University STAT 2006 Assignment 3: Statistical Inference Problems

This document presents a comprehensive solution for STAT 2006 Assignment 3, focusing on statistical inference. It covers constructing confidence intervals for standard deviation, including the optimization of interval length. The assignment also addresses hypothesis testing problems, including finding confidence intervals for proportions, determining significance levels and power functions for binomial distributions, and conducting hypothesis tests for population means and standard deviations using the Z-test. The solution demonstrates the application of the central limit theorem to determine sample sizes and critical values. Finally, the assignment explores tests related to the number of F's in a text and the fairness of a coin, providing detailed calculations and interpretations of p-values and test statistics. The document provides step-by-step solutions for each problem.

STAT 2006 Assignment 3 1
STAT 2006 Assignment 3
Institutional Affiliation
Date of Submission

STAT 2006 Assignment 3 2
1. Let X1,X2,··· ,Xn be a random sample from N(μ,σ2), then the pivotal quantity
(n− , and we can make use of its quantiles a, b to construct a
% confidence
Interval for σ. The quantiles a,b need to satisfy the constraint
Where G is the CDF of χ2 (n − 1). Obviously there are many possible choices for a, and b.
(a) Construct the 100(1 − α) % confidence interval for σ.
We use, to estimate σ =2, and we define the pivot to be

STAT 2006 Assignment 3 3
(b) (b) Show that the k is minimized when a, b also satisfy
.
Combining with the constraint above, we can numerically solve for the optimal pair of
quantiles a, b to minimize the length of the confidence interval

STAT 2006 Assignment 3 4
2. It is reported that in a telephone poll of 2000 adult, 1325 of them are nonsmokers.
Also, y1 = 650 of nonsmokers and y2 = 425 of smokers said yes to a particular question.
Let p1, p2 equal the proportions of nonsmokers and smokers that would say yes to this
question respectively
(a) Find a two-sided 95% confidence interval for p1 − p2.
By letting p1, p2 equal the proportions of nonsmokers and smokers that would say yes to
this question respectively;
p1 = 1325
2000 = 0.6625, y1 = 650
p2 =
675
2000 = 0.3375, y2 = 425
For the two-sided 95% confidence interval, Z = 1.645
Confidence Interval = (p1 – p2) ± Ɀ α
2 √ ¿ ¿
= (0.6625 – 0.3375) ± 1.645 √¿ ¿
= 0.325 ± 1.645 (0.0295)
= 0.325 ± 0.0485
= (0.276, 0.3735)
Hence the two-sided 95% confidence interval is 0.276 to 0.3735.

STAT 2006 Assignment 3 5
(b) Find a two-sided 95% confidence interval for p, the proportion of adult who
would say yes to this question.
Confidence Interval = (p1 + p2) ± Ɀ α
2 √ ¿ ¿
= (0.6625 + 0.3375) ± 1.645 √¿ ¿
= 1 ± 1.645 (0.0295)
= 1 ± 0.0485
= (0.9515, 1.0.0485)
Hence the two-sided 95% confidence interval is 0.9515 to 1.0.0485
3. Let Y be Binomial (50, p). To test H0: p = 0.08 against H1: p < 0.08, we reject H0 if and
only if Y ≤ 7.
(a) Determine the significance level α of the test
Solution
From the information given above, we consider Y to a random variable which follows a
binomial distribution with n =50 and probability p.
The null and alternative hypotheses are as follows;
Null hypothesis, Ho: P = 0.08
Alternative hypothesis H1: P = < 0.08

