`MATH 214 (NOTES)Math 214 Al Nosedal Department of Mathematics Indiana University of PennsylvaniaMATH 214 (NOTES) ­ p. 1/6&quot;Pepsi&quot; problemA market research consultant hired by the Pepsi-Cola Co. is interested in determining the proportion of IUP students who favor Pepsi-Cola over Coke Classic. A random sample of 100 students shows that 40 students favor Pepsi over Coke. Use this information to construct a 95% confidence interval for the proportion of all students in this market who prefer Pepsi.MATH 214 (NOTES) ­ p. 2/6Bernoulli Distribution1 i-th person prefers Pepsi 0 i-th person prefers Cokexi =Let p be our estimate of p. Note that p = n xi = x. If n is ^ ^ ¯ &quot;large&quot;, by the Central Limit Theorem, we know that:  x is roughly N (µ, n ), that is, ¯p is roughly N ^ p,p(1-p) nµ = E(xi ) = p  2 = V (xi ) = p(1 - p)n i=1MATH 214 (NOTES) ­ p. 3/6Interval Estimate of pDraw a simple random sample of size n from a population with unknown proportion p of successes. An (approximate) confidence interval for p is:p ± z ^ p(1 - p) ^ ^ nwhere z is a number coming from the Standard Normal that depends on the confidence level required. Use this interval only when: 1) n is &quot;large&quot; and 2) n^  5 and n(1 - p) &gt; 5. p ^MATH 214 (NOTES) ­ p. 4/6Problem 31 (page 322)A simple random sample of 400 individuals provides 100 Yes responses. a. What is the point estimate of the proportion of the population that would provide Yes responses? b. What is the point estimate of the standard error of the proportion, p ? ^ c. Compute the 95% confidence interval for the population proportion.MATH 214 (NOTES) ­ p. 5/6Solutiona. p = ^100 400= 0.25(^)(1-^) p p nb. Standard error of p = ^ c. p ± z ^(^)(1-^) p p n=(0.25)(0.75) 400= 0.02160.25 ± 1.96(0.0216) (0.2076, 0.2923)MATH 214 (NOTES) ­ p. 6/6Problem 32 (page 323)A simple random sample of 800 elements generates a sample proportion p = 0.70. ^ a. Provide a 90% confidence interval for the population proportion. b. Provide a 95% confidence interval for the population proportion.MATH 214 (NOTES) ­ p. 7/6Solutiona. p ± z ^0.70 ± 1.65(^)(1-^) p p n (0.70)(1-0.70) 8000.70 ± 1.65(0.0162) (0.6732, 0.7267) b. 0.70 ± 1.96(0.0162) (0.6682, 0.7317)MATH 214 (NOTES) ­ p. 8/6Problem 35 (page 323)A survey of 611 office workers investigated telephone answering practices, including how often each office worker was able to answer incoming telephone calls and how often incoming telephone calls went directly to voice mail. A total of 281 office workers indicated that they never need voice mail and are able to take every telephone call. a. What is the point estimate of the proportion of the population of office workers who are able to take every telephone call? b. At 90% confidence, what is the margin of error? c. What is the 90% confidence interval for the proportion of the population of office workers who are able to take every telephone call?MATH 214 (NOTES) ­ p. 9/6Solutiona. p = 281 = 0.46 ^ 611 b. Margin of error =z(^)(1-^) p p n= 1.65(^)(1-^) p p n(0.46)(0.54) 611= 1.65(0.0201) = 0.0332c. p ± z ^0.46 ± .0332 (0.4268, 0.4932)MATH 214 (NOTES) ­ p. 10/6Problem 33 (page 323)In a survey, the planning value for the population proportion is p = 0.35. How large a sample should be taken to provide a 95% confidence interval with a margin of error of 0.05? Solution.n=(Always round up).z 2  p (1 - p ) E=1.96 2 (0.35)(1 - 0.35) 0.05= 350MATH 214 (NOTES) ­ p. 11/6Determining the Sample SizeSample Size for an Interval Estimate of a Population Proportion.z n= E2p (1 - p )In practice, the planning value p can be chosen by one of the following procedures. 1. Use the sample proportion from a previous sample of the same or similar units. 