Objectives

Previously…

Central Limit Theorem (CLT)

CLT says that the sample mean (or sum) of many independent and identically distributed random variables approaches a normal distribution, regardless of the original distribution.

CLT Conditions

• Independence – Sample values must be independent
• Identical Distribution – Variables should be from the same distribution
• Finite Variance – The population must have a finite variance
• Large Sample Size – A larger sample size improves approximation

Central Limit Theorem for the Sample Mean

When we collect a sufficiently large sample of \(n\) independent observations from a population with mean \(\mu\) and standard deviation \(\sigma,\) the sampling distribution of \(\bar{x}\) will be nearly normal with \[\text{Mean} \longrightarrow \mu \text{ and } \text{Standard Error} \longrightarrow SE = \frac{\sigma}{\sqrt{n}}.\]

Evaluating the two conditions required for modeling \(\bar{x}\)

Two conditions are required to apply the Central Limit Theorem for a sample mean \(\bar{x}:\)

• Independence. The sample observations must be independent. The most common way to satisfy this condition is when the sample is a simple random sample from the population.

• Normality. When a sample is small, we also require that the sample observations come from a normally distributed population. We can relax this condition more and more for larger and larger sample sizes. This condition is obviously vague, making it difficult to evaluate, so next we introduce a couple rules of thumb to make checking this condition easier.

General rule for performing the normality check

Note, it often takes practice to get a sense for whether or not a normal approximation is appropriate.

• \(\mathbf{n < 30}:\) If the sample size \(n\) is less than 30 and there are no clear outliers in the data, then we typically assume the data come from a nearly normal distribution to satisfy the condition.
• \(\mathbf{n \geq 30}:\) If the sample size \(n\) is at least 30 and there are no particularly extreme outliers, then we typically assume the sampling distribution of \(\bar{x}\) is nearly normal, even if the underlying distribution of individual observations is not.

Normality Assesment (1/2)

Consider the four plots provided that come from simple random samples from different populations.

Are the independence and normality conditions met in each case?

Histograms of samples from two different populations.

Normality Assesment (2/2)

• The first sample has fewer than 30 observations, so we are watching for any clear outliers. With no clear outliers, the normality condition can be reasonably assumed to be met.

• The second sample has a sample size greater than 30 and includes an outlier. This is an example of a particularly extreme outlier, so the normality condition would not be satisfied.

The t-distribution (1/2)

Comparison of a \(t\)-distribution and a normal distribution.

The \(t\)-distribution is always centered at zero and has a single parameter: degrees of freedom. The degrees of freedom describes the precise form of the bell-shaped \(t\)-distribution. In general, we’ll use a \(t\)-distribution with \(df = n - 1\) to model the sample mean when the sample size is \(n\).

The t-distribution (2/2)

The larger the degrees of freedom the more closely the \(t\)-distribution resembles the standard normal distribution.

Case Study I: Mercury content in Risso’s dolphins

We will identify a confidence interval for the average mercury content in dolphin muscle using a sample of 19 Risso’s dolphins from the Taiji area in Japan.

Are the independence and normality conditions satisfied for this dataset?

• The observations are a simple random sample, therefore it is reasonable to assume that the dolphins are independent.
• The summary statistics do not suggest any clear outliers, with all observations within 3 standard deviations of the mean.
• Based on this evidence, the normality condition seems reasonable.

Case Study I: One-sample t-interval (1/2)

One-sample t-intervals

\[ \begin{aligned} \text{point estimate} \ &\pm\ t^*_{df} \times SE \\ \bar{x} \ &\pm\ t^*_{df} \times \frac{s}{\sqrt{n}} \end{aligned} \]

• We plug in \(s\) and \(n\) into the formula: \(SE = \frac{s}{\sqrt{n}} = \frac{2.3}{\sqrt{19}} = 0.528.\)
• The degrees of freedom is easy to calculate: \(df = n-1 = 19-1 = 18.\)
• For a 95% confidence level, we find the cutoff where the upper tail is equal to 2.5%: \(t^*_{18} = 2.1009.\) The area below \(t^*_{18} = -2.1009\) will also be equal to 2.5%.

