Submitted By: Name: Sagor sarkar Id: 1120090030Bus-173. Section:1 Submitted To: M. Siddique Hossain (SqH) Lecturer, North South University

Applied statistics

ASSIGNMENT # 02

1. Write a systematic not on all the topics and terminologies you have learned from chapter 8 and 9, confidence interval for the differences of population means and proportions and tests of hypothesis.

Chapter 8: Sampling distribution of sample variance

Sample variance:

Sample variance is a measure of the spread of or dispersion within a set of sample data. The sample variance is the sum of the squared deviations from their average divided by one less than the number of observations in the data set. For example, for n observations x1, x2, x3, ... , xn. The sample variance is given by

S2 = 1n−1 ∑(xi - x2 )

2

n-1 is known as degrees of freedom. x2 is the consistent and unbiased estimate of µ. S2 is the consistent and unbiased estimator of σ 2.Thus S2 serves as the basis either for constructing confidence interval estimate or testing hypothesis about σ 2.

If we can assume that the underlying population distribution is normal, then it can be shown that the sample variance and the population variance and related through a probability distribution known as the chi-square distribution.

Multiplying s2 by n-1 and dividing by σ² converts it into the chi-square random variable.

Χ2 = (n-1)s2/σ²,

which has the chi-square distribution . Thus it is possible to find probability for this variable:

α = p[(n-1)s2/σ² ≥ χ2 ]

When the number of degrees of freedom is n-1.

Chi-square distribution

The chi-squared distribution (also chi-square or χ²-distribution) with k degrees of freedom is the distribution of a sum of the squares of k independent standard normal random variables. It is one of the most widely used probability distributions in inferential statistics in hypothesis testing or in construction of confidence intervals. When there is a need to contrast it

with the noncentral chi-squared distribution, this distribution is sometimes called the central chi-squared distribution.

Important properties of the χ2 distribution,

1) χ2 distribution is a continuous probability distribution which has the value zero at its lower limit and extends to infinity in the positive direction. Negative value of χ2 is not possible since the difference between the observed and expected frequencies are always squared.

2) The exact shape of the distribution depends upon the number of degrees of freedom υ. For different values of υ, we shall have different shape of the distribution. In general, when υ is small, the shape of the curve is skewed to the right and as υ gets larger the distribution becomes more and more symmetrical and can be approximated by the normal distribution.

3) The mean of the χ2 distribution is given by the density function .E(χ2)= υ and variance is twice the density function. V(χ2)=2 υ

4) 4. As υ gets larger, χ2 approaches the normal distribution with mean υ and standard deviation √2 υ.

5) 5. The sum of independent χ2 variants is also a χ2 variant. Therefore, if χ2, is a χ2 variants with υ .d.f. and χ2 is another χ2 variants with υ2 d.f.

Chi-square distribution of sample and population variance

Given a random sample of n observation from a normally distributed population whose population variance is σ² and whose resulting sample variance is s² it can be shown that

(n-1)s²/σ²= Σ(Xi-X)²/σ²

The distribution known as the χ²(chi-square) distribution with n-1 degree of freedom.

E(χ2)= υ and Var (χ2)=z υ

The properties of the χ² distribution are to find the variance of the sampling of the sampling distribution of the sample variance when the parent population is normal.

Confidence interval for variance

Confidence interval for variance σ² is based on the sampling distribution of (n-1)s²/σ² which follows χ² distribution with υ =n-1 degrees of freedom. A 100(1- α) % confidence interval for σ² is constructed by first obtaining an interval about (n-1) s²/ σ².

Therefore, a 100(1- α) % confidence interval for (n-1) s²/σ² is given by.

p(χ²n-1,1- α/2< χ²n-1< χ²n-1,α/2)= 1- α

p(χ²n-1,1- α/2<(n-1)s²/σ²< χ²n-1,α/2)= 1- α

P[(n-1)s²/χ²n-1 ,α/2< σ²<(n-1)s²/χ²n-1,1- α/2]=1- α

Confidence intervals for variance of a normal population

A random sample of n observations from a normally distributed population with variance σ².If the observed sample variance is s², then 100(1-α) % confidence interval for the population variance is given by

(n-1) s²/ χ²n-1, α/2 < σ² < (n-1) s²/ χ²n-1, 1- α/2

Where, χ²n-1,α/2 is the number for which

P (χ²n-1> χ²n-1, α/2)= α/2

And the random variable χ²n-1 follows a chi-square distribution with (n-1) d.f.

Confidence intervals for the different between two normal population means

Dependent sample: If the values in one sample are influenced by the value in the other sample then we consider sample to be dependent. Here, d2 and sd denote observed sample mean and standard deviation of n differences di = x1i – x2i.

Let as assume that both population are normal, then 100(1- ∝)% confidence interval for the difference between means µd = µ1- µ2 is given by

d2 - tn-1,∝/2Sd/√n ≤ µd ≤ d2 + tn-1,∝/2Sd/√n

∴ CI = d2 ± ME

The standard deviation, Sd = √∑( di- d2 )2/ n -1 and ME = tn-1,∝/2Sd/√n

Independent sample, known population variance: Samples are drawn independently from the two normally distributed populations with known population so that the membership of one sample is not influenced by the membership of the other sample.

Here, the sample means are x2 1 and x2 2 and variance are σ12/n1 and σ2

2/n2, then a 100(1- ∝)% . Confidence interval for µ1- µ2 is given by,

(x2 1- x2 2) – z ∝/2√ σ12/n1 + σ2

2/n2 < µ1- µ2 < (x2 1- x2 2) + z ∝/2√ σ12/n1 + σ2

2/n2

Here the margin of error ME is given by,

ME = z ∝/2√ σ12/n1 + σ2

2/n2

Independent sample, unknown population variance: (Assumed to be equal)

Here, σ12 and σ2

2 are unknown but σ12 = σ2

2 = σ2.

Again v(x2 1- x2 2) = var(x2 1)+ var(x2 2) = σ12/n1+ σ2

2/n2

Z = (x2 1- x2 2) – (µ1- µ2)/ √σ12/n1 + σ2

2/n2

The sample means are x2 1 and x2 2 and sample variance are s12 and s2

2, then a 100(1- ∝)% . Confidence interval for µ1- µ2 is given by,

(x2 - y2 ) – tn1+ n2 – 2, ∝/2 sp√1/nx+1/ny< µx - µy< (x2 - y2 ) + tn1+ n2 – 2, ∝/2 sp√1/nx+1/ny

Independent sample, unknown population variance: (Not assumed to be equal)

Here, The sample means are x2 1 and x2 2 and sample variance are s12 and s2

2, then a 100(1- ∝)% . Confidence interval for µ1- µ2 is given by,

(x2 - y2 ) – t(ʋ, ∝/2)√sx2/nx+ sy

2/ny< µx - µy< (x2 - y2 ) + t(ʋ, ∝/2)√sx2/nx+ sy

2/n

Chapter 9: Hypothesis

Hypothesis is an assumption to be tested. In order to reach decisions it is useful to make assumptions or guesses about the population involved. Such assumptions which may or may not be true are called statistical hypothesis. They are, in fact, in general statements about the probability distribution of the population.

Types of Hypothesis:

There are mainly two kinds of hypothesis

(1) Null Hypothesis(2) Alternative Hypothesis

Null Hypothesis:

The null hypothesis, H0, represents a theory that has been put forward, either because it is believed to be true or because it is to be used as a basis for argument, but has not been proved. For example, in a clinical trial of a new drug, the null hypothesis might be that the new drug is no better, on average, than the current drug. We would write

H0: there is no difference between the two drugs on average.

Alternative Hypothesis

The alternative hypothesis, H1, is a statement of what a statistical hypothesis test is set up to establish. For example, in a clinical trial of a new drug, the alternative hypothesis might be that the new drug has a different effect, on average, compared to that of the current drug. We would write

H1: the two drugs have different effects, on average.

The alternative hypothesis might also be that the new drug is better, on average, than the current drug. In this case we would write

H1: the new drug is better than the current drug, on average.

Types of error: There are two types of error.

(1) Type 1 error(2) Type 2 error.

Type I Error

In a hypothesis test, a type I error occurs when the null hypothesis is rejected when it is in fact true; that is, H0 is wrongly rejected. For example, in a clinical trial of a new drug, the null hypothesis might be that the new drug is no better, on average, than the current drug;

H0: there is no difference between the two drugs on average.

A type I error would occur if we concluded that the two drugs produced different effects when in fact there was no difference between them.

