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Recent coffee researchHypothesis Testing

Recent coffee researchCoffee reduces the risk of diabetesHypothesis Testing

Ha: p0.137Subways FOOTLONG is not a foot longSubways FOOTLONG is a foot longH0: m=12Ha:m12Hypothesis TestingCoffee reduces the risk of diabetesHa: p0.137is normally distributed providedOR xs distribution is not heavily skewed and n > 30x is normally distributedOR xs distribution is heavily skewed and n > 50

one mean tests knownHypothesis Testings 2 is some value that only Deity knowsis normally distributed providedx is normally distributed

one mean tests knownHypothesis Testing

What do you do if Deity wont reveal to you the value of s 2?one mean tests unknownOR xs distribution is not heavily skewed and n > 30OR xs distribution is heavily skewed and n > 50The t distribution is the exact distribution if one mean tests unknown

Hypothesis Testingx is normally distributed

What do you do if Deity wont reveal to you the value of s 2?x is normally distributedThe t distribution is the approximate distribution if OR n > 30 and x is NOT heavily skewedOR n > 50 and x is HEAVILY skewedone mean tests unknown

Hypothesis Testing

What do you do if Deity wont reveal to you the value of s 2?

one proportion testis approximately normally distributed if n p0 > 5 and n (1p0) > 5Hypothesis Testing

What do you use for s 2 when you test a proportion?1960s Chips Ahoy cookie TV commercial claimHypothesis Testing

the cookies have 16 chipsH0: = 16Hypothesis TestingThe null hypothesis is assumed to be trueThe sample says the cookies do not have 16 chips when they actually do.

This error is costly because the production line will be shutdown to fix a problem that does not existRejecting a true H0 is a Type I errorthe cookies have 16 chipsH0: = 16Hypothesis TestingRejecting a true H0 is a Type I errorRejecting a true Ha is a Type II errorThe alternative hypothesis is the oppositeThe sample says the cookies have 16 chipswhen they really do not.

The error will upset Chips Ahoys customers if there are too fewOR increase Chips Ahoys costs if there are too manythe cookies do not have 16 chipsHa: < > 16The alternative hypothesis is the oppositeRejecting a true H0 is a Type I erroris rejected?Hypothesis TestingRejecting a true Ha is a Type II errorWhat conclusion is appropriate when H0the cookies have 16 chipsH0: = 16the cookies do not have 16 chipsHa: < > 16The alternative hypothesis is the oppositecannot be rejected?Hypothesis TestingRejecting a true H0 is a Type I errorRejecting a true Ha is a Type II errorWhat conclusion is appropriate when H0the cookies do not have 16 chipsHa: < > 16the cookies have 16 chipsH0: = 16We cannot conclude that Example: Chips Ahoy Chocolate Chip CookiesPerform a hypothesis test, at the 5% level of significance, to determine if Chips Ahoy cookies have an average of 16 chips per cookie.Hypothesis Testingmean tests 2 known

one mean tests unknowns 2 = p0(1 p0)mean testproportion testBottleNumber of Chipsdeviation from mean114-2.5215-1.5315-1.54170.55181.5616-0.5715-1.529192.530170.5Total4950one mean tests unknown

Hypothesis Testing

1. Determine the hypotheses.Ha: m < > 16

2. Compute the test statisticone mean tests unknownHypothesis TestingH0: m = 16 degrees of freedom.200.100.050.025.010.00528.8551.3131.7012.0482.4672.76329.8541.3111.6992.0452.4622.75630.8541.3101.6972.0422.4572.75031.8531.3091.6962.0402.4532.74432.8531.3091.6942.0372.4492.73833.8531.3081.6922.0352.4452.73334.8521.3071.6912.0322.4412.728df = 30 1 = 29a = .050 a/2 = .025Ha: m < > 16 one mean tests unknown3. Determine the critical value(s).Hypothesis Testingdegrees of freedom.200.100.050.025.010.00528.8551.3131.7012.0482.4672.76329.8541.3111.6992.0452.4622.75630.8541.3101.6972.0422.4572.75031.8531.3091.6962.0402.4532.74432.8531.3091.6942.0372.4492.73833.8531.3081.6922.0352.4452.73334.8521.3071.6912.0322.4412.728t.0250 = 2.045-t.0250 = -2.045one mean tests unknown3. Determine the critical value(s).Hypothesis Testinga = .050 Ha: m < > 16 -2.045 2.045.025 1.91t-stat0t.025 Do Not RejectH0: =164. Concludethe cookies do not have 16 chipsWe cannot conclude that one mean tests unknownHypothesis TestingExample: National Safety Council (NSC)The National Safety Council claimed that more than 50% of the accidents are caused by drunk driving. A sample of 120 accidents showed that 67 were caused by drunk driving. Perform a hypothesis test, at the 2.5% level of significance, to determine if NSCs claim is valid.

one proportion testHypothesis Testing

one proportion test2. Compute the test statisticHypothesis Testing

1. Determine the hypotheses.Z.00.01.02.03.04.05.06.07.08.09-2.2.0139.0136.0132.0129.0125.0122.0119.0116.0113.0110-2.1.0179.0174.0170.0166.0162.0158.0154.0150.0146.0143-2.0.0228.0222.0217.0212.0207.0202.0197.0192.0188.0183-1.9.0287.0281.0274.0268.0262.0256.0250.0244.0239.0233-1.8.0359.0351.0344.0336.0329.0322.0314.0307.0301.0294-1.7.0446.0436.0427.0418.0409.0401.0392.0384.0375.0367Ha: p > .5a = .0250a / 1 = .0250one proportion test3. Determine the critical value(s).Hypothesis Testing-z.0250 -1.96z.0250 1.96 1.961.28 z-stat0z.025 H0: p < .5more than 50% of accidents are caused by drunk drivingWe cannot conclude thatone proportion test4. ConcludeHypothesis TestingDo Not Reject