Question

A manufacturer claims that the mean weight of flour in its 32-ounce bags is its 32.1...

A manufacturer claims that the mean weight of flour in its 32-ounce bags is its 32.1 ounces. A T-Test is performed to determine whether the mean weight is actually less than this. The mean weight for a random sample of 45 bags of flour was 30.7 ounces with a standard deviation of 2.5 ounces. Test the claim at the 5% significance level.

a) Check the assumptions:

b) Hypotheses (State in symbols and in words): Ho: Ha:

c) Test Statistic: ___________________

d) Sketch:e) P-Value: ___________________

f) Conclusion:

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

Homework Answers

Answer #1

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 manufacturer claims that the mean weight of flour in its 32-ounce bags is 32.1 ounces....
A manufacturer claims that the mean weight of flour in its 32-ounce bags is 32.1 ounces. A T-Test is performed to determine whether the mean weight is actually less than this. The mean weight for a random sample of 45 bags of flour was 30.7 ounces with a standard deviation of 2.5 ounces. Test the claim at the 5% significance level. a) Check the assumptions: b) Hypotheses (State in symbols and in words): Ho: Ha: c) Test Statistic: d) Sketch:...
A 32 ounce can of a popular fruit drink claims to contain 20% real fruit juice....
A 32 ounce can of a popular fruit drink claims to contain 20% real fruit juice. Since this is a 32 ounce can, they are actually claiming that the can contains 6.4 ounces of real fruit juice. The consumer protection agency samples 46 such cans of this fruit drink. Of these, the mean volume of fruit juice is 6.35 with standard deviation of 0.21. Test the claim that the mean amount of real fruit juice in all 32 ounce cans...
Real Fruit Juice: A 32 ounce can of a popular fruit drink claims to contain 20%...
Real Fruit Juice: A 32 ounce can of a popular fruit drink claims to contain 20% real fruit juice. Since this is a 32 ounce can, they are actually claiming that the can contains 6.4 ounces of real fruit juice. The consumer protection agency samples 44 such cans of this fruit drink. Of these, the mean volume of fruit juice is 6.32 with standard deviation of 0.23. Test the claim that the mean amount of real fruit juice in all...
Real Fruit Juice: A 32 ounce can of a popular fruit drink claims to contain 20%...
Real Fruit Juice: A 32 ounce can of a popular fruit drink claims to contain 20% real fruit juice. Since this is a 32 ounce can, they are actually claiming that the can contains 6.4 ounces of real fruit juice. The consumer protection agency samples 42 such cans of this fruit drink. Of these, the mean volume of fruit juice is 6.34 with standard deviation of 0.21. Test the claim that the mean amount of real fruit juice in all...
A famous food manufacturer claims that the mean weight of their chocolate bars is 2.66 ounces...
A famous food manufacturer claims that the mean weight of their chocolate bars is 2.66 ounces with standard deviation 0.45 ounces. A consumer watchdog group sampled 150 chocolate bars from this company. The mean weight is 2.29 ounces. Set up a hypothesis test at the significance level 0.10 Part 1: what is the H1 statement? Part 2: what is the claim? Part 3: what is the p-value? Select one: a. Part 1 H1 μ < 2.66 Part 2 H1 is...
A company claims that the mean net weight of the contents of its brand T cereal...
A company claims that the mean net weight of the contents of its brand T cereal boxes is at least 18 ounces. Suppose you want to test whether or not the claim of the company is true. Explain briefly how you would conduct this test using a large sample. Assume that ? = 0.25 ounce.
A local chip manufacturer distributes chips in bags labeled as 150g. A group of consumers believe...
A local chip manufacturer distributes chips in bags labeled as 150g. A group of consumers believe they are being cheated. They run a test on 32 bags, measures their contents, and obtains a sample mean of 145 grams with a standard deviation of 6 ounces. Use a 0.01 significance level to test the consumer's claim that the company is cheating its customers. Null Hypothesis: Alternate Hypothesis: P-value: Conclusion: Interpretation:
Real Fruit Juice (Raw Data, Software Required): A 32 ounce can of a popular fruit drink...
Real Fruit Juice (Raw Data, Software Required): A 32 ounce can of a popular fruit drink claims to contain 20% real fruit juice. Since this is a 32 ounce can, they are actually claiming that the can contains 6.4 ounces of real fruit juice. The consumer protection agency samples 30 such cans of this fruit drink. The amount of real fruit juice in each can is given in the table below. Test the claim that the mean amount of real...
Real Fruit Juice (Raw Data, Software Required): A 32 ounce can of a popular fruit drink...
Real Fruit Juice (Raw Data, Software Required): A 32 ounce can of a popular fruit drink claims to contain 20% real fruit juice. Since this is a 32 ounce can, they are actually claiming that the can contains 6.4 ounces of real fruit juice. The consumer protection agency samples 30 such cans of this fruit drink. The amount of real fruit juice in each can is given in the table below. Test the claim that the mean amount of real...
Real Fruit Juice (Raw Data, Software Required): A 32 ounce can of a popular fruit drink...
Real Fruit Juice (Raw Data, Software Required): A 32 ounce can of a popular fruit drink claims to contain 20% real fruit juice. Since this is a 32 ounce can, they are actually claiming that the can contains 6.4ounces of real fruit juice. The consumer protection agency samples 30 such cans of this fruit drink. The amount of real fruit juice in each can is given in the table below. Test the claim that the mean amount of real fruit...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT