Question

7.2.1 Escobar et al. (A-3) performed a study to validate a translated version of the Western Ontario and McMaster Universities Osteoarthritis Index (WOMAC) questionnaire used with Spanish-speaking patients with hip or knee osteoarthritis. For the 76 women classified with severe hip pain, the WOMAC mean function score (on a scale from 0 to 100 with a higher number indicating less function) was 70.7 with a standard deviation of 14.6. We wish to know if we may conclude that the mean function score for a population of similar women subjects with severe hip pain is less than 75. Let α = .01.

7.2.6 Nine laboratory animals were infected with a certain bacterium and then immunosuppressed. The mean number of organisms later recovered from tissue specimens was 6.5 (coded data) with a standard deviation of .6. Can one conclude from these data that the population mean is greater than 6? Let α = .05. What assumptions are necessary?

7.2.13 Can we conclude that the mean maximum voluntary ventilation value for apparently healthy college seniors is not 110 liters per minute? A sample of 20 yielded the following values:

7.2.14 Suppose it is known that the IQ scores of a certain population of adults are approximately normally distributed with a standard deviation of 15. A simple random sample of 25 adults drawn from this population had a mean IQ score of 105. On the basis of these data can we conclude that the mean IQ score for the population is not 100? Let the probability of committing a type I error be .05.

Answer 1

Answer:

7.2.1

given - To test = l = 75 es 75 vs L u = 70.7 6 14.6 n = 96, test statistic 2cal = x u N Toil) acol 2017 - 75 14.6 fto zod a - 2,5675 etable value = 2.82 C l % L.o.s) decision role: ow reseat we may Ho if 121 >z Idle accept Ho. decision conclusion: here 2.5695 > 2.32 we reject Ho. mean function score for poru of similar women subjects with severe hip Pain is less than ! 15

7.2.6

We need to assume that the population follows normal distribution Ho: mu=6 Ha: mu>6 The test statistic is t=(xbar-mu)/(s/vn) = (6.5-6)/(0.6/3) =2.5 Given a=0.05, the critical value is t(0.05, df=n-1=8)=1.86 (from student t table) Since t=2.5 is larger than 1.86, we reject Ho. So, we can conclude that the population mean is greater than 6

7.2.13

Here as we are testing whether the mean maximum voluntary ventilation value for apparently healthy college seniors is not 110 litters per minute, therefore the null and the alternate hypothesis here would be given as: Now for n-1 = 19 degrees of freedom and 2 tailed tests, the p-value here is computed as: (from the t-distribution tables as) p= 2P (t19 > 0.1271) = 2*0.4501 = 0.9002 As the p-value here is very high = 0.9002 >> 0.01 which is the level of significance, therefore the test is not significant and we cannot reject the null hypothesis here. Therefore, we don't have sufficient evidence here to conclude that the mean is not equal to 110.

7.2.14

Null hypothesis: Hu= 100 Alternative hypothesis: HL:N # 100 Where His population mean of IQ score Here population standard deviation ( 0) is given, so we can used one sample z test Let's write the given information. Sample size=n=25 sample mean = T=105 level of significance (Type I error) = a=5% Let's use Minitab The command is Stat>>>Basic Statistics >>1 sample Z... Click on summarized data Sample size = 25

Mean = 105 Standard deviation = 15 then click on "Perform hypothesis test Hypothesized mean = 100 (from null hypothesis) then click on Option Confidence level = 100 - a%=100 - 5 = 95.0 Alternative " not equal" then click on OK again, click on OK We get the following output One-Sample z Test of mu = 100 vs not = 100 The assumed standard deviation = 15