STAT 2006 Assignment 3 6
Here, we reject Ho and accept H if and only if Y ≤ 7
We calculate the significance level α of the test. We make a type I error if Y ≤ 7 when in
fact p = 0.08.
Therefore the significance level of the test is as follows;
α = P(Y ≤ 7; p = 0.08)
= ∑
y=0
7
(50
y )(0.08) y (1 – 0.08) 50 – y
Since n is rather larger and p is small, these binomial probabilities can be approximately
very well by Poisson probabilities with λ = (50) (0.08) = 4
Therefore, α = P(Y ≤ 7; p = 0.08)
= ∑
y=0
7
( 8 y e−8
y ! )
Using the binomial distribution, we find that the exact value of α is 0.3032.
(b) Calculate the value of the power function if in fact p = 0.05.
4. The mean birth weight in the United States is μ = 3320 grams, with a standard
deviation of σ = 580. Let X equal the birth weight in Rwanda. Assume that the
distribution of X is N (μ, σ2). We shall test the hypothesis H0: σ = 580 against the
alternative hypothesis H1: σ < 580 at an = 0.05 significance level.

STAT 2006 Assignment 3 7
(a) What is your decision if a random sample of size n = 81 yields X¯ = 2989 and s =
516?
It is given that the population standard deviation is σ = 580
The sample size, n = 81
The sample mean, x-bar = 2989
μ = 3320 grams
The null and alternative hypotheses are given as,
Ho: μ= 580
H1: μ < 110
Given that N (μ, 100) and that alpha α = 0.01
The sampling distribution of the sample mean has mean μ and standard deviation σ/√n
The Z value is the sample mean decreased by the population mean, divided by the standard
deviation as follows;
=Ɀ x−μ
σ /√ n = 2989−3320
580/ √ 81 = -5.136
The critical values is the value in the table V which corresponds to a probability of 1 – α = 1 –
0.05 = 0.95 which from the table, z = 1.645
Then the rejection region contains all values which are larger than 1.645. If the value of the test
statistics is within the rejection region, then the null hypothesis is rejected:
= -5.136 ¿ 1.645 hence we fail to reject Ho.
(b) What is the approximate p-value of this test?

STAT 2006 Assignment 3 8
The p-values is the probability of obtaining a value that is more extreme or equal to the
standardized test statistic Z. To determine this, we the table V.
P = P (Z > -5.136) =1−P (Z≤-5.136) =1 − 0.8652 = 0.0248.
Thus the p-value of this test is 0.0248.
5. Assume that IQ scores for a certain population are approximately N (μ, 100). To test
H0: μ = 110 against H1: μ > 110
We take random sample of size n = 16 from this population and observe X¯ = 114
(a) Do we accept or reject H0 at the 1% significance level?
Solution
It is given that the population standard deviation is σ = 10
The sample size, n = 16
The sample mean, x-bar = 114
The null and alternative hypotheses are given as,
Ho: μ= 110
H1: μ > 110
Given that N (μ, 100) and that alpha α = 1% = 0.01
The sampling distribution of the sample mean has mean μ and standard deviation σ/√n
The Z value is the sample mean decreased by the population mean, divided by the standard
deviation as follows;
=Ɀ x−μ
σ /√ n = 114−110
10 /√ 16 = 1.6

STAT 2006 Assignment 3 9
The critical values is the value in the table V which corresponds to a probability of 1 – α = 1 –
0.01 = 0.99 which from the table, z = 2.3
Then the rejection region contains all values which are larger than 1.645. If the value of the test
statistics is within the rejection region, then the null hypothesis is rejected:
= 1.6 ¿ 2.3 hence we fail to reject Ho.
(b) Do we accept or reject H0 at the 5% significance level?
Solution
Given that N (μ, 100) and that alpha α = 5% = 0.05
The sampling distribution of the sample mean has mean μ and standard deviation σ/√n
The Z value is the sample mean decreased by the population mean, divided by the standard
deviation as follows;
=Ɀ x−μ
σ /√ n = 114−110
10 /√ 16 = 1.6
The critical values is the value in the table V which corresponds to a probability of 1 – α = 1 –
0.01 = 0.99 which from the table, z = 1.645
Then the rejection region contains all values which are larger than 1.645. If the value of the test
statistics is within the rejection region, then the null hypothesis is rejected:
= 1.6 ¿ 1.645 hence we fail to reject Ho.
(c) What is the p-value of this test?
The p-values is the probability of obtaining a value that is more extreme or equal to the
standardized test statistic Z. To determine this, we the table V.

