Types of Surveys - Essay Example

Week 5: One and Two Sample Tests Chapter 10: 31. A new weight-watching company, Weight Reducers International, advertises that those who join will lose, on the average, 10 pounds the first two weeks with a standard deviation of 2.8 pounds. A random sample of 50 people who joined the new weight reduction program revealed the mean loss to be 9 pounds. At the .05 level of significance, can we conclude that those joining Weight Reducers on average will lose less than 10 pounds? Determine the p-value.

Hypotheses: Null Hypothesis  H0: µ = 10Alternate Hypothesis  H1: µ ≤ 10Test Statistic:z = (9 - 10) / (2.8 / √(50)) = -2.525The p-value can be obtained from the z – table corresponding to -2.525. p - Value:p (z < -2.525) = 0.0058As the p-value is not lesser than 0.05, the null hypothesis is dropped. Hence we can conclude on the alternate hypothesis that the average weight loss at the weight – watching company is less than 10 pounds. 32. Dole Pineapple, Inc., is concerned that the 16-ounce can of sliced pineapple is being overfilled.

Assume the standard deviation of the process is .03 ounces. The quality control department took a random sample of 50 cans and found that the arithmetic mean weight was 16.05 ounces. At the 5 percent level of significance, can we conclude that the mean weight is greater than 16 ounces? Determine the p-value.Hypotheses: Null Hypothesis  H0: µ = 16Alternate Hypothesis  H1: µ > 16Test Statistic:z = (16.05 - 16) / (.03 / √(50)) = 11.785The p-value can be obtained from the z – table corresponding to 11.785. p - Value:p (z > 11.785) = 0As the p-value is not lesser than 0.

05, the null hypothesis is dropped. Hence we can conclude on the alternate hypothesis that the mean weight is greater than 16 ounces.38. A recent article in The Wall Street Journal reported that the 30-year mortgage rate is now less than 6 percent. A sample of eight small banks in the Midwest revealed the following 30-year rates (in percent):4.8, 5.3, 6.5, 4.8, 6.1, 5.8, 6.2, 5.6At the .01 significance level, can we conclude that the 30-year mortgage rate for small banks is less than 6 percent?

Estimate the p-value.Hypotheses: Null Hypothesis  H0: μ ≥ 6Alternate Hypothesis  H1: μ < 6Mean X = 5.6375Std. Dev. s = 0.63457Test statistic:t = (X - μ) / ( s / √(n)) = (5.6375 - 6) / (0.63457 / √(8)) = -1.616Degrees of Freedom = n - 1 = 8 - 1 = 7p - Value:p (t0.01,7 < -1.616) = 0.075Since the p-value is greater than 0.01, there is no evidence to drop the null hypothesis. Hence it can be concluded that the rates are greater than or equal to 6%.Chapter 11:27. A recent study focused on the number of times men and women who live alone buy take-out dinner in a month.

The information is summarized below.Statistic Men WomenSample mean 24.51 22.69Population standard deviation 4.48 3.86Sample size 35 40At the .01 significance level, is there a difference in the mean number of times men and women order take-out dinners in a month? What is the p-value?Hypotheses:H0: There is no significant difference in the mean number of times men and women order take-out dinners in a monthH1: There is significant difference in the mean number of times men and women order take-out dinners in a monthDifference (Men – Women) = 1.

82Standard Error of Difference = 0.973The z value is 1.87 and the corresponding p-value (two tailed) is found to be 0.0613.As the p – value is greater than 0.01, the null hypothesis can be accepted. Hence it can be concluded that there is no significant difference in the mean number of times men and women take – out dinners in a month.46. Grand Strand Family Medical Center is specifically set up to treat minor medical emergencies for visitors to the Myrtle Beach area. There are two facilities, one in the Little River Area and the other in Murrells Inlet.

The Quality Assurance Department wishes to compare the mean waiting time for patients at the two locations. Samples of the waiting times, reported in minutes, follow:Location Waiting TimeLittle River 31.73 28.77 29.53 22.08 29.47 18.60 32.94 25.18 29.82 26.49Murrells Inlet 22.93 23.92 26.92 27.20 26.44 25.62 30.61 29.44 23.09 23.10 26.69 22.31Assume the population standard deviations are not the same. At the .05 significance level, is there a difference in the mean waiting time?

Hypotheses: H0: There is no significant difference in the mean waiting time H­1: There is significant difference in the mean waiting time.Test Statistic: t = Degrees of Freedom = 14Difference (Little River – Murrels Inlet) = 1.772Standard Error of Difference = 1.604The t value is 1.10 and the corresponding p-value (two tailed) is found to be 0.2879.As the p – value is greater than 0.05, the null hypothesis can be accepted. Hence it can be concluded that there is no significant difference in the mean waiting time.52. The president of the American Insurance Institute wants to compare the yearly costs of auto insurance offered by two leading companies.

He selects a sample of 15 families, some with only a single insured driver, others with several teenage drivers, and pays each family a stipend to contact the two companies and ask for a price quote. To make the data comparable, certain features, such as the deductible amount and limits of liability, are standardized. The sample information is reported below. At the .10 significance level, can we conclude that there is a difference in the amounts quoted?Progressive GEICOFamily Car Insurance Mutual InsuranceBecker $2,090 $1,610Berry 1,683 1,247Cobb 1,402 2,327Debuck 1,830 1,367DuBrul 930 1,461Eckroate 697 1,789German 1,741 1,621Glasson 1,129 1,914King 1,018 1,956Kucic 1,881 1,772Meredith 1,571 1,375Obeid 874 1,527Price 1,579 1,767Phillips 1,577 1,636Tresize 860 1,188Hypotheses: H0: H­1: (one-tailed test)Level of Significance: α = 0.

1 Test Statistic: t = Degrees of Freedom = 28Difference (Progressive – GEICO) = -246.333Standard Error of Difference = 1.604The t value is -1.80 and the corresponding p-value (two tailed) is found to be 0.0827.As the p – value is lesser than 0.1, the null hypothesis can be rejected. Hence it can be concluded that there is a significant difference in the amounts quoted.Chapter 12:23. A real estate agent in the coastal area of Georgia wants to compare the variation in the selling price of homes on the oceanfront with those one to three blocks from the ocean.

A sample of 21 oceanfront homes sold within the last year revealed the standard deviation of the selling prices was $45,600. A sample of 18 homes, also sold within the last year, that were one to three blocks from the ocean revealed that the standard deviation was $21,330. At the .01 significance level, can we conclude that there is more variation in the selling prices of the oceanfront homes?Hypotheses: H0: (There is equal variation in the selling prices of the oceanfront homes) H­1: (There is more variation in the selling prices of the oceanfront homes)Level of Significance: α = 0.

05 Critical value:Test statistic:It is evident that the test statistic is greater than the critical value. Hence the null hypothesis is rejected and it is concluded that there is more variation in the selling prices of the oceanfront homes.28. The following is a partial ANOVA table.Sum of MeanSource Squares df Square FTreatment 320 2 160 8Error 180 9 20Total 500 11Complete the table and answer the following questions. Use the .05 significance level.a. How many treatments are there?

From the given data, it is evident that there are a total of 3 treatments.b. What is the total sample size?The total sample size = 12c. What is the critical value of F?The Critical Value F0.05, 2, 9 = 4.2564d. Write out the null and alternate hypotheses.Null hypothesis  H0:Alternate hypothesis  Ha:e. What is your conclusion regarding the null hypothesis?As the test statistic is greater than the critical value, the null hypothesis is rejected and it is concluded that there is a difference between the treatment means.

Chapter 17:19. In a particular market there are three commercial television stations, each with its own evening news program from 6:00 to 6:30 P.M. According to a report in this morning’s local newspaper, a random sample of 150 viewers last night revealed 53 watched the news on WNAE (channel 5), 64 watched on WRRN (channel 11), and 33 on WSPD (channel 13). At the .05 significance level, is there a difference in the proportion of viewers watching the three channels?Hypotheses: H0: P1=P2=P3 (There is no significant difference in the proportion of viewers watching the three channels) H­1: P1≠P2 ≠P3 (There is a significant difference in the proportion of viewers watching the three channels)Level of significance:α = 0.

05Critical values:Critical region:Photograph(Oi)Sample size (ni)ni*p(Oi- nip)2/ nipq5315049.50.247476415049.54.247473315049.55.5150450148.59.9949Test Statistics:As 0.051 < < 9.9949, the data does not support the null hypothesis and hence it is rejected. It is concluded that there is a significant difference in the proportion of the viewers watching the three channels. 20. There are four entrances to the Government Center Building in downtown Philadelphia. The building maintenance supervisor would like to know if the entrances are equally utilized.

To investigate, 400 people were observed entering the building. The number using each entrance is reported below. At the .01 significance level, is there a difference in the use of the four entrances?Entrance FrequencyMain Street 140Broad Street 120Cherry Street 90Walnut Street 50Total 400Hypotheses: H0: There no difference in the use of the four entrances H­1: There no difference in the use of the four entrancesLevel of significance:α = 0.01Critical value:Test statistic:where o ij observed frequencies e ij  expected frequencies Entrance observed (oi)Expected(Ei)(Oi-Ei)(Oi-Ei)^2(Oi-Ei)^2/EiMain Street14010040160016Broad Street120100204004Cherry Street90100-101001Walnut Street50100-5025002546As the chi square value (46) is greater than the critical value, the null hypothesis is rejected.

It is concluded that there is no difference in the usage of the four entrances.