2. Use a planning value of p = 0.5.MATH 214 (NOTES) ­ p. 12/6Problem 34 (page 323)At 95% confidence, how large a sample should be taken to obtain a margin of error of 0.03 for the estimation of a population proportion? Assume that past data are not available for developing a planning value for p . Solution.n=(Always round up).z 2  p (1 - p ) E=1.96 2 (0.5)(1 - 0.5) 0.03= 1068MATH 214 (NOTES) ­ p. 13/6Problem 39 (page 323)The percentage of people not covered by health care insurance in 2003 was 15.6%. A congressional committee has been charged with conducting a sample survey to obtain more current information. a. What sample size would you recommend if the committee's goal is to estimate the current proportion of individuals without health care insurance with a margin of error of 0.03? Use a 95% confidence level. b. Repeat part a) using a 99% confidence level.MATH 214 (NOTES) ­ p. 14/6Solutiona. n =z 2  p (1 - p ) E=1.96 2 (0.156)(1 - 0.156) 0.03= 563b. n =z 2  p (1 - p ) E=2.58 2 (0.156)(1 - 0.156) 0.03= 974MATH 214 (NOTES) ­ p. 15/6CHAPTER 9 HYPOTHESIS TESTSMATH 214 (NOTES) ­ p. 16/6Do you want to become a millionaire?Let's say that one of you is invited to this popular show. As you probably know, you have to answer a series of multiple choice questions and there are four possible answers to each question. Perhaps you also have seen that if you don't know the answer to a question you could either &quot;jump the question&quot; or you could &quot;ask the audience&quot;. Suppose that you run into a question for which you don't know the answer with certainty and you decide to &quot;ask the audience&quot;. Let's say that you initially believe that the right answer is A. Then you ask the audience and only 2% of the audience shares your opinion. What would you do? Change your initial belief or reject it?MATH 214 (NOTES) ­ p. 17/6Sweetening colasDiet colas use artificial sweeteners to avoid sugar. These sweeteners gradually lose their sweetness over time. Manufacturers therefore test new colas for loss of sweetness before marketing them. Trained tasters sip the cola along with drinks of standard sweetness and score the cola on a &quot;sweetness score &quot; of 1 to 10. The cola is then stored for a month at high temperature to imitate the effect of four months' storage at room temperature. Each taster scores the cola again after storage. This is a matched pairs experiment. Our data are the differences (score before storage minus score after storage) in the tasters' scores. The bigger these differences, the bigger the loss of sweetness.MATH 214 (NOTES) ­ p. 18/6Sweetening colas (cont.)Suppose we know that for any cola, the sweetness loss scores vary from taster to taster according to a Normal distribution with standard deviation  = 1. The mean µ for all tasters measures loss of sweetness, and is different for different colas. The following are the sweetness losses for a new cola, as measured by 10 trained tasters: 2.0 0.4 0.7 2.0 -0.4 2.2 -1.3 1.2 1.1 2.3. Are these data good evidence that the cola lost sweetness in storage?MATH 214 (NOTES) ­ p. 19/6Solutionµ= mean sweetness loss for the population of all tasters. 1. State hypotheses. H0 : µ = 0 vs Ha : µ &gt; 0 x-µ0 ¯  2. Test statistic. z = /n = 1.02-0 = 3.23 1/ 103. P-value. P (Z &gt; z ) = P (Z &gt; 3.23) = 0.0006 4. Conclusion. We would very rarely observe a sample sweetness loss as large as 1.02 if H0 were true. The small P-value provides strong evidence against H0 and in favor of the alternative Ha : µ &gt; 0, i.e., it gives good evidence that the mean sweetness loss is not 0, but positive.MATH 214 (NOTES) ­ p. 20/6Executives' blood pressuresThe National Center for Health Statistics reports that the systolic blood pressure for males 35 to 44 years of age has mean 128 and standard deviation 15. The medical director of a large company looks at the medical records of 72 executives in this age group and finds that the mean systolic blood pressure in this sample is x = 126.07. Is this evidence ¯ that the company's executives have a different mean blood pressure from the general population? Suppose we know that executives' blood pressures follow a Normal distribution with standard deviation  = 15.MATH 214 (NOTES) ­ p. 21/6Solutionµ= mean of the executive population. 