Here’s your next reading assignment. Read **Sections 4.1 and 4.2** in your textbook and answer the following questions by 8 a.m., Monday, February 27th. Be sure to login (using the link near the bottom of the sidebar) to the blog before leaving your answers in the comment section below.

- Suppose that a random sample of 50 bottles of a particular brand of cough syrup is selected and the alcohol content of each bottle is determined. Let μ denote the average alcohol content for the population of all bottles of the brand under study. Suppose that the resulting 95% confidence interval is (7.5, 9.4). Would a 90% confidence interval calculated from this same sample have been narrower or wider than the given interval? Explain your reasoning.
- Suppose a random sample of 30 Vanderbilt undergraduates is found to have an average height of 67 inches and a sample standard deviation of 3 inches. Construct a 95% confidence interval for the average height of the population of all Vanderbilt undergraduates and interpret its meaning.
- What’s one question you have about the reading?

1) A 90% confidence interval calculated from the same sample would be narrower. Since the confidence that μ would be within the interval decreases from 95% to 90%, it makes sense that a smaller confidence interval would result (i.e. there is a greater chance than before that the μ would lie outside the interval if our interval is smaller).

2) Assuming the heights of Vanderbilt undergraduates are normally distributed, the confidence interval is (65.90, 68.10). This was obtained by dividing the sample standard deviation (3) by the square root of the sample size (30). Then since the sample size is smaller than 50, the 95% confidence interval corresponds to a margin of error of 2*(3/sqrt30) = 1.095 yielding the confidence interval of (65.90, 68.10). This means that the probability that the population mean of the height of all Vanderbilt undergraduates lies between 65.90 and 68.10 is 95%.

3) No question

1) A 90% confidence interval would be narrower because the area between -z* and +z* would decrease.

2) 95.9 < m < 67.1. We are 95% confident that the average height of the population of all Vanderbilt undergraduates is between 65.9 inches and 67.1 inches.

3) In example 4.20, why does it say that a Z score of 2 or greater would be unusual? Why does a cutoff point starts at 2?

1. A 90% confidence interval would be smaller. A 95% confidence interval is basically the mean plus or minus 2 standard deviations, since 95% of the data of a normal distribution falls between 2 standard deviations of the mean. This means 90% of the data falls somewhere inside 2 standard deviations of the mean, so the interval is slimmer.

2. (61.12, 72.88) inches. We are 95% confident that the true mean height of Vanderbilt students is somewhere inside this interval.

3. I’m kind of confused about when you can use confidence intervals accurately and what conditions the distribution needs to have.

1) Narrower. A 95% confidence interval is associated with data falling within 2 standard errors, and a 90% interval would have less standard errors associated with it. Because 2 times the standard error is added/subracted to µ for the 95% confidence interval, less than 2 times the standard error would be added/subtracted for a 90% confidence interval, resulting in a smaller interval. (2x > (<2)x)

2) 67 – (2*3) = 61 inches

67 + (2*3) = 73 inches

95% confidence interval: 61 to 73 inches

3) More clarification on how to differentiate from sample means and point estimates

1.Narrower, since a lower confidence level captures fewer values.

2.Standard Error = 3/sqrt(30) = .547723, assuming a normal height distribution, 67+/-1.96*.547723 gives the range(65.93,68.07).

3.Is it useful to compute the standard error of parameters besides the mean? If so, how are these metrics computed?

1. Narrower. We know that to calculate the confidence interval we multiply the standard error by the z score associated with the percentile we want. So if we want a confidence interval of 90%, we would use a z score of 1.645. This gives us a smaller range.

Suppose a random sample of 30 Vanderbilt undergraduates is found to have an average height of 67 inches and a sample standard deviation of 3 inches. Construct a 95% confidence interval for the average height of the population of all Vanderbilt undergraduates and interpret its meaning.

2. CI = 67 +/- (1.96)*(3/sqrt(30))

(65.9, 68.1)

3. I didn’t understand this formula given in section 4.2.6: N(mu, sigma/sqrt(n)). I get that we are trying to approximate as normal if we are given the population standard deviation, but I don’t see what this formula is equal to.

1. 90% would be a narrower interval. The wider the interval, the more confident you get. For instance, I can say that 1 very specific value is the mean, but I would have 0% confidence, unless I actually knew the mean. Likewise, I am 100% confident the mean alcohol content is in [0,100].

2. 3/sqrt(30)=SE=.548 95%=67 +/- 2*SE=(65.9 , 67.1)

This means we are 95% confident that the average height of all vandy students is in the interval described.

3. Why does the SE not have anything to do with the population size? it is only concerned with the sample size. If we have a population of 101, and we have a sample size of 100, it seems that we should be able to have alot more confidence. Likewise, if we have a 1,000,000,000 pop and 100 sample, we should not be able to get 95% from this with the same equation.

The key is independence. If the sample observations are independent, then the SE doesn’t depend on population size. That’s usually a *good* thing, since it means we don’t have to sample some percentage of the (typically very large) population.

If the sample observations are dependent (as they would be with a sample of size 100 from a population of size 101), then the rule that SE = sigma \ sqrt(n) doesn’t hold. In that case, yes, we can have more confidence in our estimate of the population mean, but we’d have to factor in the dependence correctly.

1. A 90% confidence interval would be narrower than the 95% confidence interval because if you do not need to have as high of a confidence, your confidence interval can move closer to the mean because more values can be outside the confidence interval.

2. I am 95% confident that the mean height of all Vanderbilt undergraduates is in the interval 65.93-68.07.

1) It would be narrower because in order to have a lower confidence interval, the parameter has to be 90% plausible that it is in the range instead of 95%. So in order to be less certain that the parameter is in the range, the confidence interval has to be smaller.

2) Well this might not be completely accurate because the sample size is only 30 instead of 50 but the confidence interval is (65.92,68.07). This means that we can be 95% confident that these numbers capture the parameter.

3) What happens to the confidence interval when the data is skewed?

We’ll see (later, when we dig into the Central Limit Theorem a bit more) that the population has to be very, very skewed to cause any problem with our confidence intervals–assuming our sample sizes are big enough (50 or more).

1) The 90% confidence interval would be narrower since there would be a lower percent chance that the mean could occupy a tighter range of values. Aka, at the lowest range we are 0% confidence that the sample will equal the mean and at the highest range 100% confident that the sample could occupy any real number. Therefore lower confidence intervals have lower values.

2) Confidence = mean +- 1.96*SE = mean +- 1.96*stdDev./sqrt(n)

= 67 +- 1.96*3/sqrt(30) = (65.8, 68.1)

We are 95% confidant that the actual mean heights of students is between 65.8 inches and 68.1 inches.

3) The derivation of the confidence interval is fairly qualitative. Can you show us a more rigorous proof?

We’ll get a bit more rigorous tomorrow in class. But we’ll still have to take the Central Limit Theorem on faith (for now).

1) Suppose that a random sample of 50 bottles of a particular brand of cough syrup is selected and the alcohol content of each bottle is determined. Let μ denote the average alcohol content for the population of all bottles of the brand under study. Suppose that the resulting 95% confidence interval is (7.5, 9.4). Would a 90% confidence interval calculated from this same sample have been narrower or wider than the given interval? Explain your reasoning.

> Narrower. Take the two extremes: A 0% confidence interval is 1 point (We can never exactly hit 1 point on a continuous distribution) and a 100% confidence interval is the whole number line (We are 100% sure it is a number). This means as our interval magnitude shrinks, the confidence also shrinks.

2) Suppose a random sample of 30 Vanderbilt undergraduates is found to have an average height of 67 inches and a sample standard deviation of 3 inches. Construct a 95% confidence interval for the average height of the population of all Vanderbilt undergraduates and interpret its meaning.

> (61, 73). This means we are 95% sure that the average height falls within this interval.

3) What’s one question you have about the reading?

> Standard error

1. A 90% confidence interval would have a narrower interval. This is because if 95% of the distribution is in the previous interval, this means that the observations are within 2 standard deviations of the mean. So for 90% confidence, the standard deviation from the mean is smaller, meaning that less samples fall into a certain interval that includes the mean.

2. (63, 71) is the 95% confidence interval. This means that there is 95% confidence in the fact that the average of heights of the population of all Vanderbilt undergraduates will fall between this interval.

3. What are the ways to calculate confidence intervals if there is less than 95% confidence?

1. A 90% confidence interval for this sample would have been narrower. This is because a lower confidence level means a lower z score in the equation confidence interval = point estimate +- z * SE. Also, conceptually, a lower confidence level means you are less sure the population mean falls within the given interval which means the interval must be narrower.

2. (65.93, 68.07)

3. Why is it that a confidence level cannot be thought of as a “probability” that a given interval contains the population parameter? Isn’t that essentially what the formula calculates for and what the book demonstrates in figure 4.7?

1) A 90% confidence interval would have a narrower interval than a 95% confidence interval, because the odds of a given data value falling in the interval would be smaller.

2) For a 95% confidence interval: 67in +/- 2*(3in) = (61,73). This means that you could be 95% certain that any given student will be from 61in to 73in tall.

1. A 90% confidence interval would be narrower than the given interval. When one increases the width of the confidence interval, there is an increased chance the population parameter will fall into the interval, resulting in a higher confidence level. When one decreases the width, the opposite happens. There is a lesser chance the population parameter will fall into the interval and there will be a lower confidence level as a result. Thus, decreasing the confidence level will result in a narrower interval.

2. The confidence interval is [61.12, 72.88]. This confidence interval allows us to say “We are 95% confident that the population mean is between 61.12 and 72.88 inches.”

3. Why do we use the number 1.96 to approximate 2 standard deviations when we make our 95% confidence intervals?

1) Narrower. Since 90% is smaller than 95%, the confidence interval does not capture the actual mean as much as the 95%, therefore it is narrower.

2) (64,70) – It means that 95% chance the actual mean to be captured inside the confidence interval each time samples are taken.

3) No question.

1. Suppose that a random sample of 50 bottles of a particular brand of cough syrup is selected and the alcohol content of each bottle is determined. Let μ denote the average alcohol content for the population of all bottles of the brand under study. Suppose that the resulting confidence interval is 95% (7.5, 9.4). Would a 90% confidence interval calculated from this same sample have been narrower or wider than the given interval? Explain your reasoning.

Wider, because than only 5% more of the values would be outside 2 standard deviations of the mean.

2. Suppose a random sample of 30 Vanderbilt undergraduates is found to have an average height of 67 inches and a sample standard deviation of 3 inches. Construct a 95% confidence interval for the average height of the population of all Vanderbilt undergraduates and interpret its meaning.

(67-2*3, 67+2*3) = (61, 73)

this means 95% of the students will be between 61-73in tall.

3. What’s one question you have about the reading?

Why would conditions 2 and 3 on page 154 not matter if you know the standard deviation? It seems like it could be calculated for a distribution for which these conditions aren’t true

If the population is known to be (nearly) normal and we know the population standard deviation, then we don’t need conditions 2 and 3 to construct confidence intervals. Condition 3 comes for free in this case, since a normal population means we won’t have any significant skew. And condition 2 becomes no longer necessary–we can construct confidence intervals even for small samples.

1.) A 90% confidence interval would have been narrower, because the z-value corresponding to a 90% confidence level is smaller than 95%.

2) (67-(1.96*3), 67+( 1.96*3)) = (61.12, 72.88)

3) Is there a way to calculate confidence level if the sample distribution is not normal, but the population is?

Check out the box on page 155: If the population distribution is normal, then the sample distribution has to be normal, too.

1. “95% confident” in the context of a confidence interval means that, in a collection of samples, 95% of constructed intervals around the samples would contain the actual population mean. This definition implies that in a 90% confidence interval, only 90% of the intervals around the sampled points would contain the population mean. If the same samples contain the mean less of the time, necessarily the interval around each sample is smaller.

2. SE = stdv/sqrt(n) = 3 / sqrt(30) = .547

95% confidence interval = mean+- 2*SE = 67 +- 2(.547) = (65.906 , 68.094)

This interval suggests that the actual population mean is 95% certain to fall within these parameters.

3. When are confidence intervals used in a practical setting?

1. A 90% confidence interval calculated from this data would be narrower as 90% of the observations within 2 standard deviations would be a smaller range of values than 95% of the observations.

2. The 95% confidence interval for this data would be (61, 73). This means that of all Vanderbilt undergraduates, 95% of students are between 61 and 73 inches.

3. Some of the information regarding interpreting confidence intervals was a little confusing to me.

1. It would be narrower than the interval given. The greater the range given, the greater chance the alcohol content for a given bottle will fall in that range. Similarly, a smaller range would imply less confidence. Given this, if there is less confidence than that given for our original range, we can assume the new range is smaller.

2. The 95% confidence interval is 67 +- 2*3, or (64-70) inches. What this means, is if we were to take random sample after random sample of Vanderbilt undergraduates and compute each ones 95% confidence interval, about 95% of those would contain the actual mean (mu) within their ranges.

3. We’ve learned about the theoretical quantiles vs. sample mean graph in regards to determining whether something is a normal distribution or not, but it seems pretty… flimsy. I feel like everyone could really interpret those graphs differently or even if you do interpret them correctly, the amount of data could be so misleading that what you observe isn’t whats really going on. Are there any actual hard-and-fast rules about deciding whether or not something constitutes a normal distribution?

One can use linear regression techniques to quantify how close (or far) a probability plot is from linear. There are other, more sophisticated tests, too. See, for instance, the info on this R package: http://cran.r-project.org/web/packages/nortest/nortest.pdf.

Suppose that a random sample of 50 bottles of a particular brand of cough syrup is selected and the alcohol content of each bottle is determined. Let μ denote the average alcohol content for the population of all bottles of the brand under study. Suppose that the resulting 95% confidence interval is (7.5, 9.4). Would a 90% confidence interval calculated from this same sample have been narrower or wider than the given interval? Explain your reasoning.

1.

A 90% confidence interval would be narrower than a 95% interval. The percentages come from the standard deviation curve and predict how likely a result is to be within that range. 90% represents a smaller section of the curve, therefore implying a smaller interval.

Suppose a random sample of 30 Vanderbilt undergraduates is found to have an average height of 67 inches and a sample standard deviation of 3 inches. Construct a 95% confidence interval for the average height of the population of all Vanderbilt undergraduates and interpret its meaning.

2.

x+- Z * (o / (n)^.5)

x+- 1.96 * 3 / (30)^.5

[65.923, 68.074]

What’s one question you have about the reading?

What the difference is between population distributions and sample distributions? Are there multiple kinds of confidence intervals?

1) Narrower. A 90% interval covers a smaller range of sample means from the population mean, and therefore will have a higher lower bound and lower upper bound.

2) 67 (+/-)2*(3/sqrt(30)) = (65.9, 68.1)

This shows that a random sampling of students will have a mean between 65.9 and 68.1 95% of the time.

3) When would you use this? It seems to be mostly useful in working backwards when you already know the population mean, which will always be more precise than any other measurement you could make

1) If we’re less confident that our interval captures the mean, by simple logic we’re casting a smaller net, thus a narrower interval

2) Confidence = point estimate +/- 1.96 * SE = 67 +- 1.96*3/root(30) = (65.8, 68.1)

so if we’re operating at a 95% confidence interval, the heights of the students average between 65.8 and 68.1 inches

3) none

1) The lower confidence interval would mean narrower because there would be a lower % possibility that the mean could occupy a smaller range of values.

2) [65.9, 68.2]

95% confidence that the observed mean heights of students are between 65.9 inches and 68.2 inches.

3) Can you give more examples of using confidence intervals in practice?

1. A lower confidence interval of 90% versus a confidence interval of 95% would have a narrower interval because of the lower percentage. A lower percentage confidence interval means that the interval includes a smaller range of values, so the probability that the population mean falls within that range is also smaller.

2. Confidence is mean +/- (1.96)(SE) which equals mean +/- (1.96)(deviation)/(n)^(1/2)

67 +/- (1.96)(3)/(30)^(1/2) = (65.8, 68.1)

This means that we are 95% confident that the average height of Vanderbilt undergraduate students is between 65.8 and 68.1 inches.

3. Is the distribution of sample means always normally distributed?

A 90% interval would have a narrower interval, because a narrower interval has less chance of including the population mean.

The confidence interval would be (60,73)

I’m not sure what this means in terms of the student population

1.

The confidence interval would be narrower than the given interval. Since you have a decrease in the confidence of whether or not an observation falls within a certain range, it would mean that the size of the range must be narrower. Conversely, a bigger confidence interval would lead to more confidence because the likelihood of the observation falling within the range would be increased.

2.

confidence interval = point estimate +/- 1.96*SE = 67 +/- 1.96*[3/sqrt(30)]

confidence interval = (65.93, 68.07)

3.

Where do you come up with 1.96 for a 95% confidence level?

1) The 90% confidence interval would be narrower because you are “decreasing the size of your net”. You only 90% of the time will the interval included the true mean instead of 95% therefore the interval must be narrower.

2) SE = 3/sqrt(30) = 0.55

Confidence = estimate +/- 2*SE = 67 +/- 2*0.55

( 65.9 , 68.1 )

There is a 95% confidence that the average height of all Vanderbilt undergraduate is greater than 65.9 inches and less than 68.1 inches.

3) Why is the normal probability plot in Figure 4.8 linear?

1. A 90% confidence interval would be narrower than the 95% interval. This is because the greater the size of the interval, the more likely that the population parameter will be in the interval. To use the book’s analogy, the 95% interval would be the size of a large net. The 90% interval would still be a large net, but slightly smaller in size than the 95% interval net. Looking at a normal distribution, the 95% interval covers more area than the 90% interval.

2. The standard error of the sample is calculated by SE=σ/(n)^1/2, so the standard error is SE=3/(30)^1/2=0.548

The 95% confidence interval is calculated by mu+/- 2(SE), so the confidence interval is

=67 +/- 2(0.548)= (65.9,68.1)

3. What is the difference between all the different types of deviations and means? Is there any easy way to figure out which is which?

Having a 90% confidence interval as opposed to a 95% confidence interval means that the range should be narrower. Less random intervals will have the mean inside of the range, so this means that less values are inside the range.

Given: point estimate = 67 inches, n = 30 students, std. dev = 3 inches, confidence level = 95%.

Using a 95% confidence level and a normal distribution table, we can find that z* = 1.65. SE = std. dev/sqrt(n) = 3/sqrt(30). The lower value of the range is the point estimate – z*SE, and the upper range is point estimate + z*SE.

(66.096,67.904).

This means that if you took many 30 student samples, 95% of the time, this interval would contain the mean of the given 30 student sample.

Does the previous example fall apart since it only has a sample size of 30? How does the sample size really effect the accuracy of the confidence interval?

1. Narrower because you are trying to capture less data

2. (65.92, 68.07) this means we are 95% confident that the mean lies within that range.

3. nope

1. A 90% confidence interval would lead to a narrower interval. This is because a lower confidence interval indicates that the mean will have a smaller range. Lower values are caused by lower confidence intervals.

2. Our confidence is equal to 67 +- 1.0735. This provides a confidence range of (65.8, 68.1).

3. How do you determine the normal probability of a graph?

1)it would be narrower since the sample size is 50 so more lenient towards sample’s skew

2)67+-2*3=(61,73)

3)Why is this important?

Suppose that a random sample of 50 bottles of a particular brand of cough syrup is selected and the alcohol content of each bottle is determined. Let μ denote the average alcohol content for the population of all bottles of the brand under study. Suppose that the resulting 95% confidence interval is (7.5, 9.4). Would a 90% confidence interval calculated from this same sample have been narrower or wider than the given interval? Explain your reasoning.

A 90% confidence interval would be narrower. This is clear because you are only actually finding the mean in your range 90% of the time. If you had a wider range (like 95%) then obviously you are more likely (5% more likely) to catch the mean in that range. The narrower your interval, the less likely it contains the mean. This is common sense if you understand that each sample varies with respect to the standard error. And that the confidence intervals take into account for this variability by providing a range in which the mean could fall given the variability of a given sample mean. So, a confidence interval of 90% will definitely be narrower than that of 95% since we are “less confident” the mean will be in that range, and thus the interval covers fewer values (hence our lack of confidence).

Suppose a random sample of 30 Vanderbilt undergraduates is found to have an average height of 67 inches and a sample standard deviation of 3 inches. Construct a 95% confidence interval for the average height of the population of all Vanderbilt undergraduates and interpret its meaning.

I’m using 2 instead of 1.96 because we don’t know if a normal dist fits this.

Ans = (65.9, 68.1)

What’s one question you have about the reading?

Could you please explain what they are talking about when they say (they say this multiple times) Observations in a SRS sample consisting of less than 10% of the population are independent implies sample observations are independent??? Why choose a small sample, how do they even know this sample is independent? No details are given at all… Also, can you further explain why the three conditions needed for a sample distribution to be normal are what they are. What is the reason for them? Thanks!

Also, could you please prove why std dev of sample = std dev of pop / sqrt(n)

The 90% confidence interval would be narrower.

Lowering the confidence narrows the size of your net because a value is less likely to fall in the given interval.

SE=3/sqrt(30)=0.55

Confidence = 67 +/- 2*0.55 (65.9, 68.1)

There is a 95% confidence that the average height of all vandy students is between 65.9 and 68.1 inches.

I got a little confused about how independence works with confidence.

1. It should be narrow. By considering the extreme condition. While the confidence becomes very small, like 1%, then the confidence intervals must be very small in order to meet the requirement. Therefore, the smaller the confidence is, the smaller confidence intervals it should be.

2.It can not be achieved because the sample space is less than 50.

3. Again there is a probability plot on page 151, which says “all of the points

closely fall around a straight line”. However I remember the exercises taken in class which asked us to distinguish whether the pp comes from a normal distribution. Some plots also form a straight line, but the points are few and scattered at both ends while concentrated at middle. Then this plot doesn’t come from a normal distribution.

My question is whether there is a standard value for the proportion of the points at the ends?

1. The interval would be smaller because it is less likely that the calculated value will be within the range.

2. 58.88-61.12

3. it was clear

>>It will be narrower. It is less likely for a value to be in a smaller range, and you would therefore be less confident about it. In order to have 100% confidence for instance, you would need to include all possible values of the population in the interval.

>>Point estimate (67) +/- 2 * stdev (3) = [61, 73] inches. That means we are 95% sure that a randomly selected student will be within 61 and 73 inches tall.

>> I have an issue with the “Informal Definition of the Central Limit Theorem”. I’m a little skeptical that it’s true in all cases that a sample of 50 independent observations is enough to declare something normally distributed.

1. The 90% confidence interval would be narrower because the size of the interval depends on how many points it covers. Therefore, when confidence level drops, the size of the interval decreases accordingly.

2. SE=3/sqrt30=0.55

confidence interval = 67 +/- 1.96*SE = (65.8, 68.1)

3. Is there a formula for calculating confidence interval for arbitary percentage?

A)

To be honest, I do not understand the question. In other words, It is not clear (at least for me). If you mean by “narrower or wider”, is it inside the range (7.5 to 9.4) or out side the range? I would say that 90% confidence interval calculated suppose to be within the range (inside the range) of [7.5 to 9.4) because the range suppose to be represent the 95% of confidence interval. So, it is going to be narrower …. !

B)

–> SE = std/root(n) ===> = 3/root(30) ====> =0.54

–> confidence interval: PE +-2 *SE

===> (high limit) 67 + 2 * 0.54 ===> = 68.09

===> (low limit) 67 – 2*0.54 ===> = 65.92

–> 95% of the mean of vanderbilt students high is from 65.92 to 68.09

C)

1) The relationship between confidence interval and normal distribution.

2) The benefit of knowing confidence interval in real life.

Thanks

Anas

1. It would be narrower. If we are less confident in something, the range must be smaller (less numbers must be included).

2. The 95% confidence interval is (61 in, 73 in). That is, if we pick a Vanderbilt student at random, we can be 95% sure he or she will be between 61 in and 73 in.

3. How did they realize that 95% of a normal population will fall within two standard deviations of the mean?

1. A 90% confidence interval would have been narrower because, using the fishing analogy from the textbook, we would be less confident about catching a fish if we had a smaller/narrower net. If we want to be less sure that the value is in the interval, we narrow the range of possible values.

2. (67 – 1.96(3/sqrt(30)), 67 + 1.96(3/sqrt(30))) = (65.93, 68.07). Based on this, we can be about 95% confident that the average height of the population of all Vanderbilt undergraduates is larger than 65.93 inches but less than 68.07 inches.

3. I have no questions on the reading. I’ve seen a lot of this material in prior classes.

1. A 90% confidence interval would be narrower.

A 95% confidence interval is saying you are 95% certain the true value is within the range. The larger the range for a given variability, the more confident you can be the value will be in the range, up to the entire possible range, where you can be 100% certain.

2. SE = 3/sqrt(30) = 0.5477. 95%int = 67 +- SE or

(65.9, 68.1) inches

From this sample, there is a 95% certainty that the mean height of all Vanderbilt undergraduates students is between 65.9 and 68.1 inches (assuming the Std Dev was a good approximation, which might not be the case with only 30 samples. The range might need to be larger.)

Is there a set formula/methodology to account for possible discrepancies of the sample Std Dev and the population Std Dev from small sample size (like n=20 or n=30)?

1) The 90% confidence interval would be narrower . 90% of the time will the interval included the true mean instead of 95% therefore the interval must be narrower.

2) Confidence =67 +/- 2*3/sqrt(30) = 67 +/- 2*0.55

( 65.9 , 68.1 )

1. The Interval would be smaller because 1 90% confidence interval is a segment that contains 90% of the data. Since 90<95, the 90% interval would be smaller.

2. The 95% confidence interval is (61,73) via the equation, point estimate +/- 2*SE and that means that 95% of the vanderbilt population falls between those two heights.

3. if you are trying to estimate the sample mean is it better to take a point estimate of a large sample or many point estimates of smaller samples and making a sampling distribution as described in the reading?

1) A 90% confidence interval would be wider because there is less confidence and therefore it would contain more values.

2) Confidence interval:

I = mean +- 1.96SE = mean +- 1.96(SD/sqrt(n))= 67 +-1.96(3/sqrt(30) )

I= (65.8, 68.1)

3) Applications of confidence intervals

1. From Figure 4.9 in the textbook, you can see that a 90% confidence interval calculated from this same sample would be narrower than the 95% confidence interval. This is because we have a smaller probability of including the mean in our interval when we say that only 90% of the range of values are included rather than 95%.

2. Standard Error = SE = sigma/sqrt(n) = 3/sqrt(30) = 0.54772

Point Estimate +/- 2*SE = 67 +/- 1.0954

Confidence interval: (65.9, 68.1).

The average height of Vandy students falls between 65.9 and 68.1 inches and this is true with 95% confidence.

3. Isn’t it a little strange to have the value for the standard deviation of our data given to us in these problems ahead of time?

1. The 90% interval would be narrower since only 90% of means for new sets of data would fall in the interval.

2. 67+\- 2( 3 / (30^ .5)) = (65.9046, 68.0954)

3. I do not have questions.

The confidence interval would have been wider at 90%. It would be narrower because we are less confident that the standard error is correct for our sample.

(67-3*1.96,67+3*1.96) = (61.12, 72.88)

This means we are 95% confident that the average student will be between 61.12 inches and 72.88 inches.

1.Thinner bc if you’re less certain we are capturing less data.

2. 1.96*3 ± mean

[61,73]

3. When do we use which formulas?