Based on all student records at Camford University, students...

Based on all student records at Camford University, students spend an average of 5.5 hours per week playing organized sports. The population's standard deviation is 2.2 hours per week. Based on a sample of 121 students, Healthy Lifestyles Incorporated (HLI) would like to apply the central limit theorem to make various estimates. a. Compute the standard error of the sample mean. b. What is the chance HLI will find a sample mean between 5 and 6 hours? c. Calculate the probability that the sample mean will be between 5.3 and 5.7 hours. d. How strange would it be to obtain a sample mean greater than 6.5 hours?

Based on all student records at Camford University, students spend an average of 5.5 hours per week playing organized sports. The population's standard deviation is 2.2 hours per week. Based on a sample of 121 students, Healthy Lifestyles Incorporated (HLI) would like to apply the central limit theorem to make various estimates.
a. Compute the standard error of the sample mean.
b. What is the chance HLI will find a sample mean between 5 and 6 hours?
c. Calculate the probability that the sample mean will be between 5.3 and 5.7 hours.
d. How strange would it be to obtain a sample mean greater than 6.5 hours?

Key Concepts

Central Limit Theorem

The central limit theorem (CLT) states that the sampling distribution of the sample mean will approximate a normal distribution if the sample size is sufficiently large, regardless of the original population's distribution. This concept is crucial because it allows us to use the normal distribution to make probability calculations about sample means, even when the population distribution is unknown.

Standard Error

The standard error is a measure of the dispersion or variability of the sample mean’s distribution. It is calculated as the population standard deviation divided by the square root of the sample size. This concept is essential for understanding how much the sample mean is expected to vary from one sample to another.

Sampling Distribution of the Mean

The sampling distribution of the mean describes the probability distribution of sample means when multiple samples are drawn from a population. This concept is fundamental in statistics as it underpins the use of the normal distribution for making inferences about a population mean based on a sample mean.

Z-Score Calculation

The Z-score is a statistical measure that quantifies the number of standard deviations a data point (in this case, a sample mean) is from the population mean. This concept is used to convert values into a standard normal distribution, which then allows for the computation of probabilities associated with the sample mean.

Probability Estimation in Normal Distribution

Utilizing the properties of the normal distribution, one can estimate the probability that a sample mean falls within a specified range. This involves calculating Z-scores for the endpoints of the interval and then using the standard normal distribution to find the corresponding probabilities, which is key to understanding how likely certain outcomes are in statistical sampling.

Transcript

00:01 This is an example of a sampling distribution and how we calculate area underneath the curve.

00:08 The first thing is they mention the central limit theorem, which says if the sample size is at least 30, then we can assume that the sampling distribution is approximately normal.

00:19 So we're assuming that when we do these problems, we're dealing with a approximately normal distribution because of the central limit theorem.

00:27 Starting off the problem, they tell us to ask, or they ask us to find what is known as the standard error.

00:34 The standard error for the sampling distribution is the following formula.

00:39 You take the standard deviation from the population distribution, which they tell us in the directions is 2 .2 hours per week, and you divide it by the square root of the sample size.

00:52 So that's a formula that's on a formula sheet that will be given to you, and you just need to know that's the formula you're using in this situation.

00:58 So we're going to plug in the numbers.

01:00 We're going to get 2 .2 for the top, and the sample size they tell us is 121 stewards.

01:09 And we'll get 2 .2 over 11, and that ends up being just 0 .2.

01:16 So that is the answer for part a, the standard error for the sampling distribution, which is also the standard deviation.

01:24 So now when we answer the rest of the questions, we're not going to use the original standard deviation that they gave, which was 2 .2 hours.

01:31 We're not going to use the standard deviation for the sampling distribution, which is the 0 .2, since everything is related to this sample of 121 students.

01:42 So for part b, it says what is the chance that the hli will find a sample mean between five and six hours? it would be helpful for us to visualize what they're saying here.

01:52 We have this normal distribution.

01:54 Again, we're allowed to do that because of the central limit theorem.

01:58 We know that it is approximately normal with a mean of 5 .5.

02:03 That does not change.

02:04 That's the mean of the population.

02:06 That's the mean of our sampling distribution.

02:08 But our new standard deviation will be 0 .2.

02:12 And so that will have to make that adjustment.

02:15 So the 5 .5 is right in the middle of our normal distribution, and they want us to find the chances, the probability that's between five and six.

02:25 So five would be somewhere to the left, six a little bit to the right.

02:30 We're trying to find this area right here to find the probability.

02:34 What we need to do is calculate the z -scores for five and six.

02:38 So the formula for that again is you take the data value, you subtract the mean, you divide by the standard deviation.

02:47 For our purposes, the mean is the 5 .5, and the standard deviation will be the 0 .2.

02:55 That will not change throughout the problem.

02:57 The only thing that changes is the x value, what we're plugging in to find for that z -score.

03:03 So we're going to plug five in first.

03:05 We're going to z equals 5 minus 5 .5 over 0 .2.

03:12 And you're going to figure out that standard deviation to get an answer of negative 2 .5.

03:21 The z -score for six is going to be positive 2 .5.

03:32 And so now what we've done is we've calculated what the normalized data values are.

03:38 So we have what's called the standard normal curve, which changes the mean to zero and the standard deviation to one.

03:45 So now zero's in the middle, and we have from negative 2 .5 to positive 2 .5.

03:53 And then at this point, you either go to, there's a table that you can look for for finding probabilities of z -scores.

04:06 You can either go to a table or there's computer software you can use to figure this out.

04:15 Graphing calculators have the capability.

04:17 One way or another, you need to know how to figure out the area underneath the curve...

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