Question

can someone please explain the solutions step by step for 8.7 and 8.8

Problem 8.6 An automobile manufacturer recommends that any purchaser of one of its new cars bring it in to a dealer for a 3000-mile checkup. The company wishes to know whether the true average mileage for initial servicing differs from 3000. A random sample of 50 recent purchasers resulted in a sample average mileage of 3208 and a of 273 miles. Does the data strongly suggest that true average mileage for this checkup is something other than the récommerided value? sample standard deviation Problem 8.7 The following data were obtained on mercury residues on birds' breast mus- cles: Mallard ducks: m 16,= 6.13, s1 2.40 Blue-winged teals: n 17, 6.46, 82 1.73 Construct a 95% confidence interval for the difference between true average mercury residues 1, 2 in these two types of birds in the region of interest. Does your confidence interval indicate that =42 at a 95% confidence level? Problem 8.8 A manufacturer of a certain type of glue claims that his glue can withstand 230 units of pressure. To test this claim, a sample of size 24 is taken. The sample 191.2 units and the sample standard deviation is 21.3 units. (a) Propose a statiśtical model to test this claim and test the manufacturer's claim. (b) What is the highest claim that the manufacturer can make without rejection of this сan mean is C claim? 8.4.2 Exercise Set B Problem 8.9 Suppose that Yi, ..,Y distributed, with common mean u and common variance o2. Recall that the sample variance independent, identically sample, that is, they are are a MacBook Air

Answer 1

8.7

Minitab-Untitled File Edit Data Calc Stat Graph Editor Tools Window Help Assistant O L1M Session Two-Sample t for the Mea Two-Sample t: 0ptions Summarized dats Difference (sample 1 mean)-(sample 2 mean) Sample 2 Sample 1 Confidence level: 95.0 Sample size: 16 17 Hypothesized difference: 0.0 6.13 Sample mean: 6.46 Alternative hypothesis: Difference +hypothesized dfference 1.73 Standard deviation: 2.40 Assume equal variances C2 C3 C4 CS C6 C7 4 21 Cancel Help OH Select Graphs. Options. Helr Cancel OK 2 3 A 6 7 8 Welcome to Minitab, press F1 for help

Minitab Untitled File Edit Data Calc Stat Graph Editor Tools Window Help Assistant 0000 fix Xall T O L M Session Two-Sample T-Test and CI Method Hmean of Sample 1 H:mean of Sample 2 Difference: H-H2 Equal variances are assumed for this analysis. Descriptive Statistics Sample Sample 1 Sample 2 N Mean StDev SE Mean 16 6,13 2.40 0.60 17 6.46 1,73 0.42 Estimation for Difference Pooled 95% CI for Difference StDev Difference 2.081 1.809, 1.149) -0.330 Test Null hypothesis Alternative hypothesis Ho: H0 - Hi: H#0 T-Value P-Value DF -0.46 31 0.652

95% confidence interval is given by (-1.809,1.149)

As p-value > 0.05, confidence interval indicates that at a 95% confidence level.

E7= !7]

8.8

Here we conduct one sample Z test to test this claim.

l Minitab- Untitled File Edit Data Calc Stat Graph Editor Tools Window Help Assistant Xa T O 1M + One-Sample Z Descriptive Statistics One-Sample Z for the Mean 95% CI for u N Mean SE Mean Summarized data 24 191.20 435 (182.68, 199.72) One-Sample Z: Options Sample size: 24 mean of Sample Known standard deviation 21.3 Confidence level: ,0 191.2 Sample mean: Alternative hypothesis: Mean hypothesized mean Test Known standard deviation: 21.3 Help Of Cancel Ha 230 Null hvnnthesis Perform hypothesis test hee Hypotheszed mean: 230 14 C15 C16 C17 C1 CZ C3 CA C5 CE C18 C19 C20 C2 Graphs.. Select Options... 2 Heir Cancel 6 7 8

l Minitab Untitled Window Help File Edit Data Calc Stat Graph Editor Tools Assistant fx EEE 1 M LI Session One-Sample Z Descriptive Statistics 95% CI for u Mean SE Mean 191.20 4.35 24 (182.68, 199.72) Lmean of Sample Known standard devigtion 21.3 Test Null hypothesis Alternative hypothesis H: 230 Ho: u230 Z-Value P-Value 0.000 -8.92 3 4

As the citical value of this test i3 Te To.025 = J.96. )-96 <119).2-μ1 So, 21.3 49.452, 2 3 2 948 The highest claim is that factwrer 232. 948 units thet can make without rajection the manu this elaim