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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.
