teste de hipÓtese

Capítulo 5 – TESTE DE HIPÓTESE

Exercício 02

2) An engineering research center claims that through the use of a new computer control system, automobiles should achieve on average an additional 3 miles per gallon of gas. A random sample of 100 automobiles was used to evaluate this product. The sample mean increase in miles per gallon achieved was 2.4 and the sample standard deviation was 1.8 miles per gallon. Test the hypothesis that the population mean is at least 3 miles per gallon using 5% significance level. Find the P-value of this test, and interpret your findings.

Dados: ( one sample Z – unilateral esquerdo)

n = 100 Média = 2,4σ =1,8α: 5%

H0: µ ≥ 3H1 : µ < 3

Primeiramente, vamos achar a estatística de teste:

Sabemos que com um valor de α=0,05, Zα é -1,645.E como ZT pertence a RC, rejeita-se H0. Então, por exclusão, a média

populacional é inferior a 3.Além disso, P-value é igual a 0,000. (valor obtido através do software MINITAB)

Exercício 03

3) A beer distributor claims that a new display, featuring a life-size picture of a well-known rock singer, will increase product sales in supermarkets by an average of 50 cases in a week. For a random sample of 20 liquor weekly sales, the average sales increase was 41.3 cases and the sample standard deviation was 12.2 cases. Test at the 5% level the hypothesis that the population mean sales increase is at least 50 cases.

Dados:Aumento de 50 casos/semana (µ)n = 20 aumento da amostra: 41,3 casos/semanaσ: 12,2α: 5%


Sabemos que com um valor de α=0,05, Zα é -1,645.E como ZT > Zα, ZT Є RC, rejeita-se Ho, pois o acréscimo das vendas da

população é inferior a 50.

Exercício 04

4) In contract negotiations, a company claims that a new incentive scheme has resulted in average weekly earning of at least $400 for all customer service workers. A union representative takes a random sample of 15 workers and finds that their weekly earnings have an average of $381.25 and a standard deviation of $48.60. Assume a normal distribution. a) Test the company’s claim; b) If the same sample results had been obtained from a random sample of 50 employees, could the company’s claim be rejected at a lower significance level than in part (a)?

Trata-se de um teste de hipótese 1-sample-Z, pois σ é conhecido. E é um teste unilateral esquerdo.

a) Dados: n = 15; = 381,25; σ = 48,60

Ho: H1: ˂ 400Estatística de teste (Zt)

Tem-se, para um nível de significância de 5%, um Zcrítico = -1,645; como Zteste = -1,49 não pertence à região crítica, aceita-se a hipótese nula de que a média semanal de ganho é no mínimo 400 dólares.b) Adotando-se, agora, uma amostra de 50 empregados, temos:

Com esta nova amostra seria rejeitada a hipótese nula aceita no item a, pois Zteste agora pertence à região crítica.

Exercício 06

6) The production manager of a company has asked you to evaluate a proposed new procedure for producing its double-hung windows. The present process has a mean production of 80 units per hour with a population standard deviation of . The manager indicates that she does not want to change to a new procedure unless there is strong evidence that the mean production level is higher with the new process. Taking a random sample of n = 25 production hours the obtained sample mean was 83. Test with a significance level of if there is strong evidence that the production hour increased. Find the respective P-value. % 5

Dados: ( one sample Z – unilateral direito)

n = 25 Média = 83σ =8α: 5%

H0: µ = 80H1 : µ > 80


Sabemos que com um valor de α=0,05, Zα é 1,645.E como ZT pertence a RC, rejeita-se H0. Então, por exclusão, vemos que a

média de produção realmente aumentou, sendo pois maior do que os oitenta itens fabricados inicialmente.

Além disso, P-value é igual a 0,030. (valor obtido através do software MINITAB)

Exercício 07

7) A manufacturing process involves drilling holes whose diameter are normallydistributed with population mean of 2 inches and population standard deviation of 0.06inches. A random sample of 9 measurements had a sample mean of 1.95 inches. Use asignificance level of 5% to determine if the observed sample mean is unusual andsuggests that the drilling machine should be adjusted. (Hint: use two-sided hypothesis test).

Dados:µ = 2 polegadasn = 9 media da amostra: 1,95 polegadasσ: 0,06α: 5% (Teste bilateral)


ZT = - 2,500

Temos que Zα = -1,645, logo ZT < Zα | ZT RC.

Resposta: Como temos que ZT < Zα, a amostra nos sugere que não há necessidade de se ajustar a máquina, pois essa de adequa as exigências.

Exercício 08

8) A manufacturer of detergent claims that the contents of boxes sold weigh on average at least 16 ounces. The distribution of weight is known to be normal, with ounce. A random sample of 16 boxes yielded a sample mean weight of 15.84 ounces. Test at 5% and 10% significance levels the null hypothesis that the population mean weight is at least 16 ounces.

Trata-se de um teste de hipótese 1-sample-Z, pois σ é conhecido. E é um teste unilateral esquerdo.

a) Dados: n = 16; = 15,84; σ = 0,4

Ho: H1: < 16Estatística de teste (Zt)

Para um nível de significância de 5%, temos:Zcrítico = -1,645Portanto, como Zteste não pertence à região crítica aceita-se a hipótese nula de que o peso médio do conteúdo das caixas vendidas seria de 16 onças.Agora, para um nível de significância de 10%, temos:Zcrítico = -1,285Portanto, como Zteste pertence à região crítica não se aceita a hipótese nula de que o peso médio do conteúdo das caixas vendidas seria de 16 onças.

Exercício 10

10) A pharmaceutical manufacturer is concerned about the impurity concentration in pills, and it is anxious that this does not exceed 3%. It is known that from a particular production run, impurity concentrations follow a normal distribution with standard deviation of 0.4 %. A random sample of 64 pills was checked, and the sample mean impurity concentration was found to be 3.07%. 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%. Find the respective P-value in this case.


n = 64 Média = 0,0307σ =0,004α: 5%

H0: µ = 3%H1 : µ > 3%


Sabemos que com um valor de α=0,05, Zα é 1,645.E como ZT não pertence a RC, então, aceita-se H0. Ou seja, a concentração

média de impurezas é de 3%.Além disso, P-value é igual a 0,081. (valor obtido através do software MINITAB)

Exercício 11

11) The accounts of a corporation show that, on average, accounts payable are $125.32. An auditor checked a random sample of 16 of these accounts. The sample mean was $131.78 and the sample standard deviation was $25.41. Assume that the population distribution is normal. Test as the 5% significance level against a two-sided alternative the null hypothesis that the population mean is $125.32. Find the P-value of this test.

Dados:µ = $ 131,78n = 16 media da amostra: $125,32σ: $25,41α: 5% (Teste bilateral)- Achar P-value


ZT = - 1,017Temos que Zα = -1,645, logo ZT > Zα | ZT Є RC.

O P-value do teste é igual a 0,309 (valor obtido através do software MINITAB)

Exercício 12

12) 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.

Trata-se de um teste de hipótese 1-sample-t. E é um teste bilateral.

Dados: n = 9; = 20,36; s = 0,61

Ho: H1: Temos:

Para um nível de significância de 5%, tem-se que:

Como tteste não pertence à região crítica, aceita-se a hipótese nula de que o processo opera corretamente.

Exercício 14

14) A random sample of 172 marketing students was asked to rate on a scale from one (not important) to five (extremely important) health benefits as a job characteristic. The sample rating was 3.31 and the sample standard deviation was 0.70. Test at 1% significance level the null hypothesis that the population mean rating is at most 3.0 against the alternative that it is bigger than 3.0.


n = 172 Média = 3,31σ =0,7α: 1%

H0: µ ≤ 3H1 : µ > 3


Sabemos que com um valor de α=0,01, Zα é 2,575.E como ZT pertence a RC, rejeita-se H0. Ou seja, a taxa média do benefício do

emprego para a saúde é maior do que 3%.

teste de hipÓtese

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