The following table gives a summary of possible results of any hypothesis test:

Type II Error

In a hypothesis test, a type II error occurs when the null hypothesis H0, is not rejected when it is in fact false. For example, in a clinical trial of a new drug, the null hypothesis might be that the new drug is no better, on average, than the current drug; i.e.

H0: there is no difference between the two drugs on average.

A type II error would occur if it was concluded that the two drugs produced the same effect, i.e. there is no difference between the two drugs on average, when in fact they produced different ones. A type II error is frequently due to sample sizes being too small.

The value of 1-β is called the test of power. It represents the probability of not making type 2 error.

One tailed and two tailed tests There are three kinds of problems in tests of hypothesis :

Two Tailed Tests: A two-tailed test is a statistical hypothesis test in which the values for which we can reject the null hypothesis, H0 are located in both tails of the probability distribution.

One-sided Test A one-sided test is a statistical hypothesis test in which the values for which we can reject the null hypothesis, H0 are located entirely in one tail of the probability distribution

Left-tailed Test: A left-tailed test has rejection region in left tail.

Right –tailed Test: A right tailed test has rejection region in right tail.

One Sample t-test

A one sample t-test is a hypothesis test for answering questions about the mean where the data are a random sample of independent observations from an underlying normal distribution , where is unknown.

The null hypothesis for the one sample t-test is:

H0: µ = µ0, where µ0 is known.This null hypothesis, H0 is tested against one of the following alternative hypotheses, depending on the question posed:

H1: µ is not equal to µ H1: µ > µ H1: µ < µ

Two Sample t-test

A two sample t-test is a hypothesis test for answering questions about the mean where the data are collected from two random samples of independent observations, each from an underlying normal distribution: The null hypothesis for the two sample t-test is:

H0: µ1 = µ2

That is, the two samples have both been drawn from the same population. This null hypothesis is tested against one of the following alternative hypotheses, depending on the question posed.

H1: µ1 is not equal to µ2

H1: µ1 > µ2 H1: µ1 < µ2

2. Solve the following problems:

7:42, 7:48, 8:2, 8:4, 8:8, 8:10, 8:18, 8:20, 8:26, 8:28, 8:30, 8:40, 8:42, 8:50, 9:4, 9:8, 9:10, 9:12, 9:14, 9:20, 9:24, 9:28, 9:34, 9:38, 9:40, 9:48, 9:50, 9:56, 9:58, 9:60.

7.42 : Find the lower confidence limit for each of the following normal populations:

a. n = 21; ∝ = 0.025; s2 = 16

b. n = 16; ∝ = 0.05; s = 8

c. n = 28; ∝ = 0.01; s = 15

solution:

a. Here, n = 21, s2 = 16, ∝ = 0.025

X2n-1, 1- ∝/2 = X2

20, 0.99 = 8.260 and X2n-1, ∝/2 = X2

20, 0.01 = 37.566

(n-1)s2/ X2n-1, ∝/2 < σ 2< (n-1)s2/ X2

n-1, 1- ∝/2 = (20)(16)/37.566 < σ 2 < (20)(16)/8.260 = 8.52 < σ 2< 38.74

b. Here, n = 16, s = 8, ∝ = 0.05

X2n-1, 1- ∝/2 = X2

15, 0.975 = 6.262 and X2n-1, ∝/2 = X2

15, 0.025 = 27.488

(n-1)s2/ X2n-1, ∝/2 < σ 2< (n-1)s2/ X2

n-1, 1- ∝/2 = (15)(8)2/27.488 < σ 2 < (15)(8)2/6.262 = 34.92 < σ 2< 153.31

c. Here, n = 28, s = 15, ∝ = 0.01

X2n-1, 1- ∝/2 = X2

27, 0.995 = 11.808 and X2n-1, ∝/2 = X2

27, 0.005 = 49.645

(n-1)s2/ X2n-1, ∝/2 < σ 2< (n-1)s2/ X2

n-1, 1- ∝/2 = (27)(15)2/49.645 <σ 2 < (27)(15)2/6.262=122.37< σ 2< 514.48

7.48 : A psychologist wants to estimate the variance of employee test scores. A random sample of 18 scores had a sample standard deviation of 10.4. Find a 90% confidence interval for the population variance. What are the assumptions, if any to calculate this interval estimate?

Solution :

n = 18, s2 = 108.16, ∝ = 0.1, X217, 0.05 = 27.59 and X2

17, 0.95 = 8.67

(n-1)s2/ X2n-1, ∝/2 < σ 2< (n-1)s2/ X2

n-1, 1- ∝/2 = (17)(108.16)/27.59 < σ 2 < (17)(108.16)/8.67

= 66.6444 up to 212.078. Assume that the population is normally distributed.

8.2: A confidence interval for the difference between the means of two normally distributed populations based on the following dependent samples is desired:

Before After

6128106

8149137

a. Find the margin of error for a 90% confidence level.b. Find the UCL and the LCL for a 90% confidence level.c. Find the width of a 95% confidence interval.

Solution:

a. Two normally distribution populations based on dependent samples of n=5 observations

Paired T- test for Before and after

N Mean St.Dev. SE. MeanBefore 5 8.4000 2.6077 1.1662

After 5 10.2000 3.1145 1.3928

Difference 5 -1.80000 0.83666 0.37417

90% CI for mean difference: ( -2.59766, -1.00234)

ME = tn-1,∝/2Sd/√n = 2.132(.83666/√5) = 0.79772

b.

UCL = d2 + ME = -2.59766

LCL = d2 - ME = -1.00234

c. 95%

Width = 2[ME = tn-1,∝/2Sd/√n] = 2[2.776(.83666/√5)] = 2.07737

8.4: Solution:

Let X = without passive solar; Y = with passive solar; di = xi – yi

n = 10, ∑di = 373, d2 = 37.3, t9, o.05 = 1.833

Sd = √∑( di- d2 )2/ n -1 = √2806.1/ 9 = 17.6575

C.I. = d2 + tn-1,∝/2Sd/√n

= 37.3 + 1.833(17.6575) / √10

= 27.0649 < µx - µy < 47.5351

8.8: Assuming equal population variances, determine the number of degrees of freedom for each of the following:

a. n1 = 12, s12 =30; n2 = 14, s2

2 = 36

b. n1 = 6, s12 =30; n2 = 7, s2

2 = 36

c. n1 = 9, s12 =16; n2 = 12, s2

2 = 25

Solution:

a. degrees of freedom = nx + ny – 2 = 12+14–2 = 24

b. degrees of freedom = nx + ny – 2 = 6+7–2 = 11

c. degrees of freedom = nx + ny – 2 = 9+12–2 = 19

8.10: Assuming unequal population variances, determine the number of degrees of freedom for each of the following:

a. n1 = 12, s12 =6; n2 = 14, s2

2 = 10

b. n1 = 6, s12 =30; n2 = 10, s2

2 = 36

c. n1 = 9, s12 =16; n2 = 12, s2

2 = 25

d. n1 = 6, s12 =30; n2 = 7, s2

2 = 36

Solution:

a. υ =[(s12/n1)+(s2

2)]2/(s21/n1)2/(n1-1)+(s2

2/n2)2/(n2-1)

υ =[(6/12)+(10/14)]2/(6/12)2/(12-1)+(10/14)2/(14-1) = 24

b. υ =[(s12/n1)+(s2

2)]2/(s21/n1)2/(n1-1)+(s2

2/n2)2/(n2-1)

υ =[(30/6)+(36/10)]2/(30/6)2/(6-1)+(36/10)2/(10-1) = 11

c. υ =[(s12/n1)+(s2

2)]2/(s21/n1)2/(n1-1)+(s2

2/n2)2/(n2-1)

υ =[(16/9)+(25/12)]2/(16/9)2/(9-1)+(25/12)2/(12-1) = 19

d. υ =[(s12/n1)+(s2

2)]2/(s21/n1)2/(n1-1)+(s2

2/n2)2/(n2-1)

υ =[(30/6)+(36/7)]2/(30/6)2/(6-1)+(36/7)2/(7-1) = 11

8.18: Calculate the margin of error for each of the following:

a. n1 = 260, Ṕ1= 0.75; n2 = 200, Ṕ2= 0.68

b. n1 = 400, Ṕ1= 0.60; n2 = 500, Ṕ2= 0.68

c. n1 = 500, Ṕ1= 0.20; n2 = 375, Ṕ2= 0.25

Solution:

a. ME = z∝/2√ Ṕ1(1- Ṕ1)/ n1 + Ṕ2(1- Ṕ2)/ n2 = 1.96 √.75(1-.75)/260 + .68(1-.68)/260 =

0.083367

b. ME = z∝/2√ Ṕ1(1- Ṕ1)/ n1 + Ṕ2(1- Ṕ2)/ n2 = 1.96 √.60(1-.60)/400 + .68(1-.68)/500 =

0.063062

c. ME = z∝/2√ Ṕ1(1- Ṕ1)/ n1 + Ṕ2(1- Ṕ2)/ n2 = 1.96 √.20(1-.20)/500 + .25(1-.25)/375 =

0.056126

8.26: How large a sample is needed to estimate the population proportion for each of the following?

a. ME = 0.03; ∝ = 0.05b. ME = 0.05; ∝ = 0.05c. Compare and comment on your answers to parts (a) and (b).