1. State hypotheses. H0 : µ = 128 vs Ha : µ = 128 x-µ0 ¯  2. Test statistic. z = /n = 126.07-128 = -1.09 15/ 72 2(1 - 0.8621) = 0.27583. P-value. 2P (Z &gt; |z |) = 2P (Z &gt; | - 1.09|) = 2P (Z &gt; 1.09) = 4. Conclusion. More than 27% of the time, a simple random sample of size 72 from the general male population would have a mean blood pressure at least as far from 128 as that of the executive sample. The observed x = 126.07 is ¯ therefore not good evidence that executives differ from other men.MATH 214 (NOTES) ­ p. 22/6Tests for a population meanThere are four steps in carrying out a significance test: 1. State the hypotheses. 2. Calculate the test statistic. 3. Find the P-value. 4. State your conclusion in the context of your specific setting. Once you have stated your hypotheses and identified the proper test, you or your calculator can do Steps 2 and 3 by following a recipe. Here is the recipe for the test we have used in our examples.MATH 214 (NOTES) ­ p. 23/6Z test for a population mean µDraw a simple random sample of size n from a Normal population that has unknown mean µ and known standard deviation  . To test the null hypothesis that µ has a specified value, H0 : µ = µ0 calculate the one-sample z statisticz = x - µ0 ¯  nIn terms of a variable Z having the standard Normal distribution, the P-value for a test of H0 against Ha : µ &gt; µ0 is P (Z &gt; z ) Ha : µ &lt; µ0 is P (Z &lt; z ) Ha : µ = µ0 is 2P (Z &gt; |z |)MATH 214 (NOTES) ­ p. 24/6Problem 9 (page 356)Consider the following hypothesis test:H0 : µ = 20 Ha : µ &lt; 20A sample of 50 provided a sample mean of 19.4. The population standard deviation is 2. a. Compute the value of the test statistic. b. What is the p-value? c. Using  = 0.05, what is your conclusion?MATH 214 (NOTES) ­ p. 25/6Solutiona. Test statistic.z =x-µ0 ¯  / n=19.4-20  2/ 50b. P-value.= -2.1213c. Conclusion. Since P-value = 0.0169 &lt;  = 0.05, we reject H0 : µ = 20. We conclude that µ &lt; 20.P (Z &lt; z ) = P (Z &lt; -2.1213) = 0.0169MATH 214 (NOTES) ­ p. 26/6Problem 10 (page 357)Consider the following hypothesis test:H0 : µ = 25 Ha : µ &gt; 25A sample of 40 provided a sample mean of 26.4. The population standard deviation is 6. a. Compute the value of the test statistic. b. What is the p-value? c. Using  = 0.01, what is your conclusion?MATH 214 (NOTES) ­ p. 27/6Solutiona. Test statistic.z =x-µ0 ¯  / n=26.4-25  6/ 40= 1.4757b. P-value. c. Conclusion. Since P-value = 0.0700 &gt;  = 0.01, we CAN'T reject H0 : µ = 25. We conclude that we don't have enough evidence to claim that µ &gt; 25. (Some of us would say that we accept that µ = 25).P (Z &gt; z ) = P (Z &gt; 1.4757) = 0.0700MATH 214 (NOTES) ­ p. 28/6Problem 11 (page 357)Consider the following hypothesis test:H0 : µ = 15 Ha : µ = 15A sample of 50 provided a sample mean of 14.15. The population standard deviation is 3. a. Compute the value of the test statistic. b. What is the p-value? c. Using  = 0.05, what is your conclusion?MATH 214 (NOTES) ­ p. 29/6Problem 11 (Solution)a. Test statistic.z =x-µ0 ¯  / n=14.15-15  3/ 50b. P-value.= -2.0034c. Conclusion. Since P-value = 0.0451 &lt;  = 0.05, we reject H0 : µ = 15. We conclude that µ = 15.2P (Z &gt; |z |) = 2P (Z &gt; | - 2.0034|) = 2P (Z &gt; 2.0034) = 0.0451MATH 214 (NOTES) ­ p. 30/6Tests from confidence intervalsCONFIDENCE INTERVALS AND TWO-SIDED TESTS. A level  two-sided significance test rejects a hypothesis H0 : µ = µ0 exactly when the value µ0 falls outside a level 1 -  confidence interval for µ.MATH 214 (NOTES) ­ p. 31/6Problem 11 (again)The 95% confidence interval for µ in problem 11 is: x ± z (  ) ¯ n 3 14.15 ± 1.96(  ) 50 (13.3184, 14.9815)The hypothesized value µ0 = 15 in problem 11 falls outside this confidence interval, so we reject H0 : µ = 15.MATH 214 (NOTES) ­ p. 32/6Problem 23 (page 363)H0 : µ = 12 Ha : µ &gt; 12A sample of 25 provided a sample mean x = 14 and a ¯ sample standard deviation s = 4.32. a. Compute the value of the test statistic. b. Use the t distribution table to compute a range for the p-value. c. At  = 0.05, what is your conclusion?MATH 214 (NOTES) ­ p. 33/6Solutiona. t =x-µ0 ¯  s/ n=14-12  4.32/ 25= 2.31b. Degrees of freedom = n - 1 = 24. P-value = P (T &gt; t ) = P (T &gt; 2.31) Using t-table (on page 920), P-value is between 0.01 and 0.025. Exact P-value = 0.0147 (using a TI-83 and the function T-Test). c. Since P-value &lt;  = 0.05, we reject H0 .MATH 214 (NOTES) ­ p. 34/6Problem 24 (page 363)H0 : µ = 18 Ha : µ = 18A sample of 48 provided a sample mean x = 17 and a ¯ sample standard deviation s = 4.5. a. Compute the value of the test statistic. b. Use the t distribution table to compute a range for the p-value. c. At  = 0.05, what is your conclusion?MATH 214 (NOTES) ­ p. 35/6Solutiona. t =x-µ0 ¯  s/ n=17-18  4.5/ 48b. Degrees of freedom = n - 1 = 47. P-value = 2P (T &gt; |t |) = 2P (T &gt; | - 1.54|) = 2P (T &gt; 1.54) Using t-table (on page 920), P-value is between 0.10 and 0.20. Exact P-value = 0.1303 (using a TI-83 and the function T-Test). c. Since P-value &gt;  = 0.05, we CAN'T reject H0 .= -1.54MATH 214 (NOTES) ­ p. 36/6Another approachRecall that  = 0.05 and that we have a two-sided alternative (Ha : µ = 18) which means that we could solve this problem using a 95% confidence interval instead of the four-step procedure presented above.s x ± t ( n ) ^ 4.5 17 ± 2.012( 48 )(15.6931, 18.3068)The hypothesized value µ0 = 18 in problem 18 falls INSIDE this confidence interval, so we CAN'T reject H0 : µ = 18.MATH 214 (NOTES) ­ p. 37/6Problem 25 (page 363)H0 : µ = 45 Ha : µ &lt; 45A sample of 36 is used. Identify the p-value and state your conclusion for each of the following sample results. Use  = 0.01. a. x = 44 and s = 5.2. ¯ b. x = 43 and s = 4.6. ¯ c. x = 46 and s = 5.0. ¯MATH 214 (NOTES) ­ p. 38/6Solution aa. x = 44 and s = 5.2. ¯ P-value = P (T &lt; t ) = P (T &lt; -1.1538) = P (T &gt; 1.1538) (T distributions are symmetric) Using t-table (on page 921), P-value is between 0.10 and 0.20. Exact P-value = 0.1281 (using a TI-83 and the function T-Test). Since P-value &gt;  = 0.01, we CAN'T reject H0 .MATH 214 (NOTES) ­ p. 39/6Solution bb. x = 43 and s = 4.6. ¯ P-value = P (T &lt; t ) = P (T &lt; -2.6086) = P (T &gt; 2.6086) (T distributions are symmetric) Using t-table (on page 921), P-value is between 0.005 and 0.01. Exact P-value = 0.0066 (using a TI-83 and the function T-Test). Since P-value &lt;  = 0.01, we reject H0 . We conclude that µ &lt; 45.MATH 214 (NOTES) ­ p. 40/6Solution cc. x = 46 and s = 5.0. ¯ P-value = P (T &lt; t ) = P (T &lt; 1.2) = 1 - P (T &gt; 1.2) (T distributions are symmetric) Using t-table (on page 921), P-value is between 0.80 and 0.90. Exact P-value = 0.8809 (using a TI-83 and the function T-Test). Since P-value &gt;  = 0.01, we can't reject H0 .MATH 214 (NOTES) ­ p. 41/6Problem 27The Employment and Training Administration reported the U.S. mean unemployment insurance benefit of \$ 238 per week. A researcher in the state of Virginia anticipated that sample data would show evidence that the mean weekly unemployment insurance benefit in Virginia was below the national level. a. Develop appropriate hypotheses such that rejection of H0 will support the researcher's contention. b. For a sample of 100 individuals, the sample mean weekly unemployment insurance benefit was \$231 with a sample standard deviation of \$80. What is the p-value? c. At  = 0.05, what is your conclusion?.MATH 214 (NOTES) ­ p. 42/6Solutiona. H0 : µ = 238 vs Ha : µ &lt; 238. x-µ0 ¯ 231-238 b. t = s/n = 80/100 = -0.88Degrees of freedom = n - 1 = 99. Using t table (page 922), P-value is between 0.10 and 0.20 c. P-value &gt; 0.05, we CAN'T reject H0 . Cannot conclude mean weekly benefit in Virginia is less than the national mean.MATH 214 (NOTES) ­ p. 43/6Problem 28 (page 364)The National Association of Professional Baseball Leagues, Inc., reported that attendance for 176 minor league baseball teams reached an all-time high during the 2001 season. On a per-game basis, the mean attendance for minor league baseball was 3530 people per game. Midway through the 2002 season, the president of the association asked for an attendance report that would hopefully show that the mean attendance for 2002 was exceeding the 2001 level.MATH 214 (NOTES) ­ p. 44/6Problem 28 (cont.)a. Formulate hypotheses that could be used to determine whether the mean attendance per game in 2002 was greater than the previous year's level. b. Assume that a sample of 92 minor league baseball games played during the first half of the 2002 season showed a mean attendance of 3740 people per game with a sample standard deviation of 810. What is the p-value? c. At  = 0.01, what is your conclusion?MATH 214 (NOTES) ­ p. 45/6Solutiona.H0 : µ = 3530 vs Ha : µ &gt; 3530 x-µ0 ¯  b. t = s/n = 3740-3530 = 2.49 810/ 92 Degrees of freedom = n - 1 = 91. Using t-table (on page 922), P-value is between 0.005 and 0.01. Exact P-value = 0.007 (using a TI-83 and the function T-Test). c. Since P-value &lt;  = 0.01, we reject H0 . We conclude that the mean attendance per game has increased.MATH 214 (NOTES) ­ p. 46/6Hypotheses Tests for a ProportionTo test the hypothesis H0 : p = p0 , compute the z statistic,p0 (1-p0 ) nz =p-p0 ^In terms of a variable Z having the standard Normal distribution, the approximate P-value for a test of H0 against Ha : p &gt; p0 : is : P (Z &gt; z ) Ha : p &lt; p0 : is : P (Z &lt; z ) Ha : p = p0 : is : 2P (Z &gt; |z |) Use this test when the sample size n is so large that both np0 and n(1 - p0 ) are 10 or more.MATH 214 (NOTES) ­ p. 47/6Problem 36 (page 368)Consider the following hypothesis test:H0 : p = 0.75 Ha : p &lt; 0.75A sample of 300 items was selected. Compute the p-value and state your conclusion for each of the following sample results. Use  = 0.05. a. p = 0.68 ^ b. p = 0.72 ^ c. p = 0.70 ^ d. p = 0.77 ^MATH 214 (NOTES) ­ p. 48/6Solution az = p-p0 ^ p0 (1-p0 )/n=0.68-0.75 0.75(1-0.75)/300= -2.80Using Normal table, P-value =P (Z &lt; z ) = P (Z &lt; -2.80) = 0.0026 P-value&lt; = 0.05, reject H0 .MATH 214 (NOTES) ­ p. 49/6Solution bz = p-p0 ^ p0 (1-p0 )/n=0.72-0.75 0.75(1-0.75)/300= -1.20Using Normal table, P-value =P (Z &lt; z ) = P (Z &lt; -1.20) = 0.1151 P-value&gt; = 0.05, do not reject H0 .MATH 214 (NOTES) ­ p. 50/6Solution cz = p-p0 ^ p0 (1-p0 )/n=0.70-0.75 0.75(1-0.75)/300= -2.00Using Normal table, P-value =P (Z &lt; z ) = P (Z &lt; -2.00) = 0.0228 P-value&lt; = 0.05, reject H0 .MATH 214 (NOTES) ­ p. 51/6Solution dz = p-p0 ^ p0 (1-p0 )/n=0.77-0.75 0.75(1-0.75)/300= 0.80Using Normal table, P-value =P (Z &lt; z ) = P (Z &lt; 0.80) = 0.7881 P-value&gt; = 0.05, do not reject H0 .MATH 214 (NOTES) ­ p. 52/6Problem 35 (page 368)Consider the following hypothesis test:H0 : p = 0.20 Ha : p = 0.20A sample of 400 provided a sample proportion p = 0.175. ^ a. Compute the value of the test statistic. b. What is the p-value? c. At the  = 0.05, what is your conclusion? d. What is the rejection rule using the critical value? What is your conclusion?MATH 214 (NOTES) ­ p. 53/6Solutiona. z =p0 (1-p0 ) np-p0 ^=0.175-0.20(0.20)(0.80) 400= -1.25b. Using Normal table, P-value = 2P (Z &gt; |z |) = 2P (Z &gt;| - 1.25|) = 2P (Z &gt; 1.25) = 2(0.1056) = 0.2112 c. P-value &gt;  = 0.05, we CAN'T reject H0 .MATH 214 (NOTES) ­ p. 54/6Problem 37 (page 369)A study found that, in 2005, 12.5% of U.S. workers belonged to unions. Suppose a sample of 400 U.S. workers is collected in 2006 to determine whether union efforts to organize have increased union membership. a. Formulate the hypotheses that can be used to determine whether union membership increased in 2006. b. If the sample results show that 52 of the workers belonged to unions, what is the p-value for your hypothesis test? c. At  = 0.05, what is your conclusion?MATH 214 (NOTES) ­ p. 55/6Solutiona. H0 : p = 0.125 vs Ha : p &gt; 0.125 52 b. p = 400 = 0.13 ^z =p0 (1-p0 ) np-p0 ^=0.13-0.125(0.125)(0.875) 400= 0.30Using Normal table, P-value =P (Z &gt; z ) = P (Z &gt; 0.30) = 1 - 0.6179 = 0.3821 c. P-value = &gt; 0.05, do not reject H0 . We cannot concludethat there has been an increase in union membership.MATH 214 (NOTES) ­ p. 56/6Problem 38 (page 369)A study by Consumer Reports showed that 64% of supermarket shoppers believe supermarket brands to be as good as national name brands. To investigate whether this result applies to its own product, the manufacturer of a national name-brand ketchup asked a sample of shoppers whether they believed that supermarket ketchup was as good as the national brand ketchup.MATH 214 (NOTES) ­ p. 57/6Problem 38 (cont.)a. Formulate the hypotheses that could be used to determine whether the percentage of supermarket shoppers who believe that the supermarket ketchup was as good as the national brand ketchup differed from 64%. b. If a sample of 100 shoppers showed 52 stating that the supermarket brand was as good as the national brand, what is the p-value? c. At  = 0.05, what is your conclusion?MATH 214 (NOTES) ­ p. 58/6Solutiona. H0 : p = 0.64 vs Ha : p = 0.64 52 b. p = 100 = 0.52 ^z =p0 (1-p0 ) np-p0 ^=0.52-0.64(0.64)(0.36) 100= -2.50| - 2.50|) = 2P (Z &gt; 2.50) = 2(0.0062) = 0.0124 c. P-value = &lt; 0.05, reject H0 . Proportion differs from theUsing Normal table, P-value = 2P (Z &gt; |z |) = 2P (Z &gt;reported 0.64.MATH 214 (NOTES) ­ p. 59/6Problem 39 (page 369)The National Center for Health Statistics released a report that stated 70% of adults do not exercise regularly. A researcher decided to conduct a study to see whether the claim made by the National Center for Health Statistics differed on a state-by-state basis. a. State the null and alternative hypotheses assuming the intent of the researcher is to identify states that differ from 70% reported by the National Center for Health Statistics. b. At  = 0.05, what is the research conclusion for the following states: Wisconsin: 252 of 350 adults did not exercise regularly. California: 189 of 300 adults did not exercise regularly.MATH 214 (NOTES) ­ p. 60/6Solutiona. H0 : p = 0.70 vs Ha : p = 0.70 b. Wisconsin p = 350 = 0.72 ^ 252z =p0 (1-p0 ) np-p0 ^=0.72-0.70(0.70)(0.30) 350= 0.822P (Z &gt; 0.82) = 2(0.2061) = 0.4122 c. P-value = &gt; 0.05, do not reject H0 .Using Normal table, P-value = 2P (Z &gt; |z |) = 2P (Z &gt; |0.82|) =MATH 214 (NOTES) ­ p. 61/6TO BE CONTINUED...MATH 214 (NOTES) ­ p. 62/6`

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