Question

When people smoke, the nicotine they absorb is converted to cotinine, which can be measured.A random...

When people smoke, the nicotine they absorb is converted to cotinine, which can be measured.A random sample of 40 smokers has a mean cotinine level of 172.5 ng/ml with a sample standard deviation of 119.5 ng/ml. Use a 0.05 significance level to test the claim that the mean cotinine level of smokers is equal to 200ng/ml.

H0: μ___200 For Q23(circle one: equal to, less than , greater than, not equal to)

H1: μ___200 For Q24 (circle one:equal to, less than, greater than, not equal to)

Test Statistic t=__________ Q25 (ROUND TO 2 DECIMAL PLACES)

P-value: p =_____________ Q26 (ROUND TO4 DECIMAL PLACES)

Decision:________________________Q27 (circle one:Reject H0 , Fail to Reject H0)

Conclusion:_________________________________________________________

Q28 Circle one for the conclusion:

There is sufficient evidence to warrant the rejection of the claim.

OR

There is not sufficient evidence to warrant the rejection of the claim

Homework Answers

Answer #1

Solution :

= 200

= 172.5

s =119.5

n = 405

This is the two tailed test .

The null and alternative hypothesis is ,

H0 :    = 200

Ha :     200

Test statistic = t

= ( - ) / s / n

= ( 172.5- 200) /119.5 / 40

= -1.455

Test statistic = t = -1.455

P-value =0.1535

= 0.05  

P-value >

0.1535 > 0.05

Fail to reject the null hypothesis .

There is not sufficient evidence to warrant the rejection of the claim

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
When people smoke, the nicotine they absorb is converted to cotinine, which can be measured. A...
When people smoke, the nicotine they absorb is converted to cotinine, which can be measured. A sample of 40 smokers has a mean cotinine level of 172.5 ng/ml. Assuming the σ is known to be 119.5 ng/ml, find the following: (6 pts) a) A 90% confidence interval estimate of the mean cotinine level of all smokers. b) A 98% confidence interval estimate of the mean cotinine level of all smokers. c) A 92% confidence interval estimate of the mean cotinine...
When people smoke, the nicotine they absorb is converted to cotinine, which can be measured. A...
When people smoke, the nicotine they absorb is converted to cotinine, which can be measured. A sample of 70 smokers has a mean cotinine level of 172.3 ng/ml and a stadard deviation of 118.2 ng/ml. What is the value of the test statistic when testing the claim that the mean cotinine level of all smokers is less than 200.0 ng/ml at the significance level 5%? Select one: a. -1.64 b. 1.64 c. None of the other answers is neccessary true....
5. A sample of 50 impossible whoppersandwiches have a mean fat content of 37.6g and standard...
5. A sample of 50 impossible whoppersandwiches have a mean fat content of 37.6g and standard deviation of 4.9g . A sample of 58 regular whopper sandwiches have a mean fat content of 39.3g and standard deviation of 2.7g. Use a 0.05 significance level to test the claim that the mean fat content for the impossible whopper is less than the mean fat content for the regular whopper. (Assume that the two samples are independent simple random samples selected from...
In one study of smokers who tried to quit smoking with nicotine patch​ therapy, 39 were...
In one study of smokers who tried to quit smoking with nicotine patch​ therapy, 39 were smoking one year after treatment and 32 were not smoking one year after the treatment. Use a 0.10 significance level to test the claim that among smokers who try to quit with nicotine patch​ therapy, the majority are smoking one year after the treatment. Do these results suggest that the nicotine patch therapy is not​ effective? Identify the null and alternative hypotheses for this...
Suppose that in a random selection of 100 colored​ candies, 30​% of them are blue. The...
Suppose that in a random selection of 100 colored​ candies, 30​% of them are blue. The candy company claims that the percentage of blue candies is equal to 29​%. Use a 0.05 significance level to test that claim. a) Identify the test statistic for this hypothesis test. b) Identify the​ P-value for this hypothesis test. c) Identify the conclusion for this hypothesis test. A. RejectReject H0. There is sufficient evidence to warrant rejection of the claim that the percentage of...
Assume a significance level of α=0.1 and use the given information to complete parts​ (a) and​...
Assume a significance level of α=0.1 and use the given information to complete parts​ (a) and​ (b) below. Original​ claim: Women have heights with a mean equal to 156cm. The hypothesis test results in a​ P-value of 0.0676 a. State a conclusion about the null hypothesis.​ (Reject H0 or fail to reject H0​.) Choose the correct answer below. A. Fail to reject H0 because the​ P-value is greater than alphaα. B. Reject H0 because the​ P-value is greater than alphaα....
1) A sample of cereal boxes is randomly selected and the sugar contents (grams of sugar...
1) A sample of cereal boxes is randomly selected and the sugar contents (grams of sugar per gram of cereal) are recorded. Those amounts are summarized with these statistics: n = 16, ?̅ = 0.295g, s = 0.168g. Use a 0.05 significance level to test the claim that the mean of all cereals is less than 0.3g. Section 7-6 Problem Testing a Claim about a Standard Deviation a ) Use a 0.05 significance level to test the claim that heights...
In 1997, a survey of 880 U.S. households showed that 149 of them used email. Use...
In 1997, a survey of 880 U.S. households showed that 149 of them used email. Use a 0.05 significance level to test the cl aim that more than 14% of U.S. households used email in 1997. H0 :p____ 0.14 For Q17(circle one: equal to, less than, greater than, not equal to) H1: p___ 0.14 For Q18 (circle one: equal to, less than, greater than, not equal to) Test Statistic z =__________ Q19 (ROUND TO 2 DECIMAL PLACES) P-value: p =_____________...
19. Suppose that in a random selection of 100 colored​ candies, 30​% of them are blue....
19. Suppose that in a random selection of 100 colored​ candies, 30​% of them are blue. The candy company claims that the percentage of blue candies is equal to 28​%. Use a 0.10 significance level to test that claim. ____________________ Identify the null and alternative hypotheses for this test. Choose the correct answer below. A. H0​: p=0.28 H1​: p<0.28 B. H0​: p≠0.28 H1​: p=0.28 C. H0​: p=0.28 H1​: p≠0.28 D. H0​: p=0.28 H1​: p> ____________________ Identify the test statistic for...
You wish to test the following claim (Ha) at a significance level of α=0.10   Ho:p=0.46       Ha:p>0.46...
You wish to test the following claim (Ha) at a significance level of α=0.10   Ho:p=0.46       Ha:p>0.46 You obtain a sample of size n=648in which there are 322 successful observations. Determine the test statistic formula for this test. What is the test statistic for this sample? (Report answer accurate to three decimal places.) test statistic = What is the p-value for this sample? (Report answer accurate to four decimal places.) p-value = The p-value is... less than (or equal to) αα...