Solution:

a. n = .25(z∝/2)2/ ME2 = .25(1.96)2/ .032 = 1067.111. Take a sample of size n = 1068.

b. n = .25(z∝/2)2/ ME2 = .25(1.96)2/ .052 = 384.16. Take a sample of size n = 385.

c. In order to reduce the ME in half, the sample size must be increased by a large proportion.

8.28: A research group wants to estimate the proportion of consumers who plan to buy a scanner for their PC during the next three months.

a. How many people should be sampled so that the sampling error is at most 0.04 with a 90% confidence interval?

b. What is the sample size required if the confidence is increased to 95%, keeping the sampling error the same?

c. What is the required sample size if the research group extends the sampling error to 0.05 and wants a 98% confidence level?

Solution:

a. Z.05 = 1.645, ME = .04

n = .25(z∝/2)2/ ME2 = .25(1.645)2/ .042 = 422.8, take n = 423

b. n = .25(z∝/2)2/ ME2 = .25(1.96)2/ .042 = 600.25, take n = 601

c. n = .25(z∝/2)2/ ME2 = .25(2.33)2/ .052 = 542.89, take n = 543

8.30: the student government association at a university wants to estimate the percentage of the student body that supports a change being considered in the academic calendar of the university for the next academic year. How many students should be surveyed if a 90% confidence interval is desired and the margin of error is to be only 3% ?

Solution:

Z.05 = 1.645, ME = .03

n = .25(z∝/2)2/ ME2 = .25(1.645)2/ .032 = 751.7, take n = 752

8.40: Independent random samples from two normally distributed populations give the following results:

nx = 10, x2 = 480, sx = 30

ny = 12, y2 = 480, sy = 30

a. If we assume that the unknown population variances are equal, what is the 90% confidence interval for the difference of population means?

b. If we assume that the unknown population variances are unequal, what is the 90% confidence interval for the difference between population means?

Solution:

a. (x2 - y2 ) – tn1+ n2 – 2, ∝/2 sp√1/nx+1/ny< µx - µy< (x2 - y2 ) + tn1+ n2 – 2, ∝/2 sp√1/nx+1/ny

Sp = √(nx –1)sx2 +(ny –1)sy

2/ nx +ny –2 = √(10-1)302 +(12-1)252 /10+12-2 = 27.3633

CI = (480-520) ± 1.725(27.3633) √1/10+ 1/12 = -40 ± 20.2106

= -60.21056 to -19.7894

b. (x2 - y2 ) – t(ʋ, ∝/2)√sx2/nx+ sy

2/ny< µx - µy< (x2 - y2 ) + t(ʋ, ∝/2)√sx2/nx+ sy

2/ny

ʋ = [(sx2/nx)+(sy

2/ny)] /(sx2/nx)2/(nx –1) + (sy

2/ny)2/(ny –1)

= [(302/10)+(252/12)] /(302/10)2/(10 –1) + (252/12)2/(12 –1) = 17.606 ≈ 18

CI = (480-520) ± 1.734√(302/10)+(252/12) = -40 ± 20.669

= -60.669 to -19.331

8.42: Students in an introductory economics class are assigned to quiz sections conducted by teaching assistants. For one teaching assistant the 21 students in the quiz section obtained a mean score of 72.1 on the final examination and a standard of 11.3. For a second teaching assistant the 18 students in the quiz section obtained a mean score of 73.8 on the final examination and a standard of 10.6. Assume that these data can be regarded as independent random samples from normally distributed populations with a common variance. Find an 80% confidence interval for the difference between the population means.

