Download Ngawang Tsering CONFIDENCE INTERVAL ESTIMATES In general

Ngawang Tsering 1
CONFIDENCE INTERVAL ESTIMATES
In general, the purpose and the meaning of a confidence interval are to point out the true proportion of a
sample or the range of values to estimate the true value of a population parameter. It is an interval between where
the true value of population parameter lies within. It basically means the distance from the left threshold to right
threshold of a critical point which consists of the true value of the population parameter.
Construct a 99% confidence interval estimate for the true proportion of yellow candies.
(𝑍𝛼 = 0.01)
𝑍𝛼
2
= 0.005 (Critical value = 2.575)
𝑡𝑜𝑡𝑎𝑙 𝑦𝑒𝑙𝑙𝑜𝑤 𝑐𝑎𝑛𝑑𝑖𝑒𝑠
(p=
𝑡𝑜𝑡𝑎𝑙 𝑛𝑜. 𝑜𝑓𝑐𝑎𝑛𝑑𝑖𝑒𝑠
, p=
297
, =0.198)
1500
Point estimate= 0.198
𝑍𝛼
Margin of error (E) =
2
√
2.575
=
√
2
=.0265
(𝑝)(𝑞)
𝑛
(.198)(.802)
1500
—
Confidence interval =𝑋 ± E
=.198 ± .0265
= (.1715 < p < .2245)
Construct a 95% confidence interval estimate for the true mean number of candies per bag.
(𝑡𝛼 = 0.05)
—
(𝑀𝑒𝑎𝑛𝑋 =
𝑡𝑜𝑡𝑎𝑙 𝑛𝑜. 𝑜𝑓 𝑐𝑎𝑛𝑑𝑖𝑒𝑠
𝑛𝑜.𝑜𝑓 𝑏𝑎𝑔𝑠 𝑜𝑓 𝑐𝑎𝑛𝑑𝑖𝑒𝑠
1500
,=
25
(∑ 𝑥−𝑥 2 )
, =60)
90
Standard deviation = √
= √ =√3.75 =1.93649
𝑛−1
24
Degrees of freedom=n-1 =25-1 =24
𝑡 𝑆
Margin of error (E) = 𝛼
2 √𝑛
1.93649
=2.064
√25
=.799383
—
Confidence interval =𝑋 ± E
=60±.799383
= (59.2 < µ < 60.8)
Ngawang Tsering 2
Construct a 98% confidence interval estimate for the standard deviation of the number of candies per bag.
(𝑡𝛼 = 0.02) =
𝑡𝛼
0.02
=
2
2
=0.01
Degrees of freedom=n-1 =25-1 =24 (𝑡𝛼 = 2.492)
Standard deviation = √
=√
(𝑛−1)𝑆 2
<σ<√
(∑ 𝑥−𝑥 2 )
𝑛−1
=√
90
24
=√3.75 =1.93649
(𝑛−1)𝑆 2
2
𝑋𝑅
(25−1)1.936492
𝑋𝐿2
(25−1)1.936492
=√
<σ<√
42.980
=1.44706< σ < 2.87929
10.856
All the three interval estimates are shown above. The first one is the interval estimate of the proportion of
yellow candies from the total number of candies from class with the confidence level of 99%. The true value of
proportion (0.198) lies in between the intervals (.1715 and .0265). Second is the interval estimate of the true mean
number of candies per bag with the confidence level of 95%. The true mean number of candies (60) is within the
interval of (59.2 and 60.8). Third estimate is for the standard deviation of the number of candies per bag with the
confidence level of 98%. The standard deviation of the sample (1.94) does lie within the interval of (1.45 and 2.88).
HYPOTHESIS TESTS
Generally in statistics, hypothesis test is a procedure for testing a claim about a property of a population.
Whenever we do a hypothesis test we are making a claim or a statement for a value of population parameter. We do
the test to decide whether to accept or reject the claim. We always do the test referring to null hypothesis and
alternative hypothesis. For this, we need to calculate the value of the test statistics, choose a sampling distribution
that is relevant, find either P value or critical value, and state conclusion about the claim after you get the result
whether to accept or reject.
Use a 0.05 significance level to test the claim that 20% of all Skittles candies are red.
P=0.20, q=1-p =.80, n= 1500, x=305, α=0.05
𝑥
305
=^𝑃 = 𝑛 =1500 =0.203333
(Null hypothesis)𝐇𝟎 : P= 0.20
(Alternative hypothesis)𝐇𝟏 : P≠ 0.20
Test statistics=
=
.^𝑃 −𝑝
𝑝𝑞
√
𝑛
0.2033−0.20
√
(0.20)(0.80)
1500
Z =.322749
Ngawang Tsering 3
Z value area= .6255
P value= 1-.6255 = .3745
=.3745×2
=.746886/.75 (0.75>0.05)
“P” value is greater than α 0.05. Therefore, we fail to reject the null hypothesis.
In conclusion, we state that there is not sufficient evidence to reject the null hypothesis that 20% of all skittles
candies are red.
Use a 0.01 significance level to test the claim that the mean number of candies in a bag of Skittles is 55.
α=0.01, µ =55, n=25
Standard deviation = √
—
(𝑀𝑒𝑎𝑛𝑋 =
(∑ 𝑥−𝑥 2 )
𝑛−1
𝑡𝑜𝑡𝑎𝑙 𝑛𝑜. 𝑜𝑓 𝑐𝑎𝑛𝑑𝑖𝑒𝑠
𝑛𝑜.𝑜𝑓 𝑏𝑎𝑔𝑠 𝑜𝑓 𝑐𝑎𝑛𝑑𝑖𝑒𝑠
=√
,=
90
24
=√3.75 =1.93649
1500
25
, =60)
Degrees of freedom=n-1 =25-1 =24 (𝑡𝛼 = 2.492)
(Null hypothesis)𝐇𝟎 : µ = 55
(Alternative hypothesis)𝐇𝟏 : µ ≠ 55
Test statistics=
—
𝑋
−µ
𝑆
√𝑛
60−55
= 1.93649
√25
=12.91
T=12.9099
P value =0.00 (from T-table)
P value is less than α 0.01. Therefore, we reject the null hypothesis.
In conclusion, there is sufficient evidence in order to reject the null hypothesis that the mean number of candies in a
bag of skittles is 55. So, it supports the alternative hypothesis with sufficient evidence that is mean of candies in a
bag of skittles is not 55.
The hypothesis tests done above are the claims for proportion of candies that 20% of all skittles candies are
red and the mean number of candies in a bag of skittles is 55. We needed to do the test with null hypothesis claiming
that the proportion and mean are equal to what it is said in the description. Null hypothesis don’t expect any change
or difference from the original data. Then we need alternative hypothesis claiming that null hypothesis is not correct.
It expects changes and differences in the data from the original. So, we have performed the test and we couldn’t find
sufficient evidence to reject the first claim which is 20% of the candies is Red. But we have sufficient evidence to
reject the second claim which is mean of a bag of candies is 55.
Ngawang Tsering 4
Reflection
The conditions for doing interval estimates and hypothesis tests are that you need to know the confidence
interval or confidence level, critical point with z score that is a threshold for unusual, margin of error, point estimate,
sample size, true percentage, true mean, standard deviation. The sample should be greater than 30 to be in normal
distribution. During hypothesis, you need to do test statistics in order to find your critical value and then need to
know how to use, z, t and chi square table. My sample met these conditions because the data were there with sample
size. After that I calculated all the mean and standard deviation. The question includes the other parts such as
confidence level, proportion of the sample that we need to find. Using this data, there might not be an error if you
understand the question but if you are confused then there would be error in finding you point estimate, p value, “z”
score and mean of X. If you made error in the beginning of your findings then your whole work will be meaningless.
Sampling method could be improved by increasing your sample size “n”. From this statistical research I have learnt
how you can get a percentage or an estimate of everything in your daily life. I realized how important role statistics
can play in any aspects of your life, whether you are in business, healthcare, justice system, journalist or any other
profession. You can tell the probability of anything that’s important to you if you have the right data and the right
knowledge of performing the test on it.