Question

The average weight of a watermelon is to be estimated. A sample of 18 watermelons is...

The average weight of a watermelon is to be estimated. A sample of 18 watermelons is collected and found that the average weight is 23 pounds. If the sample standard deviation is 3 pounds  Compute the confidence interval for the true mean using a confidence level of 98%. Round answers to two place values.

______________< μ < ______________

Homework Answers

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
You measure 23 watermelons' weights, and find they have a mean weight of 76 ounces. Assume...
You measure 23 watermelons' weights, and find they have a mean weight of 76 ounces. Assume the population standard deviation is 13.5 ounces. Based on this, construct a 99% confidence interval for the true population mean watermelon weight. Give your answers as decimals, to two places _____±______ ounces
You measure 33 watermelons' weights, and find they have a mean weight of 61 ounces. Assume...
You measure 33 watermelons' weights, and find they have a mean weight of 61 ounces. Assume the population standard deviation is 14.9 ounces. Based on this, construct a 99% confidence interval for the true population mean watermelon weight. Round your answers to two decimal places. Enter your answers in the form: Sample Mean ± Margin of Error  
A supermarket wants to estimate the population mean weight of watermelons of a certain type. A...
A supermarket wants to estimate the population mean weight of watermelons of a certain type. A sample size n = 45 has sample mean = 20 pounds with sample standard deviation s = 2.2 pounds. Create a confidence interval at 90% confidence for the population mean μ.
A particular variety of watermelon weighs on average 21.3 pounds with a standard deviation of 1.07...
A particular variety of watermelon weighs on average 21.3 pounds with a standard deviation of 1.07 pounds. Consider the sample mean weight of 90 watermelons of this variety. Assume the individual watermelon weights are independent. a. What is the expected value of the sample mean weight? Give an exact answer. b. What is the standard deviation of the sample mean weight? Give your answer to four decimal places. c. What is the approximate probability the sample mean weight will be...
You measure 21 watermelons' weights, and find they have a mean weight of 41 ounces. Assume...
You measure 21 watermelons' weights, and find they have a mean weight of 41 ounces. Assume the population standard deviation is 3.2 ounces. Based on this, construct a 90% confidence interval for the true population mean watermelon weight. Give your answers as decimals, to two places _____ ± ____ ounces
You measure 28 watermelons' weights, and find they have a mean weight of 39 ounces. Assume...
You measure 28 watermelons' weights, and find they have a mean weight of 39 ounces. Assume the population standard deviation is 12.3 ounces. Based on this, what is the maximal margin of error associated with a 99% confidence interval for the true population mean watermelon weight.
You measure 42 watermelons' weights, and find they have a mean weight of 59 ounces. Assume...
You measure 42 watermelons' weights, and find they have a mean weight of 59 ounces. Assume the population standard deviation is 12 ounces. Based on this, what is the margin of error associated with a 95% confidence interval for the true population mean watermelon weight. Round your answer to two decimal places. 0.92 Incorrect ounces
Suppose that the weight of seedless watermelons is normally distributed with mean 6.6 kg. and standard...
Suppose that the weight of seedless watermelons is normally distributed with mean 6.6 kg. and standard deviation 1.4 kg. Let X be the weight of a randomly selected seedless watermelon. Round all answers to 4 decimal places where possible. What is the probability that a randomly selected watermelon will weigh more than 6 kg? What is the probability that a randomly selected seedless watermelon will weigh between 6.2 and 7.1 kg? The 90th percentile for the weight of seedless watermelons...
You measure 34 watermelons' weights, and find they have a mean weight of 44 ounces. Assume...
You measure 34 watermelons' weights, and find they have a mean weight of 44 ounces. Assume the population standard deviation is 5.3 ounces. Based on this, what is the maximal margin of error associated with a 99% confidence interval for the true population mean watermelon weight. As in the reading, in your calculations: --Use z = 1.645 for a 90% confidence interval --Use z = 2 for a 95% confidence interval --Use z = 2.576 for a 99% confidence interval....
To estimate the average weight of air-freight packages a sample of size 400 was collected. The...
To estimate the average weight of air-freight packages a sample of size 400 was collected. The sample mean for these 400 packages was found to be 85.6 and the sample standard deviation was 3.2. Construct a 95% confidence interval for the average weight of air-freight packages. Find the minimum required sample size for estimating the mean of a population to within 0.4 with 98% confidence. If you test the hypotheses Ho : μ = 80 versus H1 : μ  80, what...