Question 1207924: A random sample of n = 36 observations has a mean x = 28.7 and a standard deviation s = 3.8.
GIVEN:-
- point estimate of the population mean š = 28.7
- 95% margin of error for your estimate = 1.2413
- 90% confidence interval for š = 27.658 to 29.742
- In repeated sampling, 90% of all intervals constructed in this manner will enclose the population mean.
- (a) bound is calculated using zš¼, while the lower confidence limit in part (b) is calculated using zš¼/2.
(a) Find a 90% lower confidence bound for the population mean š. (Round your answer to two decimal places.)
(b) How many observations do you need to estimate š to within 0.6, with probability equal to 0.95? (Round your answer up to the nearest whole number.)
observations
Answer by GingerAle(43)   (Show Source):
You can put this solution on YOUR website! **a) Find a 90% lower confidence bound for the population mean š.**
* **Understand the Concept:**
* A lower confidence bound provides a minimum value for the population mean with a certain level of confidence.
* For a 90% lower bound, we're essentially saying that we are 90% confident that the population mean is greater than or equal to this value.
* **Calculate the Lower Bound:**
* Since we're given the 90% confidence interval (27.658 to 29.742), the lower bound of this interval directly represents the 90% lower confidence bound for the population mean.
* **Answer:**
* The 90% lower confidence bound for the population mean š is **27.66**.
**b) Determine the sample size needed to estimate š to within 0.6 with a 95% probability.**
* **Understand the Goal:**
* We want to find the sample size (n) that will ensure the margin of error (E) is 0.6 with a 95% confidence level.
* **Formula:**
* The margin of error (E) for a confidence interval for the mean is given by:
* E = zĪ±/2 * (Ļ / ān)
* where:
* zĪ±/2 is the critical value from the standard normal distribution for the desired confidence level (95% in this case)
* Ļ is the population standard deviation (we'll use the sample standard deviation 's' as an estimate)
* n is the sample size
* **Rearrange the formula to solve for n:**
* n = (zĪ±/2 * Ļ / E)Ā²
* **Find the critical value (zĪ±/2):**
* For a 95% confidence level, zĪ±/2 = 1.96 (from the standard normal distribution table)
* **Plug in the values:**
* n = (1.96 * 3.8 / 0.6)Ā²
* n = (12.32)Ā²
* n ā 151.75
* **Round up to the nearest whole number:**
* n = 152
* **Answer:**
* You need **152 observations** to estimate š to within 0.6 with a probability of 0.95.
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