Thursday 7 September 2017

Statistic Problem Solution

Suppose there are two types of food: Slow and Fast. Slow food is food that takes a lot of time and energy to prepare, and is possibly more “authentic.” Consider it a high-end luxury good. All other food we’ll call Fast. It’s a staple needed to live, is quick to acquire and cheap. The slow food costs $20 per unit, while the fast food costs $5 per unit. The price elasticity of demand for slow food is -2.8. The price elasticity of demand for fast food is -.20. The government is considering a tax on food. The tax on slow food is denoted ts, and the tax on fast food is denoted tf.
a. Comment on the significance of the different elasticities of demand?
b. What is the equation for the optimal (Ramsey) value of ts in terms of tf.
c. In a clearly written paragraph, comment on the relative size of ts compared to tf, and why we are seeing this result.
d. Suppose the government has selected tax levels ts and tf using the Ramsey rule. Furthermore, at those taxes, the market sells 3 million units of slow food and 300 million units of fast food., the government collects $1 billion in revenue from these taxes. What are the values of th and tp?

Answer:

The different price elasticities of the two goods help in explaining the behavior of the two goods in the market. That is the responses of the quantity demanded with change in prices. Slow food shows that 1% change in price results to decline in quantity demanded by 2.8 units while fast foods shows that 1% change in price will results to decline in quantity by 2 units.

Optimal value of fast food VF= $5 – t f
       Optimal value for slow food VS=$20 – t s

The t s in fast moving foods is low because of the low price elasticity of demand of -2.0 while the t f in slow food is high due to high price elasticity of demand of -2.8.


300 units of fast foods
3 million units of slow foods
Rev 1 billion, therefore TS= 3/303 *1Billion=9900990
T f=300/303 * 1Billion=990099010

Sunday 5 March 2017

Assignment


 
·         Make sure your answers are as complete as possible and show your work/argument.  When there are calculations involved, you should show how you come up with your answers with critical work and/or necessary tables.  You must show why you choose certain answer for true-or-false and multiple choice questions. Answers that come straight from program software packages will not be accepted.  



1.  (2 pts) True or False: In a right-tailed test, the test statistic is 1.5. If we know P(X < 1.5) = 0.96, then we reject the null hypothesis at 0.05 level of significance. (Justify for full credit)
Because P(X≥1.5)=1-P(X<1.5)=1-0.96=0.04<0.05 at 5% level of significance.

2. (2 pts) True or False: If a 99% confidence interval contains 1, thenthe 95% confidence interval for the same parameter must contain 1. (Justify for full credit)
Because 99% confidence is wider than 95% confidence interval so 99% confidence interval contains 1 but 95% confidence interval may not contain 1 for the same parameter.

3.  (2 pts) Which of the following could reduce the rate of Type I error?  (Justify for full credit)

a. Making the significant level from 0.01 to 0.05
b. Making the significant level from 0.05 to 0.01
c. Increase the β level
d. Increase the power
Because level of significance is actually the type I error in testing of hypothesis so we need to reduce the level of significance from 0.05 to 0.01 to reduce the type I error.

4. (2 pts) Three hundred students took a chemistry test. You sampled 50 students to estimate the average score and the standard deviation. How many degrees of freedom were there in the estimation of the standard deviation? (Justify for full credit)

a. 50
b. 49

c. 300
d. 299
Here the sample size n=50 so the degree of freedom is =50-1=49.
(For Questions 5&6) Mimi was the 5th seed in 2015 UMUC Tennis Open that took place in August.  In this tournament, she won 75 of her 100 serving games.
5. (2pts) Find a 90% confidence interval estimate of the proportion of serving games Mimi won. (Show work and round the answer to three decimal places)
Here sample proportion p=75/100=0.75 and sample size n=100.
90% confidence interval for the proportion is
(p-Z(1-0.9)/2*√(p(1-p)/n), p+Z(1-0.9)/2*√(p(1-p)/n))
=( p-Z0.05*√(p(1-p)/n), p+Z0.05*√(p(1-p)/n))
=(0.75-1.645*√(0.75(1-0.75)/100), 0.75+1.645*√(0.75(1-0.75)/100))
=(0.679, 0.821)




6. (5pts) According to UMUC Sports Network, Mimi wins 80% of the serving games in her 5-year tennis career.In order to determine if this tournament result isworse thanher career record of 80%.We would like to perform the following hypothesis test:
H_0:   p=0.80
H_a:   p<0.80
 (a) (2 pts) Find the test statistic. (Show work and round the answer to two decimal places)
Sample size =n=100 and sample proportion P=0.75.
Under the null hypothesis the test statistics is
Z=(P-p)/(√(P(1-P))/n)=(0.75-0.80)/(√(0.80(1-0.80))/100)=-1.25



(b) (2pts) Determine the P-value for this test.  (Show work and round the answer to three decimal places)

The P value is P(Z<-1.25)=0.106


(c) (1pt) Is there sufficient evidence to justify the rejection of   at the α=0.05 level? Explain.  
No there is no sufficient evidence because the p value corresponding to the test statistics is 0.106 which is greater than 0.05 at 5% level of significance.
7.  (5 points)  The SAT scores are normally distributed.  A simple random sample of 100 SAT scores has a sample mean of 1500 and a samplestandard deviation of 300.
(a) (1 pt) What distribution will you use to determine the critical value for a confidence interval estimate of the mean SAT score? Why?
Since the sample size n=100>30 therefore we use normal distribution which has mean 0 and variance 1 i.e, Z distribution to determine the critical value for a confidence interval estimate of the mean SAT score.
(b) (3pts) Construct a 95% confidence interval estimate of the mean SAT score.(Show work and round the answer to two decimal places)
Here given that sample size n=100, sample mean =1500 and sample standard deviation =300.
95% confidence interval estimate of the mean SAT score is
(Sample mean-Z0.05/2*(sample standard edviation)/(√sample size), sample mean-Z0.05/2*(sample standard edviation)/(√sample size))
=(1500-1.96*300/√100, 1500-1.96*300/√100)
=(1441.20, 1558.80)




(c) (1pt) Isa 99% confidence interval estimate of the mean SAT score wider than the 95% confidence interval estimate you got from part (b)? Why? [You don’t have to construct the 99% confidence interval]
Yes, because at 99% confidence interval the critical value of the test statistics is larger than 95% confidence interval.

8. (6pts) Assume the population is normally distributed with population standard deviation of 100. Given a sample size of 25, with sample mean 770, we perform the following hypothesis test.
    H_0: μ=750
H_a: μ>750
     (1 pt) Is this test for population proportion, mean or standard deviation? What distribution should you apply for the critical value?

This is the test of population mean and t distribution we use because sample size of the population is less than 30.



    (2 pts) What is the test statistic? (Show work and round the answer to three decimal places)

Under the null hypothesis the test statistics t=(sample mean-750)/(population standard deviation/√sample size)
t=(770-750)/(100/√25)=1.000




    (2 pts) What is the p-value? (Show work and round the answer to two decimal places)

The P value is P(t>1)=0.16




    (1pts) What is your conclusion of the test at the α = 0.10 level? Why? (Show work)

We accept the null hypothesis because the p value corresponding to the test statistics is 0.16 which is greater than 0.10
9.  (7 points)  Consider the hypothesis test given by          
H_0:   μ=650
H_a:   μ>650

Assume the population is normally distributed.  In a random sample of 25 subjects, the sample mean is found to be x ̅=655, and the sample standard deviation is s=28.
(a) (1 pt) Is this test for population proportion, mean or standard deviation? What distribution should you apply for the critical value?
The test is for population mean and t distribution we should apply for the critical value.
(b) (1 pt) Is the test a right-tailed, left-tailed or two-tailed test?
This is right tailed test

(c) (2 pts) Find the test statistic. (Show work and round the answer to two decimal places)

Under the null hypothesis the test statistics is t=(sample mean-650)/(sample standard deviation/√sample size)=(655-650)/(28/√25)=0.89



(d) (2pts) Determine the P-value for this test.  (Show work and round the answer to three decimal places)
The P value is P(t>0.89)=0.191





(e) (1pt) Is there sufficient evidence to justify the rejection of   at the α=0.02 level? Explain.          
No there is no sufficient evidence to justify the rejection of null hypothesis because the p value corresponding to the test statistics is 0.191 which is greater than 0.02 at 0.02 level of significance.



10. (7 pts) A new prep class was designed to improve SAT math test scores. Five students were selected at random. Their scores on two practice exams were recorded; one before the class and one after. The data recorded in the table below. We want to test if the scores, on average, are higher after the class.

SAT Math Score    Student 1    Student 2    Student 3    Student 4    Student 5
Score before the class    620    700    650    640    620
Score after the class    640    700    670    670    630


 (a) (1 pt) Which of the following is the appropriate test and best distribution to use for the test?
       (i) Two independent means, normal distribution
(ii) Two independent means, Student’s t-distribution
       (iii) Matched or paired samples, normal distribution
(iv) Matched or paired samples, Student’s t-distribution


(b) (1 pt) Letμd be the population mean for the differences of scores (scores after the class –before the class).  Fill in the correct symbol (=, ≠, ≥, >, ≤, <) for the null and alternative hypotheses.
Here we first take the difference of the paired samples and then we find the mean, standard deviation, based on that we have to select those two hypotheses.
Here we take the difference math score before –after the class and take the average of it and hypothesis will be as follows:
(i)  H0:  μd> 0

(ii) Ha: μd≤0


(c) (2 pts) What is the test statistic? (Show work and round the answer to three decimal places)

Under the null hypothesis the test statistics is t=(sample 1 mean-sample 2 mean-0)/(√((sample 1 variance)/(sample 1 size)+(sample 2 variance)/(sample 2 size)))
=(646-662)/(√(1080/5+770/5))=-0.832




(d)  (2 pts) What is the p-value? (Show work and round the answer to three decimal places)
The P value is P(|t|>0.832)=0.215
We have to find the P value from the T table with 5+5-2=8 degree of freedom.



(e) (1 pt) What is your conclusion of the test at the α = 0.05 level? Why? (Show work)

We accept the null hypothesis because the p value corresponding to the test statistics is 0.215>0.05 at 5% level of significance.


Assignment



1. For a distribution of scores, X = 40 corresponds to a zscore of z = +1.00, and X = 28 corresponds to a zscore of z = -0.50. What are the values for the mean and standard deviation for the distribution? (Hint: Sketch a distribution and locate each of the zscore positions.)
Solution: we know Z=(x-µ)/𝞼  then we get x=µ+𝞼Z…..(1)
For X=40 and Z=1 we get from (1) ,  40=µ+𝞼…………(2)
For X=28 and Z=-0.50 we get from (1), 28=µ-0.50*𝞼…………(3)
Subtracting (3) from (2)  we get 40-28=µ+𝞼- µ+0.50*𝞼=1.50𝞼

ð  12=1.50𝞼

ð  𝞼=8.

ð  Putting 𝞼=8 in equation (2) we get µ=40-8=32

ð  Hence the mean µ=32 and standard deviation 𝞼=8


3. For a normal distribution,
          a.  What z-score separates the highest 10% from the rest of the scores?
Solution:Z-scores separates the highest 10% from the rest of the scores is 1.282
          b.  What z-score separates the highest 30% from the rest of the scores?
Solution: Z-scores separates the highest 30% from the rest of the scores is 0.5244.
          c.  What z-score separates the lowest 40% from the rest of the scores?
Solution: Z-scores separates the highest 40% from the rest of the scores is -0.253
          d.  What z-score separates the lowest 20% from the rest of the scores?
Solution: Z-scores separates the highest 20% from the rest of the scores is -0.8416
4. A population consists of the following N = 10 scores: 0, 6, 4, 3, 12, 6, 7, 5, 1, 11
Your task is to enter the data for this variable into SPSS, use the descriptives command to do a z-transformation of the whole distribution into a standardized distribution, and then get the frequencies with mean and standard deviation for both the original raw scores distribution and the standardized distribution. Attach the printout with the frequencies to this homework.

Statistics

esteem
Zscore(esteem)
N
Valid
10
10
Missing
0
0
Mean
5.5000
.0000000
Std. Error of Mean
1.22247
.31622777
Median
5.5000
.0000000
Mode
6.00
.12934
Std. Deviation
3.86580
1.00000000
Variance
14.944
1.000
Skewness
.404
.404
Std. Error of Skewness
.687
.687
Kurtosis
-.357
-.357
Std. Error of Kurtosis
1.334
1.334
Range
12.00
3.10414
Minimum
.00
-1.42273
Maximum
12.00
1.68141
Sum
55.00
.00000





esteem

Frequency
Percent
Valid Percent
Cumulative Percent
Valid
.00
1
10.0
10.0
10.0
1.00
1
10.0
10.0
20.0
3.00
1
10.0
10.0
30.0
4.00
1
10.0
10.0
40.0
5.00
1
10.0
10.0
50.0
6.00
2
20.0
20.0
70.0
7.00
1
10.0
10.0
80.0
11.00
1
10.0
10.0
90.0
12.00
1
10.0
10.0
100.0
Total
10
100.0
100.0







Zscore(esteem)

Frequency
Percent
Valid Percent
Cumulative Percent
Valid
-1.42273
1
10.0
10.0
10.0
-1.16405
1
10.0
10.0
20.0
-.64670
1
10.0
10.0
30.0
-.38802
1
10.0
10.0
40.0
-.12934
1
10.0
10.0
50.0
.12934
2
20.0
20.0
70.0
.38802
1
10.0
10.0
80.0
1.42273
1
10.0
10.0
90.0
1.68141
1
10.0
10.0
100.0
Total
10
100.0
100.0



 
 

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