96>critical region>1.96
We would fail to reject the null hypothesis because the (p-value) 0.0418>.a (.025) and the
test statistic -1.73 is not less than -1.96 so it is not in the critical region of being less than1.96 Therefore, there is insufficient evidence to reject the claim.
To use a .01 significance level to test the claim that the mean number of candies per bag
is 55 we would need the following information to compute.
𝑥̅ = 60.2
μ = 55
s = 3.127
n = 31
a=.01
H0: μ = 55
H1: μ ≠ 55
𝑥̅ −𝜇
t= 𝑠
√𝑛
t=
60.2−55
3.127
√31
= 9.259= test statistic
P-value < .01
Significance level =.01
Critical values= -2.750 and 2.750
-2.750>critical region>2.750
Our rejection region is ±2.750 and our T.S. is 9.259 which is greater that the critical
value rejection which means it is in the rejection region. So we would reject the H0
because there is sufficient evidence to reject the claim.
Conditions for testing a claim about p:
1- Must be SRS
2- Binomial distribution
3- np≥ 5 and nq≥ 5 and both must be satisfied.
Conditions for testing a claim about 𝜇 with σ unkown:
1- Must be SRS
2- Either or both, normally distributed and n>30
Conditions were for both tests because they were both SRS, and np > 5 and nq > 5,
conditions for binomial distribution are met, and the sample is greater than 30. n>5