Using R to find the critical \(t^*\) with \(df=18\)

cl <- 0.95 # confidence level
lt <- (1-cl)/2 # lower tail probability
df <- 18 # degrees of freedom
qt(lt, df) # t-star

## [1] -2.100922

Case Study I: One-sample t-interval (2/2)

\[ \begin{aligned} \bar{x} \ &\pm\ t^*_{18} \times SE \\ 4.4 \ &\pm\ 2.1009 \times 0.528 \\ 4.4 \ &\pm\ 1.1093 \\ \end{aligned} \] \[(3.29,5.51)\]

We are 95% confident the average mercury content of muscles in Risso’s dolphins is between 3.29 and 5.51 \(\mu\)g/wet gram, which is considered extremely high.

Case Study II

Every year, the US releases to the public a large data set containing information on births recorded in the country. This data set has been of interest to medical researchers who are studying the relation between habits and practices of expectant mothers and the birth of their children. We will work with a random sample of 1,000 cases from the data set released in 2014.

Here are four examples in the data set.

Case Study II: Baby Weights - Smoker vs Non-Smoker

We would like to know, is there convincing evidence that newborns from mothers who smoke have a different average birth weight than newborns from mothers who don’t smoke?

Here is the summary statistics for the dataset.

Case Study II: CLT Conditions

Conditions:

• The data come from a simple random sample, the observations are independent, both within and between samples.
• Both groups over 30 observations, we inspect the data for any particularly extreme outliers and find none.

Since both conditions are satisfied, the difference in sample means may be modeled using a \(t\)-distribution.

Case Study II: Examining the Distributions (1/2)

The top panel represents birth weights for infants whose mothers smoked during pregnancy. The bottom panel represents the birth weights for infants whose mothers who did not smoke during pregnancy.

Case Study II: Examining the Distributions (2/2)

Case Study II: One Sample t-test (1/4)

Consider one group (smoking) from the data. It is known that a newborn baby has an average weight of \(7.5\) lbs. We want to test whether the average weight for the smoking group is less than the average using a one sample t-test.

Is the data (smoking group) a convincing evidence to support the claim of the average weight to be less than \(7.5\) lbs?

Case Study II: One Sample t-test (2/4)

• Null Hypothesis \(H_0\): The average weight of the smoking group is \(7.5\) lbs. \[\mu = 7.5\]
• Alternative Hypothesis \(H_A\): The average weight of the smoking group is less than \(7.5\) lbs. \[\mu < 7.5\]
• The null value is \(\mu_0 = 7.5\). The point-estimate is \(\bar{x} = 6.68\) and the sample standard deviation is \(s = 1.5966\).
• We set the significance value \(\alpha = 0.01\).

Case Study II: One Sample t-test (3/4)

• Compute the standard error \[ \begin{aligned} SE & = \frac{s}{\sqrt{n}} \\ & = \frac{1.5966}{\sqrt{114}} \\ SE & = 0.1495 \end{aligned} \]
• Compute the T statistic \[ \begin{aligned} t & = \frac{\bar{x} - \mu_0}{SE} \\ & = \frac{6.68 - 7.5}{0.1495} \\ t & = -5.4850 \end{aligned} \]

Case Study II: One Sample t-test (4/4)

• Degrees of freedom is \(df = n - 1 = 114 - 1 = 113\).
• The p-value is \(1.27 \times 10^{-07} \approx 0\).

Using R to find the p-value

df <- 113 # degrees of freedom
t <- -5.48495 # test statistic
pt(t,df) # p-value

## [1] 1.278671e-07

Conclusions:

• Since the p-value is less than significance value of \(0.01\) (the p-value is really small), we can conclude that the data is a strong evidence that the average weights for the smoking group is not equal to \(7.5\) lbs.
• Since the T statistic is negative, we can say that the average weights is less than the null value.

Activity: Determine Confidence Intervals for One Mean

1. Make sure you have a copy of the M 3/24 Worksheet. This will be handed out physically and it is also digitally available on Moodle.
2. Work on your worksheet by yourself for 10 minutes. Please read the instructions carefully. Ask questions if anything need clarifications.
3. Get together with another student.
4. Discuss your results.
5. Submit your worksheet on Moodle as a .pdf file.

References