Question

Consider: Sugar content in a particular brand of cranberry sauce.  A random sample was selected in order...

Consider: Sugar content in a particular brand of cranberry sauce.  A random sample was selected in order to estimate the true sugar content.

Sample results:

Sample size: 76

Sample Mean: 24.7 g

Sample standard deviation: 1.3 g.

a. Provide the 90% Confidence Interval for the Mean amount of sugar per serving.

b. The sample result of ?̅= 24.7 ? can be used as a point estimate for the mean amount of sugar per serving. At the 90% Confidence Level, what is the Margin of Error associated with this estimate?

Math   Statistics-And-Probability   
0 0
Add a commentTranscribed image text
Next >

Homework Answers

Answer #1

Answer:-

From the above information

Page one

Page two

Dear student,
I am waiting for your feedback. I have given my 100% to solve your queries. If you are satisfied by my given answer. Can you please like it☺
  
Thank You!!!

0 0
Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Log In

Sign Up

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
A study was conducted to estimate the average cold cereal sugar content per serving. It was...
A study was conducted to estimate the average cold cereal sugar content per serving. It was determined that the sample mean was 6.95 g. Confidence intervals were obtained for 90% 95% and 99% confidence levels and listed below. Which interval would be the 90% confidence interval? (5.8030, 8.0970) Not enough info (5.9921, 7.9079) (5.4243, 8.4757)
a sugar content in one cup serving of a cereal was measured for a sample of...
a sugar content in one cup serving of a cereal was measured for a sample of 140 servibgs. the acerage was 11.9 abs the standard deviation was 1.1 g. find ta 95% cobfidence intefval for the mean sugar content. what is the confidence level of the interval (11.81, 11.99) how large a sample is neededso that a 95% confidence interval specifies the mean within +~0.1?
For a particular brand of SUV, you want to estimate the miles per gallon (mpg). The...
For a particular brand of SUV, you want to estimate the miles per gallon (mpg). The manufacturer says that σ = 1.75 mpg. Find the sample size if you want a margin of error of ± 0.55 and 90% confidence?
A random sample of 50 respondents is selected, and each respondent is instructed to keep a...
A random sample of 50 respondents is selected, and each respondent is instructed to keep a detailed record of time spent engaged viewing content across all screens (Internet, video on a computer video on mobile phone) in a particular week. The sample mean content viewing time per week is 42 hours with a standard deviation of 3.2 hours. Construct a 99% confidence interval estimate for the mean content viewing time per week in this area. What is the margin of...
Suppose that a random sample of 50 bottles of a particular brand of cough syrup is...
Suppose that a random sample of 50 bottles of a particular brand of cough syrup is selected and the alcohol content of each bottle is determined. Let μ denote the average alcohol content for the population of all bottles of the brand under study. Suppose that the resulting 95% confidence interval is (7.6, 9.2). (a) Would a 90% confidence interval calculated from this same sample have been narrower or wider than the given interval? Explain your reasoning. The 90% would...
A sample of 100 of a particular brand of tires found the mean life span to...
A sample of 100 of a particular brand of tires found the mean life span to be 50,000 miles . The population standard deviation is 800 miles . (a) Construct a 90% confidence interval containing the mean life span of all tires. (b) How large must the sample be in order to be 90% confident that the mean lifespan of tires is within 500 miles of the sample mean ? (c) How could we reduce the margin of error?
The following data represents the calorie-content per 1/2-cup serving for 16 randomly selected flavours of Jen...
The following data represents the calorie-content per 1/2-cup serving for 16 randomly selected flavours of Jen and Berry's Ultra-Creamy Ice Cream: 270, 150, 170, 140, 160, 160, 290, 190, 190, 160, 170, 150, 110, 180, 170, 185 It is also known that the samples are form the normal population with the standard deviation 45. (a) What is the point estimate of mean for calorie-content per 1/2-cup serving? Calculate margin of error when ?=0.05 (b) Construct a 95% confidence interval for...
Suppose we know that the population standard deviation for wages (σ) is $15. A random sample...
Suppose we know that the population standard deviation for wages (σ) is $15. A random sample of n = 900 results in a sample mean wage of $50. (10 points) Provide 95% confidence interval estimate of µ. Interpret the result. What is the value of the margin of error? Provide 90% confidence interval estimate of µ. What is the value of the margin of error? Provide 99% confidence interval estimate of µ. What is the value of the margin of...
A random sample of 49 lunch customers was selected at a restaurant. The average amount of...
A random sample of 49 lunch customers was selected at a restaurant. The average amount of time the customers in the sample stayed in the restaurant was 40 minutes. From past experience, it is known that the population standard deviation equals 10 minutes. a. Compute the standard error of the mean. b. Construct a 95% confidence interval for the true average amount of time customers spent in the restaurant. c. With a .95 probability, what sample size would have to...
A manufacturer of a certain brand of rice cereal claims that the average content of Saturated...
A manufacturer of a certain brand of rice cereal claims that the average content of Saturated fat is 1.5 grams per serving. You have taken a poll and the sample of 30 elements yields an average of 1.7 grams, so you want to test 98% confidence level that the mean differs from 1.5 grams with one deviation population of 0.2 g. Perform a hypothesis test and conclude with the p-value.
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT