Question

For a test market, find the sample size needed to estimate the true proportion of consumers satisfied with a certain new product within 0.04 at the 90 percent confidence level. Assume you have no strong feeling about what the proportion is?

SolutionAssume p = q = 0.5 e = 0.04Z = 1.64

Food tiger, a local grocery store, sells generic garbage bags and has received quite a few complaints about the strength of these bags. It seems that the generic bags are weaker than the name-brand competitors bags and therefore, break more often. John C. Tiger, VP in charge of purchasing is interested in determining the average maximum weight that can put into one of the generic bags without its breaking. If the standard deviation of garbage breaking weight is 1.2lb, determine the number of bags that must be tested in order for Mr. Tiger to be 95 percent confident that the sample average breaking weight is within 0.5 lb of the true average. Question

Question = 1.2Z= 1.96e = 0.5

QuestionJill Johnson, product manager for Southern Electrics smoke alarm, is concerned over recent complaints from consumer groups about the short life of the device. She has decided to gather evidence to counteract the complaints by testing a sample of the alarms. For the test, it costs $4 per unit in the sample. Precision is desirable for presenting persuasive statistical evidence to consumer groups, so Johnson figures the benefits she will receive for various sample sizes are determined by formula Benefit . If Johnson wants to increase her sample until the cost equals the benefit, how many units should she sample? The population standard deviation is 265.

Solution

Cost = Benefit

Thus, she should sample at least 25 detectors.

*Example

Suppose a hospital uses large quantities of packaged doses of a particular drug. The individual dose of this drug is 100 cubic centimeters (100cc). The action of the drug is such that the body will harmlessly pass off excessive doses. On the other hand, insufficient doses do not produce the desired medical effect and they interfere with patient treatment. The hospital has purchased this drug from the same manufacturer for a number of years and known that the population standard deviation is 2 cc. The hospital inspects 50 doses of this drug at random from a very large shipment and finds the mean of these doses to be 99.75 cc. If the hospital sets a 0.10 significance level and asks us whether the dosages in this shipment are too small, how can we find the answer?

*SolutionH0: = 100 The mean of the shipments dosages is 100ccH1: < 100 The mean is less than 100cc = 0.10, = 2, n = 50 = 99.75Critical z 1.28.

So hospital should accept null hypothesis because observed mean of sample is not significantly lower than our hypothesized mean of 100 cc. On the basis of this sample of 50 doses, hospital should conclude that the does in the shipment are sufficient.

*Question From 1980 until 1985, the mean price per earnings (P/E) ratio of the approximately 1800 stocks listed on the New York Stock Exchange was 14.35 and the standard deviation was 9.73. In a sample of 30 randomly chosen NYSE stocks, the mean P/E ratio in 1985 was 11.77. Does this sample present sufficient evidence to conclude (at 0.05 level of significance) that in 1986 the mean P/E ratio for NYSE stocks had changed from its earlier value?

*Solution Ans: = 9.73 n = 30 = 0.025H0: = 14.35, H1: 14.35The limits of the acceptance region are zCRIT = 1.96Since ,

The calculated value of z is less than critical value so null hypothesis is not rejected.

The mean P / E ratio in 1986 is not significantly different from its previous value.

*QuestionBefore 1973 oil embargo and subsequent increases in the price of crude oil, gasoline usage in the United States had grown at a seasonally adjusted rate of 0.57 percent per month, with a standard deviation of 0.01 percent per month. In 15 randomly chosen months between 1975 and 1985, gasoline usage grew at an average rate of only 0.33 percent per month. At a 0.01 level of significance, can you conclude that the growth in the use of gasoline had decreased as a result of the embargo and its consequences?

*SolutionHere n = 15 = 0.10 = 0.01H0: = 0.57 H1: < 0.57Critical z value is 2.33

so we should reject H0. The rate of growth has decreased significantly, and we infer that was because of the oil embargo and its consequences.

*A member of public interest group concerned with environmental pollution asserts at a public hearing that fewer than 60 percent of the industrial plants in this area are complying with air pollution standards Attending this meeting is an official of the Environmental Protection Agency who believes that 60 percent of the plants are complying with the standards: she decides to test that hypothesis at the 0.02 significance level. The official makes a through search of the records in her office. She samples 60 plants from a population of over 10000 plants and find that 33 are complying with air pollution standards. Is the assertion by the member of the public interest group a valid one?Question

*SolutionH0: p = 0.6 Proportion of plants complying with the air pollution standards is 0.6H1: p < 0.6 Proportion complying with the standards is less than 0.6 = 0.02, p = 0.6 q = 0.4, n = 60 = 33/60 = 0.55 Critical z value is 2.05

Sample proportion lies within acceptable region. Therefore EPA official should accept the null hypothesis that the true proportion of complying plants is 0.6

*QuestionFrom a total of 10200 loans made by a state employees credit union in the most recent 5 year period, 350 were sampled to determine what proportion was made to woman. This sample showed that 39 percent of the loans were made to woman employees. A complete census of loans 5 years ago showed that 41 percent of the borrowers then were woman. At the significant level of 0.02, can you conclude that the proportion of loans made to women has changed significantly in the past five years?

*SolutionHere n = 350 = 0.02H0: p = 0.41 H1: p 0.41The critical value of z is 2.33

The null hypothesis is not rejected The proportion of loans made to women has not changed significantly.

*QuestionSome financial theoreticians believe that the stock markets daily prices constitute a random walk with positive drift. If this is accurate, then the Dow Jones Industrial Average should show a gain on more than 50 percent of all trading days. If the average increased on 101 of 175 randomly chosen days, what do you think about the suggested theory? Use a 0.01 level of significance.

*SolutionHere n = 175, 0.01 H0: p = 0.5 H1: p> 0.5The critical value of z is 2.33

so we do not reject H0. The data do not provide significant support to the theory.

*A personnel specialist of a major corporation is recruiting a large number of employees for an overseas assignment. During the testing process, management asks how things are going and she replies, Fine. I think the average score on the aptitude test will be around 90. When management reviews 20 of the test results compiled, it finds that the mean score is 84, and the standard deviation of this score is 11. If the management wants to test her hypothesis at the 0.10 level of significance, what is the procedure?

Question

*SolutionAns: H0: = 90 True population mean score is 90H1: 90 True population mean score is not 90 = 0.10, n = 20 = 84, s = 11Critical t value is 1.729.

The sample mean falls outside the acceptance region. Therefore management should reject the null hypothesis (the personal specialists assertion that the true mean score of the employees being tested is 90).

*QuestionRealtor Elain Snyderman took a random sample of 12 homes in a prestigious suburb of Chicago and found the average appraised market value to be $780000 and the standard deviation was $49000. Test the hypothesis that for all homes in the area, the mean appraised value is $825000 against the alternative that it is less than $825000. Use the 0.05 level of significance.

*SolutionHere s = 49000, n= 12, = 0.05H0: = 825000, H1: < 825000The critical t value at 11 degrees of freedom is 1.796

so we should reject H0. The average appraised value of homes in the area is significantly less than $825000.

*QuestionA television documentary on overheating claimed that Americans are about 10 pounds overweight on average. To test this claim, eighteen randomly selected individuals were examined, their average excess weight was found to be 12.4 pounds, and the sample standard deviation was 2.7 pounds. At the significance level of 0.01, is there any reason to doubt the validity of the claimed 10 pound value?

*SolutionHere s = 27, n = 18, = 0.01H0: = 10, H0: 10The critical value of t is 2.898

we should reject H0. The claim does not appear to be valid.

QuestionA sample of 32 money-market mutual funds was chosen on January, 1996 and the average annual rate of return over the past 30 days was found to be 3.23 percent, and the sample standard deviation was 0.51 percent; A year earlier, a sample of 38 money-market funds showed an average rate of return of 4.36 percent, and the sample standard deviation was 0.84 percent. Is it reasonable to conclude (at = 0.05) that money-market interest rates declined after 1995?

Solution Sample 1 (1995): s1=0.84, n1=38Sample 2 (1996): s2=0.51, n2=32H0:1= 2, H1:1> 2

The critical z value is 1.64 We reject H0 and conclude that money market rates declined significantly after 1995

Example: Concerning the sensitivity of a manager at a personal computer manufacture to the needs of their Spanish speaking employees. The company has been investigating two education programmes for increasing the sensitivity of its managers. Original programme consisted of several informal question answer sessions with leaders of the Spanish-speaking community. Over past few years, a programme involving formal classroom contact with professional psychologists and sociologist has been developed. New programme is considerably more expensive, and the president wants to know at the 0.05 level of significance whether this expenditure has resulted in greater sensitivity. Table contains the data resulting from a sample of the manager trained in both programme

Data from Sample of Two Sensitivity ProgramsProgram sampledMean sensitivity after this programNumber of managers observedEstimated standard deviation of sensitivity after this programFormal92%1215%Informal84%1519%

H0:1= 2 there is no difference in sensitivity level achieved by the two program.H1:1> 2 the new program results in higher sensitivity levels.At =0.05, the t distribution with 12 + 15 2 = 23 degree of freedom, the critical value t is 1.708.

Null hypothesis is not rejected implies there is no difference between the sensitivities achieved by the two programs. Companys expenditures on formal instructional program have not produced significantly higher sensitivities among its managers.

Solution

QuestionA sample of 30 year conventional mortgage rates at 11 randomly chosen banks in California yielded a mean rate of 7.61 percent and a standard deviation of 0.39 percent. A similar sample taken at 8 randomly chosen banks in Pennsylvania had a mean rate of 7.43 percent, and a standard deviation of 0.56 percent. Do these samples provide evidence to conclude (at = 0.10) that conventional mortgage rates in California and Pennsylvania come from populations with different means?

Solution Sample 1 (California): sc=0.39, nc=11,Sample 2 (Pennsylvania): sp=0.56, np=8,H0: c= pH1: c p=0.01Critical t value at 17 degrees of freedom is 1.7401

Do not reject H0. The California and Pennsylvania mean mortgage rates are not significantly different.

QuestionA health spa has advertised a weight reducing program and has claimed that the average participant in the program loses more than 17 pounds. A somewhat overweight executive is interested in the program but skeptical about the claims and asks for some hard evidence. The spa allows him to select randomly the records of 10 participants and record their weights before and after the program. These data are recorded in the table.The overweight executive wants to test at the 5 percent significant level the claimed average weight loss of more than 17 pounds.

Weights Before and After a Reducing ProgramBf189202220207194177193202208233Af170179203192172161174187186204

H0: 1-2=17 average weight loss is only 17 pound.H1: 1-2>17 average weight loss exceeds 17 pound.H0: d=17H1: d>17Difference weight is 19, 23, 17, 15, 22, 16, 19, 15, 12, 29At = 0.05 and 9 degree of freedom the critical t value is 1.833

So executive can reject the null hypothesis and conclude that the claimed weight loss in the program is legitimate.

Solution

QuestionDonna Rose is a production supervisor on the disk-drive assembly line at Winchester Technologies. Winchester recently subscribed to an easy listening music service at its factory, hoping that this would relax the workers and lead to greater productivity. Donna is skeptical about this hypothesis and fears the music will be distracting, leading to lower productivity. She sampled weekly production for the same six workers before the music was installed and after it was installed. Her data are given below. At = 0.02, has the average production changed at all?

Employee 1 2 3 4 5 6Week without music 219 205 226 198 209 216Week with music 235 186 240 203 221 205

SolutionEmployee 1 2 3 4 5 6Week without music 219 205 226 198 209 216Week with music 235 186 240 203 221 205Changes in product (x) 16-1914512-11

H0:AFTER= BEFORE and H1: AFTER BEFOREAt = 0.02 and 5 degrees of freedom, critical value of t is 3.365

Do not reject H0. The music does not have significant effect on productivity.

ExampleConsider the case of a pharmaceutical manufacturing company testing two new compounds intended to reduce blood pressure levels. The compounds are administered to two different sets of laboratory animals. In group one, 71 of 100 animals tested respond to drug 1 with lower blood pressure levels. In group two, 58 of 90 animals tested respond to drugs 2 with lower blood pressure levels. The company wants to test at the 0.05 level whether there is a difference between the efficacies of these two drugs. How should we proceed with this problem?

Solutionn1 = 100n2 = 90H0: p1=p2 no difference between two drugsH1: p1p2 difference between themAt = 0.05, the critical value of z is 1.96.

The null hypothesis accepted and conclude that these two new compounds produce effects on blood pressure that are not significantly different.

QuestionSuppose that for tax purposes, a city government has been using two methods of listing property. The first requires the properties owner to appear in person before a tax lister, but the second permits the property owner to mail in a tax form. The city manager thinks the personal appearance method produces far fewer mistakes than the mail in listings. Ten percent of the personal appearance forms contain errors, whereas the mail-in forms contain error is 13.3 percent. Sample size of personal appearance forms is 50 and Sample size of mail-in forms is 75. Level of significance for hypothesis testing is 0.15

n1 = 5n2 = 75H0: p1= p2 and H1: p1< p2At = 0.15 the critical value of z is 1.04.

Null hypothesis is not rejected implies there is no difference between the two methods of tax listing. Therefore if mailed in listing is considerably less expensive to the city, the city manager should consider increasing the use of this method.

Solution

QuestionMr. George Mc Mohan, president of National General Health Insurance Company, is opposed to national health insurance. He argues that it would be too costly to implement, particularly since the existence of such a system would among other effects, tend to encourage people to spend more time in hospitals. George believe that the length of stays in hospitals are dependent on the types of health insurance that people have. He asked Donna McClish, his staff statistician, to check the matter. Donna collected data on a random sample of 660 hospital stays and summarized them in Table. (Test at 0.01 significance level)

Hospital stay by types of insurance coverage & length of stay length of stay < 5 days5 10 > 10 daysTotalFraction of cost covered insurance50%40100190330Total110220330660

Chi-square value of 24.315 is not within acceptable region as chi-square is 13.277 (0.01 significant level with 4 degrees of freedom). Thus Donna must inform Mr. McMahan that the evidence supports his belief that length of hospital stay and insurance coverage are dependent on each other.Solution

Calculation of Expected Frequencies and Chi-Square RowColumnfofe(= (RT X CT)/n)fo - fe(fo - fe)2(fo - fe)2 / fe114030(180 x 110)/660101003.333127560(180 x 220)/660152253.750136590(180 x 330)/660-256256.944213025(150 x 110)/6605251.000224550(150 x 220)/660-5250.500237575(150 x 330)/660000.000314055(330 x 110)/660-152254.09132100110(330 x 220)/660-101000.90933190165(330 x 330)/660256253.788Chi-Square24.316

Question

A newspaper publisher, trying to pinpoint his markets characteristics, wondered whether newspaper readership in the community is related to readers educational achievement. A survey questioned adults in the area on their level of education and their frequency readership. The results are shown in the following table.At 0.10 significance level, does frequency of newspaper readership in the community differ according to readers level of education?

Level of educational achievementFrequency of ReadershipProfessional or PGCollege graduateHigh schoolBellow high schoolTotalNever 1017112159Sometime12238548Morning or evening353816796Both editions281961366Total85974146269

Solution

Critical Chi-square is 14.648, reject null hypothesis . So frequency of readership differs according to education

RowColumnfofefo - fe(fo - fe)2(fo - fe)2 / fe111018.643-8.64374.703574.007121721.275-4.27518.276420.85913118.9932.0074.0297950.448142110.08910.911119.045111.799211215.167-3.16710.03170.661222317.3095.69132.39261.8712387.3160.6840.4678760.0642458.208-3.20810.292411.254313530.3354.66521.766210.718323834.6173.38311.444010.331331614.6321.3681.8715050.12834716.416-9.41688.667785.401412820.8557.14551.050762.448421923.799-4.79923.032860.96843610.059-4.05916.479371.638441311.2861.7142.9369550.26032.856

QuestionAs the head of a department of a consumer research organization you have the responsibility for testing and comparing the life time of four brands of electric bulbs. Suppose you test the life time of three electric bulbs of each of the four brands. The data is shown below, each entry representing the life time of an electric bulb, measure in hundred of hours. Can we infer that the mean lifetimes of the four brands of electric bulbs are equal. Take 95% as the confidence level

BrandABCD202524231923202021212220

SolutionNull and alternative hypothesis is given bellow:H0: A = B= C= DH1: A B C D

nA = nB = nC = nD = 3

BrandABCD202524231923202021212220

SolutionSSC (sum of squares between columns) =

= 3 (20-21.5)2 + 3 (23-21.5)2 + 3 (22 21.5)2 + 3 (21-21.5)2 = 15

SSE (sum of squares within columns) =

= (20-20)2 + (19-20)2 + (21-20)2 + (25-23)2 + (23-23)2 + (21-23)2 + (24-22)2 + (20-22)2 + (22-22)2 + (23-21)2 + (20-21)2 + (21-21)2 = 24

MSC (mean square) =

MSE (mean square) =At 95% confidence level, the critical value obtained from the F table is F0.05,3,8 = 4.07. Calculated value of F is 1.67 which is lesser than critical value i.e. 4.07, so the null hypothesis is not rejected.

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