Chapter 9.4 Homework

Consider the following hypotheses: H0: p ≥ 0.29 HA: p < 0.29 Compute the p-value based on the following sample information. (You may find it useful to reference the appropriate table: z table or t table) (Round "z" value to 2 decimal places. Round intermediate calculations to at least 4 decimal places and final answers to 4 decimal places.) Find the P-value: a. x = 21; n = 124 b. x = 50; n = 265 c. p¯ = 0.20; n = 48 d. p¯ = 0.20; n = 383

A) Given: p̂= 21/124= 0.1694 p=0.29 n= 124 z = (p̂ -p)/ √p (1-p)/n = (0.1694-0.29)/ √0.29 (1-0.29)/124 = -0.1206/ 0.0407 = -2.96 The p-value = P(Z ≤ − 2.96) = 0.0015 (from Z-table) B) Given: p̂= 50/265= 0.1887 p=0.29 n= 265 z = (p̂ -p)/ √p (1-p)/n =(0.1887-0.29)/ √0.29(1-0.29)/265 = -0.1013/ 0.0279 =-3.63 The p-value = P(Z ≤ − 3.63) = 0.0001 (from the Z-table) C) Given: p̂= 0.20 p=0.29 n= 48 z = (p̂ -p)/ √p (1-p)/n = 0.20-0.29/ √0.29(1-0.29)/48 = -0.09/0.0655 = -1.37 The p-value = P(Z ≤ − 1.37) = 0.0853 (from the Z-table) D) Given: p̂= 0.20 p=0.29 n= 383 z = (p̂ -p)/ √p (1-p)/n = 0.20-0.29/ √0.29(1-0.29)/383 =-0.09/0.0232 = -3.88 The p-value = P(Z ≤ − 3.88) = 0.0001 (from the Z-table)

You would like to determine if the population probability of success differs from 0.70. You find 62 successes in 80 binomial trials. Implement the test at the 1% level of significance. (You may find it useful to reference the appropriate table: z table or t table) a. Select the null and the alternative hypotheses. b. Calculate the sample proportion. (Round your answer to 3 decimal places.) c. Calculate the value of test statistic. (Round intermediate calculations to at least 4 decimal places and final answer to 2 decimal places.) d. Find the p-value. e. At the 1% significance level, What is the conclusion? f. Interpret the results at α = 0.01.

A) H0: p = 0.70; HA: p ≠ 0.70 B)SAMPLE PROPORTION=0.775 p̂= 62/80= 0.775 C) TEST STATIATIC= 1.46 Given: p̂=0.775 p=0.70 n= 80 z = (p̂ -p)/ √p (1-p)/n = 0.775-0.70/ √0.70(1-0.70)/ 80 = 0.075/ 0.05 = 1.46 D) p-value ≥ 0.10 =TDIST(1.46,79,2) (USING EXCEL) =p-value = 2P(Z ≥ 1.46) = 0.1442 E) Do not reject H0 since the p-value is greater than significance level. The p-value = 2P(Z ≥ 1.46) = 0.1442 F) We cannot conclude that the population probability of success differs from 0.70.

The margarita is one of the most common tequila-based cocktails, made with tequila mixed with triple sec and lime or lemon juice, often served with salt on the glass rim. A common ratio for a margarita is 2:1:1, which includes 50% tequila, 25% triple sec, and 25% fresh lime or lemon juice. A manager at a local bar is concerned that the bartender uses incorrect proportions in more than 50% of margaritas. He secretly observes the bartender and finds that he used the correct proportions in only 10 out of 30 margaritas. Test if the manager's suspicion is justified at α = 0.05. Let p represent incorrect population proportion. (You may find it useful to reference the appropriate table: z table or t table) a. Select the null and the alternative hypotheses. b. Calculate the sample proportion. (Round your answer to 3 decimal places.) c. Calculate the value of test statistic. (Round intermediate calculations to at least 4 decimal places and final answer to 2 decimal places.) d. Find the p-value. e. At the 5% significance level, what is the conclusion?

A) H0: p ≤ 0.50; HA: p > 0.50 B) Sample Proportion= 0.667 The bartender uses correct proportions only in 10 out of 30, so the incorrect proportions are used in 20 out of 30. x=20 n= 30 p̂=20/30= 0.667 C) TEST STATISTIC= 1.83 z = (p̂ -p)/ √p (1-p)/n = 0.667-0.50/ √0.50(1-0.50)/ 30 =0.167/0.0913 =1.83 D) 0.025 ≤ p-value < 0.05 = p-value = P(Z ≥ 1.83) = 0.9664 = 1-0.9664 =0.0336 E) The manager's suspicion is justified since the p-value is less than the significance level. p-value = 0.0336 < α = 0.05.

You would like to determine if more than 50% of the observations in a population are below 10. At α = 0.05, conduct the test on the basis of the following 20 sample observations: (You may find it useful to reference the appropriate table: z table or t table) 8 12 5 9 14 11 9 3 7 8 12 6 8 9 2 6 11 4 13 10 a. Select the null and the alternative hypotheses. b. Calculate the sample proportion. (Round your answer to 2 decimal places.) c. Calculate the value of test statistic. (Round intermediate calculations to at least 4 decimal places and final answer to 2 decimal places.) d. Find the p-value. e. At the 5% significance level, What is the conclusion? f. Interpret the results at α = 0.05.

A) H0: p ≤ 0.50; HA: p > 0.50 B)SAMPLE PROPORTION=0.65 In the sample, 13 observations are less than 10 x= 13 n= 20 p̂= 13/20= 0.65 C) TEST STATISTIC= 1.34 z = (p̂ -p)/ √p (1-p)/n = 0.65-0.50/ √0.50(1-0.50)/ 20 =0.15/0.11 =1.34 D) 0.05 ≤ p-value < 0.10 The p-value = P(Z ≥ 1.34)= 0.9099 (from Z-table) =1- 0.9099 = 0.0901 E)Do not reject H0 since the p-value is greater than significance level. P-value= 0.0901 ≥ α = 0.05. F) We cannot conclude that the population proportion is greater than 0.50.

Research shows that many banks are unwittingly training their online customers to take risks with their passwords and other sensitive account information, leaving them more vulnerable to fraud (Yahoo.com, July 23, 2008). Even web-savvy surfers could find themselves the victims of identity theft because they have been conditioned to ignore potential signs about whether the banking site they are visiting is real or a bogus site served up by hackers. Researchers at the University of Michigan found design flaws in 78% of the 214 U.S. financial institution websites they studied. Is the sample evidence sufficient to conclude that more than three out of four financial institutions that offer online banking facilities are prone to fraud? Use a 5% significance level for the test. (You may find it useful to reference the appropriate table: z table or t table) a. Select the null and the alternative hypotheses. b. Calculate the sample proportion. (Round your answer to 2 decimal places.) c. Calculate the value of test statistic. (Round intermediate calculations to at least 4 decimal places and final answer to 2 decimal places.) d. Find the p-value. e. At the 5% level of significance, what is the conclusion?

A) H0: p ≤ 0.75; HA: p > 0.75 The claim is that 3 out of 4 (0.75) financial institutions that offer online banking facilities are prone to fraud (right tail) B) Sample Proportion= 0.78 x=78%*214= 167 n= 214 p̂=167/214= 0.78 C) TEST STATISTIC= z = (p̂ -p)/ √p (1-p)/n = 0.78-0.75/ √0.75(1-0.75)/ 214 = 0.03/0.0296 =1.01 D) p-value ≥ 0.10 p-value = P(Z ≥ 1.01)=0.8438 =1-0.8438 = 0.1562 E) Do not reject H0; the claim is not supported by the data. p-value= 0.1562 > α = 0.05 The sample evidence does not support the claim that more than 75% of financial institutions that offer online banking facilities are prone to fraud.

A television network is deciding whether or not to give its newest television show a spot during prime viewing time at night. For this to happen, it will have to move one of its most viewed shows to another slot. The network conducts a survey asking its viewers which show they would rather watch. The network will keep its current lineup of shows unless the majority of the customers want to watch the new show. The network receives 827 responses, of which 428 indicate that they would like to see the new show in the lineup. (You may find it useful to reference the appropriate table: z table or t table) a. Set up the hypotheses to test if the television network should give its newest television show a spot during prime viewing time at night. b-1. Calculate the value of the test statistic. (Round intermediate calculations to at least 4 decimal places and final answer to 2 decimal places.) b-2. Find the p-value. c. At α = 0.01, what should the television network do?

A)H0: p ≤ 0.50; HA: p > 0.50 x=428 n=827 p̂=428/827=0.5175 B-1) TEST STATISTIC= 0.01 z = (p̂ -p)/ √p (1-p)/n = 0.5175-0.50/ √0.50(1-0.50)/ 827 =0.0175/ 0.0174 =1.01 B-2) p-value ≥ 0.10 =p-value = P(Z ≥ 1.01)=0.8438 =1-0.8438 = 0.1562 C) Do not reject H0, the television network should keep its current lineup. P-value= 0.1562 > α = 0.01 At the 1% significance level, the percentage of individuals who want to watch the new show is not more than 50%. Thus, the television network should keep its current lineup.

In order to conduct a hypothesis test for the population proportion, you sample 300 observations that result in 111 successes. (You may find it useful to reference the appropriate table: z table or t table) H0:p≥ 0.41;HA:p< 0.41. a-1.Calculate the value of the test statistic.(Negative value should be indicated by a minus sign. Round intermediate calculations to at least 4 decimal places and final answer to 2 decimal places.) a-2. Find the p-value. a-3. At the 0.10 significance level, What is the conclusion? a-4. Interpret the results at α = 0.10 H0: p = 0.41; HA: p ≠ 0.41. b-1. Calculate the value of the test statistic. (Negative value should be indicated by a minus sign. Round intermediate calculations to at least 4 decimal places and final answer to 2 decimal places.) b-2. Find the p-value. b-3. At the 0.10 significance level, What is the conclusion? b-4. Interpret the results at α = 0.10

A-1) Given: p̂= 111/300= 0.37 p=0.41 n= 300 z = (p̂ -p)/ √p (1-p)/n = (0.37-0.41)/√0.41(1-0.41)/ 300 = -0.04/ 0.03 = -1.41 TEST STATISTIC: -1.41 A-2) 0.05 p-value < 0.10 =TDIST(1.41,299,1) (USING EXCEL) = 0.0793 A-3) Reject H0 since the p-value is smaller than significance level. p-value = P(Z ≤ −1.41) = 0.0793 < 0.10 = α A-4) We conclude that the population proportion is less than 0.41. B-1) Given: p̂= 111/300= 0.37 p=0.41 n= 300 z = (p̂ -p)/ √p (1-p)/n = (0.37-0.41)/√0.41(1-0.41)/ 300 = -0.04/ 0.03 = -1.41 TEST STATISTIC: -1.41 B-2) p-value 0.10 =TDIST(1.41,299,2) (USING EXCEL) = 0.1586 B-3) Do not reject H0 since the p-value is greater than significance level. p-value = 2P(Z ≤ −1.41) = 0.1586 > 0.10 = α B-4) We cannot conclude that the population proportion differs from 0.41.