Question 1191500: 1)Sleep duration of college students. In Example 5.4, the daily sleep duration among college students was approximately Normally distributed with mean μ=7.13 hours and standard deviation σ=1.67 hours. You plan to take an SRS of size n=60 and compute the average total sleep time.
(a)What is the standard deviation for the average time?
(b)Use the 95 part of the 68 - 95 - 99.7 rule to describe the variability of this sample mean.
(c)What is the probability that your average will be below 6.9 hours?
2)Determining sample size. Refer to the (Q1). You want to use a sample size such that about 95% of the averages fall within ±10 minutes (0.17 hour) of the true mean μ=7.13.
(a)Based on your answer to part (b) in (Q1), should the sample size be larger or smaller than 60? Explain.
(b)What standard deviation of x¯ do you need such that approximately 95% of all samples will have a mean within 10 minutes of μ?
(c)Using the standard deviation you calculated in part (b), determine the number of students you need to sample.
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Here's the solution:
**1) Sleep duration of college students:**
**(a) Standard deviation for the average time:**
The standard deviation of the sample mean (also called the standard error) is calculated as:
Standard error = σ / √n = 1.67 / √60 ≈ 0.216 hours
**(b) 95 part of the 68-95-99.7 rule:**
The 68-95-99.7 rule states that for a normal distribution:
* Approximately 68% of the data falls within 1 standard deviation of the mean.
* Approximately 95% of the data falls within 2 standard deviations of the mean.
* Approximately 99.7% of the data falls within 3 standard deviations of the mean.
In this case, the mean is 7.13 hours, and the standard error is 0.216 hours. Therefore, approximately 95% of all sample means will fall within 2 standard errors of the mean:
7.13 ± (2 * 0.216) = 7.13 ± 0.432
So, about 95% of sample means will be between 6.698 hours and 7.562 hours.
**(c) Probability that the average will be below 6.9 hours:**
1. **Calculate the z-score:**
z = (x - μ) / (σ / √n) = (6.9 - 7.13) / 0.216 ≈ -1.06
2. **Find the probability:** Use a z-table or calculator to find the probability associated with a z-score of -1.06. P(z < -1.06) ≈ 0.1446
Therefore, there's approximately a 14.46% chance that the average sleep time in your sample will be below 6.9 hours.
**2) Determining sample size:**
**(a) Larger or smaller sample size?**
You want a smaller margin of error (10 minutes or 0.17 hours) than the one implied by the 95% range in part (b) of Q1 (which was about 0.432 hours). A smaller margin of error requires a *larger* sample size.
**(b) Required standard deviation of x¯:**
You want 95% of the sample means to fall within ±0.17 hours of the mean. Using the 95% rule, this means that 2 standard deviations of the sample mean should equal 0.17 hours:
2 * (standard deviation of x¯) = 0.17
Standard deviation of x¯ = 0.17 / 2 = 0.085 hours
**(c) Number of students to sample:**
We know that the standard deviation of the sample mean is σ / √n. We want to find n, so we set up the equation:
0.085 = 1.67 / √n
√n = 1.67 / 0.085 ≈ 19.65
n = (19.65)² ≈ 386.12
Since you can't have a fraction of a student, round up to the nearest whole number. Therefore, you need to sample 387 students.
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