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Unformatted text preview: Paul Tucker STAT 280 HW 7
8.02 a. b. c. alternative hypothesis. The statement refers to a range of values (greater than 100), in particular value. null hypothesis. The statement predicts a particular value. alternative hypothesis. The statement refers to a range of values (less than 0.5), instea value. 8.04 Assuming GPA notation has not changed (graded on 4point scale) the two hypotheses would be: H0: x = 2.80 Ha: x 2.80 where x denotes the population mean. 8.08 a. & b. 0.38 does not give strong evidence against the null hypothesis, but 0.001 does. 8.11 H0: p = 0.50 z = 1.04 a. b. c. d. Ha: p > 0.50 Ha: p 0.50 for z = 1.04, pvalue is (1  0.8508) = .1492 for z = 1.04, pvalue is 2(.1492) = .2984 Ha: p < 0.50 for z = 1.04, pvalue is 0.8508 None of the above values give strong evidence against H0, because none o very close to 0. Therefore the null hypothesis is plausible. Ha: p > 0.50 Ha: p 0.50 8.12 H0: p = 0.50 z = 2.50 a. for z = 2.50, pvalue is (1  0.9938) = .0062 for z = 2.50, pvalue is 2(.0062) = .0124 b. Ha: p < 0.50 for z = 2.50, pvalue is .9938 That p is not 0.50, but higher seems to have strong evidence. The pvalue null and greaterthanpnull are very close to 0. 8.20 n = 2232 p = 0.557 a. the variable is the whether one voted for Clinton or not, the parameter is the proportion of those that voted for Clinton. The null hypothesis states that .5 polutation voted for Clinton, and the alternative hypothesis states that at lea Clinton b. The null hypothesis states that .50 of the polutation voted for Clinton, and t hypothesis states that at least .50 voted for Clinton. At least 15 positive res failures were received as responses. If the sample was randomly selected our testing requirement tstat: (0.5570.5) / {[(.5)(.5)/2232]^.5} c. H0: p = 0.50 Ha: p 0.50 d. Ha: p 0.50 pvalue for z = 5.0 (closest zscore in normal table) is 0 multiplied times 2 = e. This pvalue is tiny. I mean way way small. This provides strong evidence reject the null hypothesis. If the pvalue were less than the significance level of .05, then we would ha strong enough evidence to reject the null hypothesis with 95% confidence. is the case here. Ha: 0.50 Ha: > 0.50 Ha: < 0.50 8.26 H0: = 100 t = 2.40 n = 20 a. b. c. the pvalue is on somewhere on the interval .02 < pvalue < the pvalue is on somewhere on the interval .01 < pvalue < the pvalue is on somewhere on the interval .95 < pvalue < 8.27 H0: = 100 t = 1.20 n= 5 8.28 H0: = 0 Ha: > 0 x = 3.0 n = 29 t = 2.21 pvalue = 0.036 for Ha: 0 a. b. larger, the t value of 1.20 is less extreme. It lies somewhere more than .10 because it decreases the standard error overall by making its denominator l a. b. c. for t = 2.21, and Ha: > 0, the pvalue is between 0.025 and .01. this is sm value for Ha: 0 We could interpret this pvalue as strong evidence to reject the null hypothe on the level of significance that we seek. graphically, rejecting the null in favor of Ha: > 0 puts us on the righttail o curve, as far from < 0 as possible. Comparing > 0 to < 0 will yield ev statistical evidence to reject < 0. 8.39 = .05 tstat = 2.58 pval = .01 8.43 a. b. 8.44 a. b. c. d. a. b. If H0 is true, the test was designed for a .05 probability of Type I error. If the test resulted ina decision error, it was a Type I by definition, rejecting the null hypothesis (that no disease is present) when in fact ther is a false positive. We have falsely attributed disease where there is none. we failed to reject the null hypothesis, but therapy would actually have an effect if given a Type I error would be where a patient is incorrectly diagnosed with prostate cancer. then undergo treatment (potentially hazardous) for no reason. a Type II error incorrectly diagnoses a man as not having prostate cancer when he doe an error, a patient would forego needed treatment. Type II The quote describes the conditional on the basis that you tested positive, the Type I co based on your state of disease. The first is P(NoPos), the second P(PosNo). Both ha numerator [P(Pos&No)], but have a different denominator when calculating. 8.49 the expected outcome of 60 significance tests is that .05 will result as "significant," or 3 total. When the m only the 3, they are interpreting normal variation of data as practically significant when it is not. 8.53 H0: p = 0.50 Ha: p > 0.50 n = 100 = .05 a. b. area of rejection is p > {.50 + 1.645([(.5)(.5)/100]) = .58225} c. P(Type II) = {pval of z = (.5882.60)/.048989795 = .36232036}, or .3594 8.56 H0: p = 1/3 Ha: p > 1/3 n = 116 = .05 8.57 H0: p = 1/3 Ha: p 1/3 n = 116 suppose p =.50 a. b. se for p =.35 is ((.35)(.65)/116)^.5 = .044285516 suppose p = .35, P(Type II) = pval of zscore {(.405.35)/.044285516 = 1.2 the closer a parameter is to the null hypothesis, the more their two distribut overlap [the P(Type II)] a. b. se0 = ((.333)(.666)/116)^.5 = .043764 area of rejection is p is outside of .333 1.96(se0) or {.247222, .418778} se for p = .50 is ((.50)(.50)/116)^.5 = .046424 P(p < .247) = pval for the z score of {(.247.50)/.046424 = 5.44979} is less c. P(p > .419) = pval for the z score of {(.419.50)/.046424 = 1.74958} is .959 Type II occurs from {.247222, .418778}, which is .04 of the actual distributio 8.58 H0: p = 0.50 a. Ha: p > 0.50 power = .98 at p = 2/3 b. the power of .98 is the probability of correctly rejecting H0 when it is false. If probability of correct predictions by TT practitioners was 2/3, there was a 9 data such that the significance test performed would reject H0. a Type II error here would occur if this experiment failed to reject the null hy TT practice corresponded with random guessing when in fact it is more acc this tells us that the test has only a .02 chance of reporting a Type II error c. 8.98 a. choose a smaller when the there might be serious side effects of rejectin hypothesis (such as in adopting a new drug). This will ensure that there is v evidence to reject the null (reject the old drug) in favor of the alternative. using a smaller increases the chances of Type II errors b. of values (greater than 100), instead of a of values (less than 0.5), instead of a particular wo hypotheses would be: othesis, but 0.001 does. 8508) = .1492 2) = .2984 nce against H0, because none of the values are is is plausible. 9938) = .0062 2) = .0124 e strong evidence. The pvalues for both notpo 0. nton or not, the parameter is the population he null hypothesis states that .50 of the tive hypothesis states that at least .50 voted for olutation voted for Clinton, and the alternative Clinton. At least 15 positive results and 15 sample was randomly selected, then this meets 570.5) / {[(.5)(.5)/2232]^.5} = 5.36 score in normal table) is iplied times 2 = 0 This provides strong evidence to e level of .05, then we would have ypothesis with 95% confidence. This on the interval .02 < pvalue < .05 on the interval .01 < pvalue < .025 on the interval .95 < pvalue < .98 lies somewhere more than .10 erall by making its denominator larger etween 0.025 and .01. this is smaller than the p idence to reject the null hypothesis, depending > 0 puts us on the righttail of the distribution aring > 0 to < 0 will yield even stronger probability of Type I error. ase is present) when in fact there is no disease where there is none. d actually have an effect if given. iagnosed with prostate cancer. A man would eason. ng prostate cancer when he does. With such ou tested positive, the Type I conditional is the second P(PosNo). Both have the same tor when calculating. gnificant," or 3 total. When the media stresses nificant when it is not. 5)/100]) = .58225} 89795 = .36232036}, or .3594 re {(.405.35)/.044285516 = 1.241941} = .8925 esis, the more their two distribution curves will 96(se0) or {.247222, .418778} .50)/.046424 = 5.44979} is less than .000000287 or about 0 .50)/.046424 = 1.74958} is .9599 ich is .04 of the actual distribution ly rejecting H0 when it is false. If the actual titioners was 2/3, there was a 98% chance of ed would reject H0. riment failed to reject the null hypothesis that ssing when in fact it is more accurate. ce of reporting a Type II error e serious side effects of rejecting the null). This will ensure that there is very strong ug) in favor of the alternative. Type II errors ...
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This homework help was uploaded on 04/22/2008 for the course STAT 280 taught by Professor Thomas during the Spring '08 term at Rice.
 Spring '08
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