Here's your penultimate reading assignment. Read **Section 6.1 **in your textbook and answer the following questions by 8 a.m., Friday, April 13th. 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 10 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). If the sample size had been 50 instead of 10 but the sample mean and sample standard deviation were the same, would the 95% confidence interval constructed from this larger sample have been narrower or wider than the given interval from the smaller sample? Explain your reasoning.
- What's one question you have about the reading?

1. Narrower, since the T curve at DF 9 is wider

2. (64.856,69.144) We must assume the data is independent and nearly normal.

3. None

1. The confidence interval for the larger sample would be narrower because increasing n decreases the standard error and because when working with z you multiply by 1.96 for a 95% confidence interval, which is less than 2.10.

2. We have to assume that this is a random sample of less than 10% of the population (which it is). We also need to assume that the height of Vanderbilt students is normally distributed.

x +/- t9 * SE = 67 +- 2.10 * 3/sqrt(10) = (65.008, 68.992)

3. I'm still confused as to what degree of freedom means. I understand that it determines what your distribution looks like and that n-1 is how many degrees of freedom you have, but what exactly is it?

1. The confidence interval would be narrower because the larger sample size would decrease the standard error.

2. 95% confidence interval: 64.854 - 69.146. To construct this confidence interval, you have to assume that the population is nearly normal and the observations are independent.

1) Narrower since the standard error becomes smaller and the t confidence interval factor also become smaller for larger samples.

2) df = 9, t factor = 2.26 for 95% confidence interval.

SE = 3/sqrt(10) = 0.95. Interval = 67 +/- (2.26)(0.95) = (64.85,69.15)

Assumptions:

1)Observations are independent

2)Observations comes from a nearly normal distribution.

3)No question.

1) The confidence interval will become narrower, since the standard error is inversely proportional to sample size and the width is proportional to the standard error in the t model.

2) If we assume the distribution is nearly normal and the samples are independent, we can use the t-distribution.

SE = 3/10^.5 = .95

With df = 9, two tails, and p = .050, using table C.2 we get t* = 2.23

95% CE = 67 +- 2.23*.95 =(64.88, 69.12)

3) Can we use t-values for intermittent values of p using some kind of formula?

1.

The 95% confidence level would have been narrower. This is because increasing the sample size n would decrease the standard error size. This would in turn make your confidence interval range narrower.

2.

confidence interval = xbar +/- t*df x (SE)

SE = 3/sq(10) = 0.95

df = n-1 = 9 -> t* = 2.26 in table C.2

Confidence Interval = 67 +/- 2.26*0.95 = (64.85, 69.15)

3.

Can you explain the degrees of freedom concept and where it came from?

1) With a larger sample size, the 95% confidence interval would be larger, because the standard error would be smaller.

2) df = 10-1 = 9

t = 2.26

67 +/- 3/sqrt(10)*2.26 = (64.86, 69.14)

Assume the observations are independent and the population is nearly normal.

1) The CI would become smaller because there is a large number of samples, so you can be more confident.

2) Use t-distribution( meets criteria)

SE = .95

With df = 9, p = .050, t = 2.23

95% CI =(64.9, 69.1)

3) None

1) It would probably be narrower since a sample size of 50 is large enough to use the normal distribution, which is narrower than the t-distribution. Therefore, at the same 95% confidence interval, the interval would be narrower. Furthermore, the SE would also decrease, resulting in a narrower confidence interval.

2) n=10, df=9, xbar=67, s=3

Conf. Interval: 67 +- (2.262)*(3/sqrt(10))

= (64.854, 69.146)

Assumptions: 10 observations are independent and are from a normal population.

3) No questions.

1. Narrower. the standard error and sample size are inversely proportional. Increasing the sample size would then decrease the standard error which correlates to a narrower confidence interval.

2. Assume nearly normal distribution, independent samples

SE = 3/10^.5 = .95, df = 9, 2 tails, p = .050

Table C.2: t* = 2.23

95% CE = 67 +/- 2.23*.95 = (64.88, 69.12)

3. What if we used a small sample size but ran the test/survey several times. Do we now consider the sample size to be large again?

1) It would be narrower

2) That the observations are independent and the population distribution is nearly normal

3) None

1) The confidence interval will become narrower. Because in the t model, the standard error is inversely related to sample size and the width is proportional to the standard error.

2)

SE = 3/10^.5 =0 .95

df = 9, two tails, p =0 .05, t* = 2.23

95% CE = 67 +- 2.23*.95 =(64.88, 69.12)

1) Narrower. The std error decreases as the sample size increases. A smaller std error would lead to a narrower confidence interval.

2) Since we do not know the population mean, we must assume that it is the same as the average of our sample.

3) How are we supposed to verify that the variables are not related to each other? Would not fully evaluating all the variables beforehand lead to inaccurate results? Can we see a demo of how to construct this? The book just kinda assumes we have the data already.

1. It would be narrower because if you were still using degrees of freedom n=50 would have a dof = 49 while n=10 only as a dof = 9. The greater the dof, the closer to normal the curve is.

2. n = 10

dof = 9

u = 67

s = 3

SE = s/root(n) = 3/root(10) = .949

t* = 2.252

67 ± 2.252*.949

(64.86, 69.14)

3. How did statisticians come up with this stuff?

1. Due to the fact that the sample size is larger, the confidence interval will probably be smaller from extrapolation of the central limit theorem. The larger the sample size, the more normal, and closer in to the mean the samples are (from Figure 6.2).

2. 67+/- 1.96*3/sqrt(10) = 67 +/- 1.86 = (65.14, 68.86)

An assumption is that the undergraduates were a relatively accurate representation of all students, also with an accurately defined variance.

3. Everything seems good, now. I'll probably have questions later, though.

1. The confidence interval will be smaller. This because you can be more accurate with a higher sample size.

2. (64.9, 69,1). We assume it is normal.

3. The reading was straight forward

1. As the sample size gets larger, the confidence interval becomes narrower. Not only does the standard error get smaller because it is divided by the square root of the sample size, the t distribution is narrower because there is now 49 degrees of freedom rather than 9.

Suppose a random sample of 10 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. What assumptions are required to construct this confidence interval?

2. Assuming the observations are independent and the distribution is nearly normal...

d.o.f=10-1=9

t*=2.26 (two-tailed)

SE=3/sqrt(10)=.95

Confidence interval: ( 64.856, 69.144 )

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

It was unclear to me whether to use one or two tails for the confidence interval.

1. With a bigger sample size will come a smaller standard error. A smaller standard error will lead to a smaller confidence, Therefore a sample size of 50 will have a smaller confidence interval than a sample size of ten.

2. [64.86, 69.14] The assumptions necessary are that the observations are independent and the population distribution is nearly normal.

3. What is the deal with t scores... very confused..

1) The interval would be narrower. When finding SE we are dividing by n, so larger n will give smaller SE and thus narrower interval.

2) (64.86, 69.14). We need to make sure the observations are independent and nearly normal.

3) What do "degrees of freedom" actually mean?

1. Narrower, since the standard error would be smaller.

2. SE = s/sqrt(n) = 0.9487. For a 95% confidence interval, t*18 = 2.10. So, the interval is (65.01,69.00). We must assume the observations are independent and nearly normal.

3. How did the t distribution get its name?

The interval would now be narrower. Since you are not changing the confidence level, the sample mean, or the sample standard deviation, this means that only the sample size and the t-value is changing. The standard error, in this case, will only be affected by the sample size change. Since the size increases, the standard error goes down. The same applies with the t-value. t for 9 degrees of freedom in a 95% confidence interval is 2.26, while for 49, it is 2.01. Since both t* and SE go down, the interval narrows.

(64.86,69.14). We're assuming the observations are independent and that the population distribution (for height) is nearly normal, and that a t distribution will fit the sample well.

Why do the degrees of freedom increase as it becomes more normal, and does this mean a normal distribution has infinite degrees of freedom?

1) Narrower.

With 10 observations, df=9, and t* = 2.26. Holding the sample mean and standard deviation constant, the standard error of the small sample has a margin of error of .715 times the standard deviation, and the large sample has a margin of error of .277 times the standard deviation, resulting in a narrower interval.

2) Assume independent observations and nearly normal data. 10 observations => df = 9, t* = 2.26. SE = 3/(sqrt(10)) = .949

67-2.26(.949) = 64.855 inches

67+2.26(.949) = 69.145 inches

95% CI: 64.855 inches < mu < 69.145 inches

3) Just need more practice with the t-table

1. A higher sample size means a higher number for df (degrees of freedom), which means a higher t-value, thus a wider confidence interval.

2. SE = .949

t(df) = 2.68

Interval = (64.46, 69.54)

Assumptions include that the data is random and nearly normal, and that the Vanderbilt undergraduate population consists of more than 100 people.

3. Would using t-scores for values above n = 50 give data similar to the usual confidence intervals?

1. Narrower. As the sample size gets larger the degrees of freedom would get smaller making the 95% confidence interval narrower.

