Question

In order to evaluate the effectiveness of a new type of plant food that was developed for tomatoes, a study was conducted in

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

The margin of error is computed below: ME = t. SE(F) As the population standard deviation is unknown, so a t-test statistic wCI = F ME(F) = 56 +1.571 =(56-1.571, 56+1.571) = (54.429,57.571) Lower limit = 54.429| Upper limit = 57.571 The hypothesis to

Add a comment
Know the answer?
Add Answer to:
In order to evaluate the effectiveness of a new type of plant food that was developed...
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
  • In order to evaluate the effectiveness of a new type of plant food that was developed...

    In order to evaluate the effectiveness of a new type of plant food that was developed for tomatoes, a study was conducted in which a random sample of n = 76 plants received a certain amount of this new type of plant food each week for 14 weeks. The variable of interest is the number of tomatoes produced by each plant in the sample. The table below reports the descriptive statistics for this study: Descriptive statistics for the tomato food...

  • A research team is interested in the effectiveness of hypnosis in reducing pain. The responses from...

    A research team is interested in the effectiveness of hypnosis in reducing pain. The responses from 8 randomly selected patients before and after hypnosis are recorded in the table below (higher values indicate more pain). Construct a 90% confidence interval for the true mean difference in pain after hypnosis. Perceived pain levels 'Pre' and 'Post' hypnosis for 8 subjects Pre 11.3 8.1 8.4 11.1 13.4 15.1 10.6 10.1 Post 9.9 9.1 3.1 8.5 7.0 9.9 10.9 7.7 Difference a) Fill...

  • A food safety guideline is that the mercury in fish should be below 1 part per...

    A food safety guideline is that the mercury in fish should be below 1 part per million (ppm). Listed below are the amounts of mercury (ppm) found in tuna sushi sampled at different stores in a major city. Find the margin error for 95% confidence interval and construct a 95% confidence interval estimate of the mean amount of mercury in the population. Does it appear that there is too much mercury in tuna sushi? Mecury(ppm) .50 .079 .0.09 .89 1.29...

  • Question 4 A company has developed a new type of light bulb, and wants to estimate...

    Question 4 A company has developed a new type of light bulb, and wants to estimate its mean lifetime. A simple random sample of 12 bulbs had a sample mean lifetime of 651 hours with a sample standard deviation of 43 hours. It is reasonable to believe that the population is approximately normal. Find the lower bound of the 95% confidence interval for the population mean lifetime of all bulbs manufactured by this new process. Round to the nearest integer....

  • 1. A food safety guideline is that the mercury in fish should be below 1 part...

    1. A food safety guideline is that the mercury in fish should be below 1 part per million (ppm). Listed below are the amounts of mercury (ppm) found in tuna sushi sampled at different stores in a major city. Find the margin error for 95% confidence interval and construct a 95% confidence interval estimate of the mean amount of mercury in the population. Does it appear that there is too much mercury in tuna sushi? Mecury(ppm) .50 .079 .0.09 .89...

  • a. Assume that a sample is used to estimate a population proportion p. Find the margin...

    a. Assume that a sample is used to estimate a population proportion p. Find the margin of error M.E. that corresponds to a sample of size 343 with 292 successes at a confidence level of 99.8%. M.E.= b. You measure 46 textbooks' weights and find they have a mean weight of 79 ounces. Assume the population standard deviation is 7.5 ounces. Based on this, construct a 90% confidence interval for the true population mean textbook weight. Give your answers as...

  • A new kind of typhoid shot is being developed by a medical research team. The old...

    A new kind of typhoid shot is being developed by a medical research team. The old typhoid shot was known to protect the population for a mean time of 36 months, with a standard deviation of 3 months. To test the time variability of the new shot, a random sample of 17 people were given the new shot. Regular blood tests showed that the sample standard deviation of protection times was 1.5 months. Using a 0.05 level of significance, test...

  • A simple random sample of 900 elements generates a sample proportion 0.72 a. Provide the 90%...

    A simple random sample of 900 elements generates a sample proportion 0.72 a. Provide the 90% confidence interval for the population proportion (to 4 decimals). b. Provide the 99% confidence interval for the population proportion (to 4 decimals). How large a sample should be selected to provide a 95% confidence interval with a margin of error of 3? Assume that the population standard deviation is 50, Round your answer to next whole number. Sales personnel for Skillings Distributors submit weekly...

  • problems 4, 5, 6, 11 and 13 If the population standard deviation was doubled to 10.4...

    problems 4, 5, 6, 11 and 13 If the population standard deviation was doubled to 10.4 and the level of confidence remained at 90%, what would be the new margin of error and confidence interval Margin of error, E. Confidence interval: 20.11<x<34.31 O Did the confidence interval increase or decrease and why? increase 4. Definition of Confidence Intervals (Section 6.1) Circle your answer, True of False. • A 99% confidence interval means that there is a 99% probability that the...

  • 1/The effectiveness of a blood-pressure drug is being investigated. An experimenter finds that, on average, the...

    1/The effectiveness of a blood-pressure drug is being investigated. An experimenter finds that, on average, the reduction in systolic blood pressure is 56.7 for a sample of size 664 and standard deviation 11.2. Estimate how much the drug will lower a typical patient's systolic blood pressure (using a 95% confidence level). Enter your answer as a tri-linear inequality accurate to one decimal place (because the sample statistics are reported accurate to one decimal place). ( 65.6742 ) < μμ <(68.3258...

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