Question

(CO7) A state Department of Transportation claims that the mean wait time for various services at...

(CO7) A state Department of Transportation claims that the mean wait time for various services at its different location is less than 6 minutes. A random sample of 16 services at different locations has a mean wait time of 9.5 minutes and a standard deviation of 7.3 minutes. At α=0.05, can the department’s claim be supported assuming the population is normally distributed?

No, since p of 0.037 is less than 0.05, reject the null. Claim is null, so is not supported

Yes, since p of 0.037 is less than 0.07, reject the null. Claim is alternative, so is supported

Yes, since p of 0.037 is less than 0.05, fail to reject the null. Claim is null, so is supported

No, since p of 0.037 is less than 0.05, fail to reject the null. Claim is alternative, so is not supported

(CO7) A used car dealer says that the mean price of a three-year-old sport utility vehicle in good condition is $18,000. A random sample of 20 such vehicles has a mean price of $18,450 and a standard deviation of $1050. At α=0.08, can the dealer’s claim be supported assuming the population is normally distributed?

No, since the test statistic of 1.92 is close to the critical value of 1.24, the null is not rejected. The claim is the null, so is supported

No, since the test statistic of 1.92 is in the rejection region defined by the critical value of 1.85, the null is rejected. The claim is the null, so is not supported

Yes, since the test statistic of 1.92 is in the rejection region defined by the critical value of 1.46, the null is rejected. The claim is the null, so is supported

Yes, since the test statistic of 1.92 is not in the rejection region defined by the critical value of 1.85, the null is not rejected. The claim is the null, so is supported

Homework Answers

Answer #1

Ans:

1)

(claim)

test statistic:

t=(9.5-6)/(7.3/SQRT(16))

t=1.918

df=16-1=15

p-value=tdist(1.918,15,1)=0.037

As,p-value<0.05,reject the null hypothesis.

Yes, since p of 0.037 is less than 0.05, reject the null. Claim is alternative, so is supported

2)

test statistic:

t=(18450-18000)/(1050/SQRT(20))

t=1.92

df=20-1=19

critical t value(two tailed)=tinv(0.08,19)=+/-1.85

Reject H0,if t<-1.85 or t>1.85

As,test statistic falls in rejection region,we reject the null hypothesis.

No, since the test statistic of 1.92 is in the rejection region defined by the critical value of 1.85, the null is rejected. The claim is the null, so is not supported

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