Name:__________________ Per:____________________

Directions: SHOW YOUR WORK. Include the items discussed in class (such as original claim, the null hypothesis, the alternative hypothesis, draw a graph (shade the graph), calculate test statistic, critical values, etc. and write a conclusive statement regarding the null and claim. Identify whether H 0 or H1 is the claim by writing claim next to the correct box. Write RTN (Reject the null) or FRTN (Fail to reject the null) by each test. CV= Critical Value TS= Test Statistic 1. A researcher wants to check the claim that convicted burglars spend an average of 17.7 months in jail. She takes a random sample of 12 such cases from court files and finds that X = 20.2 months and s = 6.8 months. Test the null hypothesis that = 17.7 at the 0.05 significance level. Graph: with , CL, CV’s and TS marked H0 : µ H1 : µ df = n = CV: TS : t Graph:

RTN or FRTN? Conclusion:

2. A manufacturer makes steel rods that are supposed to have a mean length of 40 cm. A retailer suspects that the bars are running short. A sample of 40 bars is taken and their mean length is determined to be X = 41 cm. Test the retailer’s claim that <40 cm at the 1 percent significance level. = 3.5 cm Graph: with , CL, CV’s and TS marked H0 : µ H1 : µ TS: z= n = CV : z Graph:

RTN or FRTN? Conclusion:

3. In a sample of 164 children selected randomly from one town, it is found that 46 of them suffer from asthma. At the 0.05 significance level, test the claim that the proportion of all children in the town who suffer from asthma is 22%. Graph: with , CL, CV’s and TS marked H0 : p H1 : p x n = TS : z CV : z 2

p value =

^

p

Graph:

RTN or FRTN? Conclusion: 4. A researcher claims that the amounts of acetaminophen in a certain brand of cold tablets have a mean different from the 600 mg claimed by the manufacturer. Test this claim at the 0.02 level of significance. The mean acetaminophen content for a random sample of n= 49 tablets is 602.2mg. = 4.1 mg Graph: with , CL, CV’s and TS marked H0 : µ H1 : µ n = X TS : z CV : z 2

p value = Graph:

RTN or FRTN? Conclusion:

α=

5. A manufacturer makes ball bearings that are supposed to have a mean weight of 40 g. A retailer suspects that the mean weight is actually less than 40g. The mean weight for a random sample of 16 ball bearings is 39.5 g with a standard deviation of 4.8 g. At the 0.05 significance level, test the claim that the mean is less than 40g. Graph: with , CL, CV’s and TS marked

H0 : µ n = CV : t

H1 : µ df = TS : t

Graph:

RTN or FRTN? Conclusion:

6. A light-bulb manufacturer advertises that the average life for its light bulbs is 950 hours. A random sample of 15 of its light bulbs resulted in the following lives in hours: 995 590 510 539 739 917 571 555 916 728 664 693 708 887 849 At the .10 significance level, does the data provide evidence that the mean life for the company’s light bulbs differs from the advertised mean? Graph: with , CL, CV’s and TS marked H0 : µ H1 : µ df = n = s= X TS : t CV : t 2

Graph:

RTN or FRTN? Conclusion:

7. According to a recent poll, 51% of Americans would vote for the incumbent president. If a random sample of 100 people results in 45% who would vote for the incumbent, test the claim that the actual percentage is 51%. Use a 0.10 significance level. Graph: with , CL, CV’s and TS marked

H0 : p n = CV : z

H1 : p x TS : z

p value =

α=

Graph:

RTN or FRTN? Conclusion:

8. In test of a computer component, it is found that the mean time between failures is 837 hours. A modification is made which is supposed to increase reliability by increasing the time between failures. Tests on a sample of 36 modified components produce a mean time between failures of 883 hours. At the 0.01 level of significance, test the claim that for the modified components, the mean time between failures is greater than 837 hours. Graph: with , CL, CV’s and TS marked = 41 hours H0 : µ H1 : µ n = X TS :z CV : z α= p value = Graph:

RTN or FRTN? Conclusion: ______________________________________________________ ______________________________________________________________________________ ______________________________________________________________________________

9. A large software company gives job applicants a test of programming ability and the mean for that test has been 150 in the past. 25 job applicants are randomly selected from one large university and they produce a mean score and standard deviation of 173 and 10, respectively. Use a 0.05 level of significance to test the claim that this sample comes from a population with a mean score greater than 150. Graph: with , CL, CV’s and TS marked H0 : µ H1 : µ df = n = s= X TS : t CV : t Graph:

RTN or FRTN? Conclusion

10. A brochure claims that the average maximum height a certain type of plant is 0.6m. A gardener suspects that this estimate is not accurate locally due to soil conditions. A random sample of 42 mature plants is taken. The mean height of the plants in the sample is 0.64 m. Test the claim made in the brochure at the 1 percent level of significance. Graph: with , CL, CV’s and TS marked = .22m

H0 : µ n = CV : z p value =

H1 : µ

X TS : z α=

Graph:

RTN or FRTN? Conclusion:

11. To judge certain safety features of a car, an engineer must know whether the reaction time of drivers to a given emergency situation has a standard deviation of 0.012 second. What can she conclude at the 0.05 level of significance if she gets s = 0.014 for a sample of size n = 25? Graph: with , CL, CV’s and TS marked H0 : H1 : df = n = s= α=

CV : CV :

2 ? 2 ?

= =

TS : 2 =

RTN or FTRN? Conclusion: Graph: 12. The specifications for the mass production of certain springs require that the standard deviation of their compressed lengths should not exceed 0.045 cm. If a random sample of size n = 35 from a certain production lot yields s = 0.058 and the probability of a Type I error is not to exceed 0.01, does this constitute evidence for the null or for the alternative hypothesis? Graph: with , CL, CV’s and TS marked H0 : H1 : df = n = s= α= 2 TS : 2 = CV : =

?

RTN or FTRN? Conclusion:

Graph: