Question
number 12
Ifand a sample a e have higher 10s Use the rd dev In October 1945, tbe Ci ed 1487 randomly sam s them iomatou veity eaee thon use them against us? way to protect ourselves soding yes. Did a ig lop a way to proteet itself from atomic theDid a majority of Americans feel the r countries tried to u = (.05 level of significance anufacturer of high-strength, low res that the standard deviation of yield lev 10457 Use the ir strength A h. Sleepy? between ttwe night. In a sur ea 7000 pounds per square inch (psi). The ager selected a sample of 20 steel beams no he mean a their yield s me that yield strengths are normally p control n of theare sleepmg 75 n of yield hevidence suggest that the standard strength exceeds 7000 psi at the =0.01 level 7500 psi the 001 le e 17. Text w he propc Su if they te als A pharmaceutical company manufactures 12 Pharma milligr estire at must not exceed 5 mg. The quality-control manager the res acen am (mg) pain reliever. Company specifications d deviation of the amount of the active that)DPha gdandom sample of 30 tablets from a certain batch and e sample standard deviation is 7.3 me, Assume that s s that the t of the active ingredient is normally distributed. Determine w ative ingredi whether the standard deviation of the amount of the ent is greater than 5 mg at the = 0.05 level of nicance Course Redesign Pass rates for Intermediate Algebra at a the course, faculty of a community college develop a mastery- trough a computer program. The instructor serves as a learning unity college are 52.6%. In an effort to improve pass rates amm 1% d learning model where course content is delivered in a lab nentor for the students. Of the 480 students who enroll in the nastery-based course, 267 pass at is the variable of interest in this study? What type of variable is it? b) At the 001 level of significance, decide whether the sample enden c SCthe mastery-based learning model suggests the
Explanation / Answer
This is a variance test so you don´t need x-bar or mu
H0: 5
H1: > 5
n=30, s=7.3, =0.05
Statistic = (n-1)s^2/5^2 = 29(7.3)^2/25=61.81
Critical value = chi-inv(0.05,29)=42.56 from standard chisquare table
Rejection region: t statst > critical value so
Decision: Since the statistic value is in the rejection region we can reject H0, there is sufficient evidence to say that the standard deviation is greater than 5
