Question

Let μ be the true average nicotine content of a brand of cigarettes whose nicotine content may reasonably be assumed Normally distributed with σ = 0.2. Suppose the observed sample mean of the nicotine content from 32 randomly selected cigarettes: = 1.4. Find the p-value associated with the following hypothesis: Ho : μ = 1.5 vs. HA : μ 1.5 What do you conclude?

0 0
Add a comment Improve this question Transcribed image text
Answer #1

Given that, sample size ( n ) = 32

sample mean (ar x) = 1.4

population standard deviation (sigma) = 0.2

The null and the alternative hypotheses are,

Ho : μ = 1.5 vs. HA : μ 1,5

Test statistic is,

x-μ 1.4-1.5 Vn 7t y 32

p-value = 2 * P(Z < -2.83) = 2 * 0.0023 = 0.0046

p-value = 0.0046

Here, p-value = 0.0046 < significance level (alpha) = 0.05

So, we reject the null hypothesis ( H0).

Add a comment
Know the answer?
Add Answer to:
Let μ be the true average nicotine content of a brand of cigarettes whose nicotine content...
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. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
  • The nicotine content in cigarettes of a certain brand is Normally distributed with a standard deviation...

    The nicotine content in cigarettes of a certain brand is Normally distributed with a standard deviation of σ = 0.1 milligrams. The brand advertises that the mean nicotine content of their cigarettes is μ = 1.5, but you are suspicious and plan to investigate the advertised claim by testing the hypotheses H0 : μ = 1.5 versus Ha : μ > 1.5 at the 5% significance level. You will do so by measuring the nicotine content of 15 randomly selected...

  • (1 point) The nicotine content in cigarettes of a certain brand is normally distributed with mean...

    (1 point) The nicotine content in cigarettes of a certain brand is normally distributed with mean (in milligrams) μ and standard deviation σ=0.1. The brand advertises that the mean nicotine content of their cigarettes is 1.5 mg. Now, suppose a reporter wants to test whether the mean nicotine content is actually higher than advertised. He takes measurements from a random sample of 15 cigarettes of this brand. The sample yields an average of 1.4 mg of nicotine. Conduct a test...

  • 3. The nicotine content in cigarettes of a certain brand is normally distributed with mean (in...

    3. The nicotine content in cigarettes of a certain brand is normally distributed with mean (in milligrams) 4. The brand advertises that the mean nicotine content of their cigarettes is 1.5, but measurements on a random sample of 100 cigarettes of this brand gave a mean of r = 1.53 and standard deviation s=0.1. Is there sufficient evidence in the sample to suggest that the mean nicotine content is actually higher than advertised? Use a = 0.05. (Hint: follow the...

  • A cigarette manufacturer claims that the average nicotine content of their new brand XXX is at mo...

    A cigarette manufacturer claims that the average nicotine content of their new brand XXX is at most 1.35 mg. It would be unwise to reject the manufacturer's claim without strong contradictory evidence. An appropriate problem formulation is to test A random sample of 100 brand XXX cigarettes are tested and the sample standard deviation and sample mean are found to be 0.07 mg and 1.361 mg, respectively. Assuming the central limit theorem applies and s ≈ σ, find the p-value...

  • (1 point) The nicotine content in cigarettes of a certain brand is normally distributed with mean...

    (1 point) The nicotine content in cigarettes of a certain brand is normally distributed with mean (in milligrams) u and standard deviation o= 0.1. The brand advertises that the mean nicotine content of their cigarettes is 1.5 mg. Now, suppose a reporter wants to test whether the mean nicotine content is actually higher than advertised. He takes measurements from a SRS of 20 cigarettes of this brand. The sample yields an average of 1.4 mg of nicotine. Conduct a test...

  • (1 point) The nicotine content in cigarettes of a certain brand is normally distributed with mean...

    (1 point) The nicotine content in cigarettes of a certain brand is normally distributed with mean (in milligrams) u and standard deviation o= 0.1. The brand advertises that the mean nicotine content of their cigarettes is 1.5 mg. Now, suppose a reporter wants to test whether the mean nicotine content is actually higher than advertised. He takes measurements from a SRS of 20 cigarettes of this brand. The sample yields an average of 1.55 mg of nicotine. Conduct a test...

  • The nicotine content in a certain brand of king-size cigarettes has a normal distribution with a...

    The nicotine content in a certain brand of king-size cigarettes has a normal distribution with a mean content of 1.8 mg and a standard deviation of 0.2 mg. Find the probability that the nicotine content of a randomly chosen cigarette of this kind is (a) more than 1.95 mg; (b) between 1.95 mg and 2.35 mg.

  • (2 points) For each problem, select the best response. a) The nicotine content in cigarettes of...

    (2 points) For each problem, select the best response. a) The nicotine content in cigarettes of a certain brand is normally distributed with mean (in milligrams) u and standard deviation o=0.1. The brand advertises that the mean nicotine content of their cigarettes is 1.5, but you believe that the mean nicotine content is actually higher than advertised. To explore this, you test the hypotheses H. :p=1.5, H iu > 1.5 and you obtain a P-value of 0.052. Which of the...

  • Question 1 A study was conducted to estimate μ, the mean number of weekly hours that...

    Question 1 A study was conducted to estimate μ, the mean number of weekly hours that U.S. adults use computers at home. Suppose a random sample of 81 U.S. adults gives a mean weekly computer usage time of 8.5 hours and that from prior studies, the population standard deviation is assumed to be σ = 3.6 hours. A similar study conducted a year earlier estimated that μ, the mean number of weekly hours that U.S. adults use computers at home,...

  • im so grateful for all your help! this has helped me in so many ways especially...

    im so grateful for all your help! this has helped me in so many ways especially with all the things going on right now. thank you so much !! 1 2 3 4 (1 point) 35 people are randomly selected and the accuracy of their wristwatches is checked, with positive errors representing watches that are ahead of the correct time and negative errors representing watches that are behind the correct time. The 35 values have a mean of 94sec and...

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT