Question

R3. 41: Hardness of a rubber product doesn’t follow the Normal distribution, but the mean and...

R3. 41:

Hardness of a rubber product doesn’t follow the Normal distribution, but the mean and standard deviation of a sample of 36 are 70 and 6 Shore A (the hardness scale). What is the 95% confidence interval for population mean? (Z0.025 = 1.96, Z0.05 = 1.645)

  1. A. The distribution is not Normal, so we cannot calculate the confidence interval.
  2. B. 68.04 < μ < 71.96
  3. C. 68.355 < μ < 71.345
  4. D. The population standard deviation is not known so we cannot calculate the confidence interval.

Homework Answers

Answer #1

B. 68.04 < μ < 71.96

Confidence interval for Population mean is given as below:

Confidence interval = Xbar ± Z*σ/sqrt(n)

From given data, we have

Xbar = 70

σ = 6

n = 36

Confidence level = 95%

Critical Z value = 1.96

(by using z-table)

Confidence interval = Xbar ± Z*σ/sqrt(n)

Confidence interval = 70 ± 1.96*6/sqrt(36)

Confidence interval = 70 ± 1.9600

Lower limit = 70 - 1.96 = 68.04

Upper limit = 70 + 1.96 = 71.96

Confidence interval = (68.04, 71.96)

B. 68.04 < μ < 71.96

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
1). A random variable X has a normal distribution with mean 132 and variance 36. If...
1). A random variable X has a normal distribution with mean 132 and variance 36. If x = 120, its corresponding value of Z is? 2).Fill the blank. In the standard normal distribution, z0.05 = 1.645 means that 95% of all values of z are below ____ and ____% are above it.
A) a sample of 101 body temperatures of randomly selected people has a mean of 98.20...
A) a sample of 101 body temperatures of randomly selected people has a mean of 98.20 degrees Fahrenheit and a standard deviation of 0.62 degrees Fahrenheit. Construct a 95% confidence level confidence interval for the population mean body temperature. B) Use the above result to test the claim that the population mean body temperature is 98.6 degrees Fahrenheit with significance level 0.05. Critical Values: z0.005 = 2.575, z0.01 = 2.325, z0.025 = 1.96, z0.05 = 1.645 When d.f.=6: t0.005 =...
Question 1 A researcher wishes to find a 90% confidence interval estimate for an unknown population...
Question 1 A researcher wishes to find a 90% confidence interval estimate for an unknown population mean using a sample of size 25. The population standard deviation is 7.2. The confidence factor z α 2for this est Group of answer choices 1.28 1.645 1.96 Question 2 A 90% confidence interval estimate for an unknown population mean μ is (25.81, 29.51). The length of this CI estimate is Group of answer choices 1.96 1.85 3.7 1.645 Question 3 Data below refers...
Use the standard normal distribution or the​ t-distribution to construct a 99​% confidence interval for the...
Use the standard normal distribution or the​ t-distribution to construct a 99​% confidence interval for the population mean. Justify your decision. If neither distribution can be​ used, explain why. Interpret the results. In a random sample of 13 mortgage​ institutions, the mean interest rate was 3.59​% and the standard deviation was 0.42​%. Assume the interest rates are normally distributed. Select the correct choice below​ and, if​ necessary, fill in any answer boxes to complete your choice. A. The 99​% confidence...
1) Which of the following statements is correct about the standard normal distribution and t distribution?...
1) Which of the following statements is correct about the standard normal distribution and t distribution? a) standard normal distribution is used to eliminate population mean, while t distribution is used to estimate population standard deviation b) t distribution is always centered at 0, while the center of a standard normal destruction depends on the sample mean c) none of the 3 answer choices d) standard normal distribution curve is higher than t distribution curve at the tails 2) If...
A random variable ?x has a Normal distribution with an unknown mean and a standard deviation...
A random variable ?x has a Normal distribution with an unknown mean and a standard deviation of 12. Suppose that we take a random sample of size ?=36n=36 and find a sample mean of ?¯=98x¯=98 . What is a 95% confidence interval for the mean of ?x ? (96.355,99.645)(96.355,99.645) (97.347,98.653)(97.347,98.653) (94.08,101.92)(94.08,101.92) (74.48,121.52)
The rockwell hardness of steel beams has a normal distribution with average 75 rockwells and with...
The rockwell hardness of steel beams has a normal distribution with average 75 rockwells and with a standard deviation of 3. a) The manufacturer wishes to find the specification of the steel beams so that 98% of the beams comply with the minimum hardness. this is only 2% of the time it does not comply with the minimum hardness. b) If a client wants to buy 2500 steel beams and wants the beams to have a hardness greater than 73...
Assume a stock's monthly returns follow a normal distribution with a mean of 8% and standard...
Assume a stock's monthly returns follow a normal distribution with a mean of 8% and standard deviation of 8%. A) What is the lower bound of the 68.26% confidence interval on realized monthly returns? B) What is the upper bound of the 68.26% confidence interval?
For Central Limite Theorem, if n>30, we say the sampling distribution is normal. However, most of...
For Central Limite Theorem, if n>30, we say the sampling distribution is normal. However, most of the time, with population standard deviation unknown, we still have to use t value to compute a confidence interval. But I wonder for normal distribution(z distribution), even though we do not know population sd, why cannot we use z value directly to compute confidence interval, as it has stated in central limit theorem that the distribution is normal.
A sample of n = 16 is to be taken from a distribution that can reasonably...
A sample of n = 16 is to be taken from a distribution that can reasonably be assumed to be Normal with a standard deviation σ of 100. The sample mean comes out to be 110. 1. The standard error of the mean, that is, the standard deviation of the sample mean, is σx¯ = σ/√ n. What is its numerical value? 2. The 97.5 percentile, 1.96, of the standard Normal distribution is used for a 95% confi- dence interval....