Solution:

nx = 21, x2 = 72.1, sx = 11.3, t37, 0.100 = 1.303 (here, I use df = 40)

ny = 18, y2 = 73.8, sy = 10.6

(x2 - y2 ) – tn1+ n2 – 2, ∝/2 sp√1/nx+1/ny< µx - µy< (x2 - y2 ) + tn1+ n2 – 2, ∝/2 sp√1/nx+1/ny

Sp = √(nx –1)sx2 +(ny –1)sy

2/ nx +ny –2 = √(21-1)11.32 +(18-1)10.62 /21+18-2 = 10.9839

CI = (72.1-73.8) ± 1.303(10.9839) √1/21+ 1/18 = -1.7 ± 4.5971= -6.2971 to 2.8971

8.50: A newspaper article reported that 400 people in one state were surveyed and 75% were opposed to a recent court decision. The same article reported that a similar survey of 500 people in another state indicated opposition by only 45%. Construct a 95% confidence interval of the difference in population proportions based on the data.

Solution:

Ṕ1 = xn =

300400

= 0.75, Ṕ2 = xn =

225500

= 0.45

CI = (Ṕ1- Ṕ2)± z∝/2√ Ṕ1(1- Ṕ1)/ n1 + Ṕ2(1- Ṕ2)/ n2

= (.75- .45) ± 1.96 √.75(1-.75)/400 + .45(1-.45)/500

= 0.30 ± .06085

= 0.23915 up to 0.36085

9.4: During 2000 and 2001 many people of Europe objected to purchasing food that was generally modified, produced by farmers in the United State. The U.S. farmers argued that there was no scientific evidence to conclude that these products were not healthy. The Europeans argued that there still might be a problem with these foods.

a. State the null and alternative hypothesis from the perspective of the Europeans.b. State the null and alternative hypothesis from the perspective of the U.S. farmers.

Solution:

a. European perspective:

H0: Genetically modified food stuffs are not safe.

H1: They are safe.

b. U.S. farmer perspective:

H0: Genetically modified food stuffs are safe.

H1: They are not safe.

9.8: Using the results from the above two exercises, indicate how the critical value x2 c is influenced by sample size. Next indicate how the critical value is influenced by population variance, σ2.

Solution:

The critical value x2 c is farther away from the hypothesized value the smaller the sample size n. This is due to the increase in the standard error with a smaller sample size.

The critical value x2 c is farther away from the hypothesized value the larger the population variance. This is due to the increased standard error with a larger population variance.

9.10: A random sample of size n= 25 is obtained from a population with variance σ2, and the sample mean is computed to be x2 = 70. Consider the null hypothesis H0: µ = 80 verse the alternative hypothesis H1: µ ≤ 80. Compute the p-value for the following options.

a. The population variance is σ2 = 225.b. The population variance is σ2 = 900.c. The population variance is σ2 = 400.d. The population variance is σ2 = 600.

Solution:

A random sample of n = 25, variance = σ2 and the sample mean is = 70.

Consider the null hypothesis H0:µ = 80 versus the alternative H1:µ≤80.

Compute the p-value

a. σ2 = 225. z = x2 - µ0 /σ/√n = 70 – 80/15/√25 = -3.33. p-value = P(zp< -3.33) = .0004

b. σ2 = 900. z = x2 - µ0/σ/√n = 70 – 80/30/√25 = -1.67. p-value = P(zp< -1.67) = .0475

c. σ2 = 400. z = x2 - µ0/σ/√n = 70 – 80/20/√25 = -2.50. p-value = P(zp< -2.50) = .006

d. σ2 = 600. z = x2 - µ0/σ/√n = 70 – 80/24.4949/√25 = -2.04. p-value = P(zp< -3.33) = .0207

9.12: A company that receives shipments of batteries tests a random sample of nine of them before agreeing to take a shipment. The company is concerned that the true mean lifetime for all batteries in the shipment should be at least 50 hours. From past experience it is safe to conclude that the population distribution of lifetime is normal with standard deviation 3 hours. For one particular shipment the mean lifetime for a sample of nine batteries was 48.2 hours. Test at the 10% level the null hypothesis that the population mean lifetime is at least 50 hours.

Solution:

H0:µ ≥ 50; H1:µ < 50; reject H0 if Z10< -1.28

Z = 48.2 –50

3/√ 9 = -1.8, therefore, reject H0 at the 10% level.

9.14: Test the hypothesis

H0 : µ ≤ 100

H1 : µ > 100

Using a random sample of size n = 25, a probability of types I error to 0.05 and the following sample statistics.

a. x2 = 106; s = 15

b. x2 = 104; s = 10

c. x2 = 95; s = 10

d. x2 = 92; s = 18

Solution:

Test H0: µ ≤ 100; H1: µ > 100, using n = 25 and alpha = .05

a. x2 = 106, s = 15. Reject if x2 - µ0/s/√n > tn-1,α/2,106 – 100/15/√25 = 2.00. Since 2.00 is

greater than the critical value of 1.711, there is sufficient evidence to reject the null hypothesis.

b. x2 = 104, s = 10. Reject if x2 - µ0/s/√n > tn-1,α/2, 104-100/10/√25 = 2.00. Since 2.00 is

greater than the critical value of 1.711, there is sufficient evidence to reject the null hypothesis.

c. x2 > 92, s>18. Reject if x2 - µ0/s/√n > tn-1,α/2, 95-100/10/√25 = - 2.50 is less than the

critical value of -1.711, there is sufficient evidence to reject the null hypothesis.

d. x2 > 92, s>18. Reject if x2 - µ0/s/√n > tn-1,α/2, 95-100/18/√25 = - 2.22 is less than the

critical value of -1.711, there is sufficient evidence to reject the null hypothesis.

9.20: A random sample of 170 people was provided with a forecasting problem. Each sample member was given, in two ways, the task of forecasting the next value of a retail sales variable. The previous 20 values were presented both as numbers and as points on a graph. Subjects were asked to predict the next value. The absolute forecasting errors were measured. The sample then consisted of 170 differences in absolute forecast errors. The sample mean of these differences was -2.91 and the sample standard deviation was 11.33. Find and interpret the p-value of a test of the null hypothesis that the population mean difference is 0 against the alternative that it is negative.

Solution:

H0:µ > 0; H1:µ = 0

t > -2.91- 0/11.33/√170 = -3.35, p-value is < .005. Reject H0 at any common level of alpha.

9.24: A process that produces bottles of shampoo, when operating correctly, produces bottles whose contents weigh, on average, 20 ounces. A random sample of nine bottles from a single production run yielded the following content weights ( in ounces):

21.4 19.7 19.7 20.6 20.8 20.1 19.7 20.3 20.9

Assuming that the population distribution is normal, test at the 5% level against a two-sided alternative the null hypothesis that the process is operating correctly.

Solution:

H0:µ=20; H1:µ ≠ 20, Reject H0 if │t8,.05/2│>2.306

t > 20.3556 -20/.6126/√9 = 1.741, Therefore, do not reject H0 at the 5% level.

9.28: Solution:

a. n = 400. Reject H0 if Ṕ = Ṕc > P0, zα√P0(1-P0)/n = .25 + 1.88√(.25)(1-.25)/400 = 0.2907

b. n = 225. Reject H0 if Ṕ = Ṕc > P0, zα√P0(1-P0)/n = .25 + 1.88√(.25)(1-.25)/225 = 0.30427

c. n = 625. Reject H0 if Ṕ = Ṕc > P0, zα√P0(1-P0)/n = .25 + 1.88√(.25)(1-.25)/625 = 0.28256

d. n = 900. Reject H0 if Ṕ = Ṕc > P0, zα√P0(1-P0)/n = .25 + 1.88√(.25)(1-.25)/900 = 0.2771

9.34: A random sample of 50 university admissions officers was asked about expectations in application interviews. Of these sample members, 28 agreed that the interviewer usually expects the interviewee to have volunteer experience doing community projects. Test the null hypothesis that one-half of all interviewers have this expectation against the alternative that the population proportion is bigger than one-half. Use ∝ = 0.05.

Solution:

H0:P >.5; H1:P ≤.5;

z > .56 -.5/√(.5)(.5)/50 = .85, p-value = 1- Fz(.85) = 1 - .8023 = .1977

Therefore, reject H0 at alpha level in excess of 19.77%

9.38: Solution:

What is probability of Type II error if the actual proportion is

a. P =.52

β = P(.46≤ Ṕ ≤.54 │P = .52) = P[.46-P1/√P1(1- P1)/n ≤ z ≤ .54- P1/√ P1(1- P1)/n]

= P[.46-.52/√.52(1-.52)/600 ≤ z ≤ .54-.52/√.52(1-.52)/600]

= P(-2.94 ≤ z ≤ .98) = .4984 +.3365 = .8349

b. P =.58

β = P(.46≤ Ṕ ≤.54 │P = .58) = P[.46-P1/√P1(1- P1)/n ≤ z ≤ .54- P1/√ P1(1- P1)/n]

= P[.46-.58/√.58(1-.58)/600 ≤ z ≤ .54-.58/√.58(1-.58)/600]

= P(-5.96 ≤ z ≤ -1.99) = .5000 +.4767 = .0233

c. P=.53

β = P(.46≤ Ṕ ≤.54 │P = .53) = P[.46-P1/√P1(1- P1)/n ≤ z ≤ .54- P1/√ P1(1- P1)/n]

= P[.46-.53/√.53(1-.53)/600 ≤ z ≤ .54-.53/√.53(1-.53)/600]

= P(-3.44 ≤ z ≤ .49) = .4997 +.1879 = .6876

d. P =.48

β = P(.46≤ Ṕ ≤.54 │P = .48) = P[.46-P1/√P1(1- P1)/n ≤ z ≤ .54- P1/√ P1(1- P1)/n]

= P[.46-.48/√.48(1-.48)/600 ≤ z ≤ .54-.48/√.48(1-.48)/600]

= P(-.98 ≤ z ≤ 2.94) = .3365 +.4984 = .8349

e. P =.43

β = P(.46≤ Ṕ ≤.54 │P = .43) = P[.46-P1/√P1(1- P1)/n ≤ z ≤ .54- P1/√ P1(1- P1)/n]

= P[.46-.43/√.43(1-.43)/600 ≤ z ≤ .54-.43/√.43(1-.43)/600]

= P(1.48 ≤ z ≤ 5.44) = .5000 +.4306 = .0694

9.40: A pharmaceutical manufacturer is concerned that the impurity concentration in pills does not exceed 3%. It is known that from a particular production run impurity concentrations follow a normal distribution with standard deviation 0.4%. A random sample of 64 pills from a production run was checked and the sample mean impurity concentration was found to be 3.07%.

a. Test at the 5% level the null hypothesis that the population mean impurity concentration is 3% against the alternative that it is more than 3%.

b. Find the probability of a 5% level test rejecting the null hypothesis when the true mean impurity concentration is 3.10%.

Solution:

a. H0 is rejected when x −3

.4/√64 > 1.645 or when x2 > 3.082. Since the sample mean is 3.07%

which is less than the critical value, the decision is do not reject the null hypothesis.

b. The β = P(z < 3.082−3.1

.4 /√ 64 ) = 1 – Fz(.36) = .3594. Power of the test = 1 – β = .6406

9:48: At the insistence of a government inspector a new safety device is installed in an assembly-line operation. After the installation of this device a random sample of 8 days’ output gave the following results for number of finished components produced:

618 660 638 625 571 598 639 582

Management is concerned about the variability of daily output and views as undesirable any variance above 500. Test at the 10% significance level the null hypothesis that the population variance for daily output does not exceed 500.

Solution:

Ho : σ2 ≤ 500, H1 : σ2 > 500 Reject H0 if X2(7.10) >12.02

X2 > (n-1)s2/σ2 > 7(933.982)

500 =13.0757, therefore, Reject H0 at the 10% level.

9.50: Solution:

Ho : σ2 = 300, H1 : σ2 ≠ 300

X2 = 29(480)

300 = 46.4, p-value = 0.214 Reject H0 at the 5% level.

9.56: Solution:

a. False. The significance level is the probability of making a Type I error falsely rejecting te null hypothesis when in fact the null is true.

b. True.c. True.d. False. The power of the test is the ability of the test to correctly reject a false null

hypothesis.e. False. The rejection is farther away from the hypothesized value at the 1% level than it is

at the 5% level. Therefore, it is still possible to reject at the 5% level but not at the 1% level.

f. True.g. False. The p-value tells us the strength of the evidence against the null hypothesis.

9.58: Solution:

a. α = P(Z < 776 – 800120/√100

) = P(Z < -2) = .0228

b. β = P(Z > 776 – 740120/√100

) = P(Z > 3) = .0014

c. i) Smaller ii) Smaller;d. i) Smaller ii) Larger;

9.60: Solution:

H0: P = .5; H1: P ≠ .5

z = .4808−.5

√ (.5)(.5)/104 = .39, p-value = 2[1 – Fz(.39)] = 2[1 - .6517] = .6966

Therefore, reject H0 at levels in excess of 69.66%

THE END