STAT 2006 Assignment 3 10
P = P (Z > 1.60) =1−P (Z≤1.60) =1 − 0.9452 = 0.0548.
Thus the p-value of this test is 0.0548.
6. The following text was shown to a large class of students for 30 seconds, and they
were told to report the number of F’s that they found:
IN FINANCIAL TRANSACTIONS, SIMPLE INTEREST IS OFTEN USED FOR
FRACTIONS OF AN INTEREST PERIOD FOR CONVENIENCE.
Let p equal the proportion of students who find 6F’s. We shall test the null hypothesis
H0: p = 0.5 against H1: p < 0.5
(a) Given a sample size of n = 230, define a critical region with an approximate
significance level of α = 0.05.
Part a Here, we have to use one sample z test for population proportion. H0: p = 0.5 versus H1:
p < 0.5. This is a lower tailed or left tailed test. (One tailed test) We are given Level of
significance = α = 0.05.
Given that n = 230 and α = 0.05.
(b). If y = 110 students report that they found 6F’s, what is your conclusion?
Ɀ =1.645 110
230 = 0.4523
Thus the student were correct.
c. What is the p-value of this test?
The p-values is the probability of obtaining a value that is more extreme or equal to the
standardized test statistic Z. To determine this, we the table V.
P = P (Z > 1.23) =1−P (Z≤1.23) =1 − 0.8452 = 0.1548.

STAT 2006 Assignment 3 11
7. In 1000 tosses of a coin, 560 heads and 440 tails appear. Using direct calculation or
normal approximation, test whether the coin is fair, at the 5% significance level.
Solution
First, this sampling from the Bernoulli (p) population and it is reasonable to assume that the
coin is fair.
Let X ∼ Binomial (n =1000, p) denote the number of heads in 1000 throws – we can easily
verify that X is a sufficient statistic for p.
Null hypothesis H0: p = ½
Alternative hypothesis H1: p > 1/2
We then derive a test using only a test statistic W(X1… Xn) = W(X), a function of the sample,
although we do not know its distribution entirely, we know its distribution exactly under the
null hypothesis.
This test statistic is often obtained through a point estimator (for example, X/n, the sample
proportion) of the parameter of interest, or a sufficient statistic (for example, X, the total
number of successes) of the parameter of interest (in the given example, it is p).
So, subsequently, we compute the p-value, and make decisions based on the p-value.
In our example, such a test statistic could be the total number of heads when the null hypothesis
is true (H0: p = 1/2), that is: X0 ∼ Binomial (1000, p = 1/2).

STAT 2006 Assignment 3 12
The observed test statistic value is:
XO = 560
If the null hypothesis is true, we would expect about 500 heads. If the alternative hypothesis is
true, we would expect to see more than 500 heads. The more heads we observed, the more
evidence we have in supporting the alternative hypothesis.
Thus by the definition, the p-value would be the probability of observing 560 or more heads,
when the null hypothesis is true, i.e.
P(X ≥ 560 | H0: p = 1/2) = ∑
X =560
1000
(1000
x )(0.5) x (0.5) 1000 – x
≈ 0.0000825
Alternatively
Using the normal approximation we have X0 ⩪ N (500, 250) yielding the large sample p-value
of:
P(X0 ≥ 560 | H0: p = 1/2) = P ( Xo−500
√ 250 ≥ 560−500
√ 250 | p = ½
≈ P (Z0 ≥ 3.795) ≈ 0.0000739
8. For a random sample X1,··· ,Xn of Bernoulli(p) variables, it is desired to test
H0: p = 0.49 against H1: p = 0.51