2. We must assume the distribution is nearly normal. The SE = 0.95 and df= 9 which leads to a t* of 2.26. 95% CI = 67 +/- 2.26*0.95 = (64.85, 69.15)

3. The section on a difference of two means was a little confusing to me.

1. The 95% confidence interval would be smaller because increasing the sample size decreases the SE. A smaller standard error leads to a narrower confidence interval.

2. CI: (64.86,69.14)

3. Why does the t-table go up past 50 if at n=50 you can use a standard normal probability table?

1) The confidence interval constructed from this larger sample would have been about the same size as the given interval from the smaller sample. For the smaller sample size, we will be using the t-distribution, and for the larger sample size, we will be using the normal distribution, but both should yield similar results.

2) (63.92, 70.08)

3) The t-distribution seems kind of magical. What are the nuts and bolts behind it?

1. As "n" increases, two things happen: standard error decreases and the degrees of freedom increases. As standard error decreases, the confidence interval will become more narrow. As the degrees of freedom increases, the multiplication factor "t" will decrease, also causing the confidence interval to become narrower. Overall, the confidence interval becomes narrower.

2.

n = 10

xBAR = 67 inches

s = 3 inches

α = 0.05

df = n - 1 = 9

tdf = 2.262

SE = s/sqrt(n) = 0.9487

The confidence interval is thus set up as:

XBAR +/- tdf*SE

67 +/- 2.262*0.9487

67 +/- 2.1460

(64.85, 69.15)

It is assumed that there are more than 100 Vanderbilt undergraduates (a reasonable assumption) and thus our sample composes less than 10% of the target population. It also assumes that the height of Vanderbilt undergraduate students is a nearly normal distribution.

3. Is there a surefire way to identify whether a distribution is nearly normal, having only a small amount of data to examine in the first place?

1) Narrower. As sample size increases the degrees of freedom goes up making the t-value decrease. This t distribution describes the bell shape of the graph. The higher you go in degrees of freedom with decreasing t value the more normal the bell looks. The smaller sample size > low degree of freedom > high value brings a wider interval. Our larger sample size provides a narrower interval.

2) df = 10-1 = 9

t = 2.262

67+/-(2.262)(3/root(10)) = (64.85 , 69.15)

We assume the observations are independent of each other and that they come from a nearly normal distribution

3) I am not sure why we don't just stick with the normal distraction.

1) Narrower because since the sample size is larger the degree of freedom will be smaller.

2) 69 +- 2.262(.3) = [68.32, 69.67] The assumptions that you have to make are that the observations are independent and that they come from a nearly normal distribution.

3) How are the graphs for the t distributions created?

1. The confidence interval will be narrower because SE will decrease when sample size increases.

2. SE = 3/EXP(0.5) = 0.95

t = 2.23

95% confidence interval = 67 +/- 2.23*0.95 = (64.88, 69.12)

3. What will be a good condition to apply t distribution.

1. Smaller, t*(df=49) is smaller than t*(df=9) and SE would decrease if n increased. Since the confidence interval is μ+/- t*(SE), the interval would decrease

2.

Assumptions:

1. Observations are independent.

2. population is nearly normal.

t*(df=9, .05)=2.26

SE = 3/(10^.5)

67+/-2.26*3/(10^.5)

(64.86, 69.14)

1. Suppose that a random sample of 10 bottles ... would the 95% confidence interval constructed from this larger sample have been narrower or wider than the given interval from the smaller sample? Explain your reasoning.

With the larger sample size comes a smaller confidence interval. As you test more and more of the population, you are progressively less and less likely to have sampled many non-representative data. This sureity increase leads us to be more confident about our estimates, leading to a smaller confidence interval.

2. Suppose ... Construct a 95% confidence interval for the average height of the population of all Vanderbilt undergraduates. What assumptions are required to construct this confidence interval?

mean = 67, std dev = 6, n = 10

T* = 2.262

mean +- T* (s/(n^.5))

Interval:

[64.853, 69.1460]

Assumptions:

Data is from random sample of the population

Samples are independent

Population is roughly normally distributed

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

Can't T-tests be used for large samples within a population you don't know the standard deviation of? It would seem

1) narrower because the SE term of the confidence interval will get smaller.

2) (65.01,68.99) the assumptions we must make are that observations are independent and that the population is normally distributed.

3) none

1. The confidence interval will be narrower. Because when t is small, the both of tails are longer than the corresponding tails of a normal distribution. Therefore, the confidence interval of a normal distribution is narrower.

2. We have to assume that the distribution is nearly normal and the samples are independent.

SE = 0.95

df = 9, according to the table we get the confidence interval(64.88, 69.12)

3. According to the table on p245, the duration of two probability is large. Whether I can say that the estimation in t table is not as accurate as normal distribution?

Suppose that a random sample of 10 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). If the sample size had been 50 instead of 10 but the sample mean and sample standard deviation were the same, would the 95% confidence interval constructed from this larger sample have been narrower or wider than the given interval from the smaller sample? Explain your reasoning.

Narrower because a larger sample implies we are more confident in our confidence interval since we have a larger percent of the population in our sample. Also the SE is inversely related to conf int.

Suppose a random sample of 10 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. What assumptions are required to construct this confidence interval?

We must verify independence, SRS, and normal distribution.

What's one question you have about the reading?

No questions. Reading was clear.

1) The CI would will become narrower because the standard error decreases when the sample size increases. When the standard error decreases the width of the CI decreases also in the t-model.

2) Assuming the samples are independent and the distribution is nearly normal, the t-distribution can be applied.

SE = 3/10^(1/2) = 0.95

df = 9

p = 0.050

two tails

t* = 2.23

95% CI = 67 +/- 2.23(0.95) = (64.88 , 69.12)

3) Why are the CI different for t-distribution?

1. It would have made the 95% confidence interval narrower, because if the sample size is larger, it makes the SE smaller, which in turn makes the value being added or subtracted to/from the sample mean smaller.

2. We assume that there are no extreme outliers and that the observations are independent since they represent less than 10% of the Vanderbilt undergraduate population. SE = 3/sqrt(10) = 0.949. Since this is a 95% confidence interval, we use the column of the t-table where two tails is 0.050 and df = 10 - 1 = 9. This corresponds to a t* = 2.26. 67 + 2.26 * 0.949 = 69.14, and 67 - 2.26 * 0.949 = 64.86, so the confidence interval is (64.86, 69.14).

3. I have seen this material before, and so I don't have any questions.

1) The confidence interval will become much more narrow, since the standard error is inversely proportional to the sample size. The width is proportional to the standard error in the t model.

2)SE = 3/10^.5 = .95

With df = 9 and p = .050, using table C.2 we get t* = 2.23

95% CE = 67 +- 2.23*.95 =(64.88, 69.12)

3) How often can we use t values instead of t values when attempting to answer these questions?

1. narrower because a larger sample size means a more accurate representation of the population, so we're more confident that the sample mean is close to the population mean.

2. assuming the samples are independent

SE = 3/sqrt(10) = 0.95

df = 9

p = 0.05

t*df = 2.262

67 +- 2.262*0.95 = (64.89, 69.11)

A)

It is going to be narrower. As known from an equation, As sample size increase, the standard error decreases. And, when standard error decrases the width decreass too. For that, when sample size increases the width decreases. So then, we are going to have narrower width.

B)

===> SE = 3/10^.5 = .95

==> 95% CE = 67 +- 2.2*.95 ===> (64, 69)

I would assume that the samples are independent. Also, the distribution is a normal distribution. These are the most important assumations

C)

Thanks

1.In the formula for calculating confidence intervals you are dividing by a larger number, so you will add a smaller number on either side of the sample mean-- thus the interval will be narrower

2.67+/-1.96*3/sqrt(10)=(65.1406,68.8594)

3.One of the criteria for using the methods set out in 6.1 is that the distribution be nearly normal. What is the easiest way to verify this?

1. we would expect the confidence interval to be narrower because we are taking a larger sample size, which would decrease the size of standard error.

2. we assume that our data is independent and nearly normal. the SE will be 3(10^1/2) = .95. CI is 65 +- 2.23*.95

1. Smaller, as the Standard Error will be lower, and that is in the denominator of the z score equation.

2. 67+/-3(2.26)=(60.2,73.8). Must assume that we have independence in observations and a nearly normal population.

3. Is there a mathematical formula for t values, or do we need to always use the tables.

1. The confidence interval would have been narrower because if the number of samples goes up, the standard error goes down, and the interval shrinks as a result.

Suppose a random sample of 10 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. What assumptions are required to construct this confidence interval?

2. 67 +/- 2.262 * 3/(sqrt(10)) => 67 +/- 2.146 => (64.854, 69.146).

Here we are assuming that the samples are independent and that the population distribution is nearly normal.

3. How is it at all feasible to make an estimate based on say two samples, so a degree of freedom of 1? I guess I just dont believe that any sort of estimate would be realistic.