We discuss inference for the mean for numerical data in this chapter. The frame work and ideas are like that of chapter 5 (Inference for proportion), but we introduce and use a new t-distribution.

• The sampling distribution for \(\bar{x}\)

• t-distribution

• Find probability under t-distribution (for P-value)

• Find critical t-value for given probability of tail (for \(𝑡_{\frac{\alpha}{2}})\)

• Confidence interval for population mean

• One sample mean t-tests

Central Limit Theorem for the Sample Mean

For random samples of large size n (\(𝑛\ge 30\)) from a population with mean 𝜇 and standard deviation 𝜎, the sampling distribution of the sample mean \(\bar{x}\) approaches a normal distribution with the mean equal to the population mean 𝜇 and the standard deviation equal to the population standard deviation divided by the square root of the sample size: that is, \(\bar{x}\) is approximately \(𝑁(\mu, \frac{\sigma}{\sqrt{n}}).\)

For a population with normal distribution, no matter what sample size is, \(\bar{x} \sim 𝑁(\mu, \frac{\sigma}{\sqrt{n}})\)

Use a Webapp https://istats.shinyapps.io/sampdist_cont/

1. choose a (continuous) population – the graph on the right displays the probability distribution, the mean and standard deviation. Try

   1. Real population data (show data)
   2. Skewed (then chose skewness)
   3. Bell-shaped
   4. Build your own

2. choose sample size 𝑛 and the number of simulations

3. Then click “Draw samples” (see next slide for the result)

1. The conditions for CLT to be true:

   • The sample observations must be independent: the sample is a simple random sample from population;
   • For any population if the sample size is at least 30 and there are no particularly extreme outliers;
   • Or if the population is normal then the sample size can be smaller than 30

2. Use CLT (for inference about population mean):

   • Use \(\bar{x}\) as the point estimate for \(\mu\);

   • Construct the confidence interval in the similar way as we do in the proportion case: \(\bar{x} \pm 𝑧_{\frac{\alpha}{2}} \times 𝑆.𝐸.\) , where \(𝑆.𝐸.= \frac{\sigma}{\sqrt{n}} \approx \frac{s}{\sqrt{𝑛}}\) if 𝜎 is unknown;

3. For hypothesis testing, use the same framework and terms as in the case of proportion, use z test statistic

Why we need to use t-distribution.

• The sampling distributions of sample means is nearly normal for any sample size if the population distribution is nearly normal. While this is helpful special case, it is usually difficult to verify normality.

• When the standard deviation is unknown (almost always), and the sample size is small, the CLT cannot be applied. Or even when the size is large, we still have error if we use 𝑠 for 𝜎.

Introducing t-distribution

• For small sizes, and when \(\sigma\) is unknown, keeping the intendance condition, then for sample mean \(\bar{x}\) , the variable \(𝑇=\frac{\bar{x}−\mu}{𝑠/√𝑛}\) is a Student’s t-distribution with 𝒏−𝟏 degrees of freedom, shortened as: \[𝑻\sim𝒕_{𝒅𝒇}\]
• The t-distribution was introduced by British chemist W.S. Gosset who first published his work in 1908 under the pseudonym “Student”.

• The t distribution is \(\color{blue}{\text{centered at 0}}\), it also has a bell shape (therefore symmetric, like the standard normal (𝑧) distribution), but \(\color{blue}{\text{its tails are thicker than that of the standard normal model}}\).

• Extra thick tails are helpful for resolving our problem with a less reliable estimate the standard error (since n is small).

  

• The t- distribution has an associated parameter: degrees of freedom, df, 𝒅f=𝒏−𝟏.

• The smaller (larger) df, the thicker (thinner) the tails.

• As \(df\) increases (i.e. n increases) , the t distribution approaches the standard normal distribution.

x = seq(-6, 6, length.out = 1000)
plot(x, dnorm(x), type = "l", ylab = "Density") #standard normal
lines(x, dt(x, df=1), col = "red", lty=2) #t(df=1)
lines(x, dt(x, df=2), col = "blue", lty=3) #t(df=2)
lines(x, dt(x, df=10), col = "orange", lty=5) #t(df=10)

Solution

• \(SE = \frac{s}{\sqrt{n}}= \frac{2.3}{\sqrt{19}}= 0.527\)

• for \(df\)= n-1=19-1 = 18, and 95% confidence level, \(t_{0.025}= 2.1009\)

qt(0.025,18,lower.tail = FALSE)
qt(0.975,18)

• \(M.E = t_{\frac{\alpha}{2}} \times S.E= 2.1009 \times 0.527 =1.1072\)

• \(CI: \bar{x} \pm t_{\frac{\alpha}{2}} \times S.E = 4.4 \pm 2.1009 \times 0.527= (3.2928,5.5072)\)

Practice: Find 90% CI using sample with n=25, \(\bar{x}= 0.287, s=0.069\).

(Reminder: CI:\(\bar{x} \pm t_{\frac{\alpha}{2}} \times S.E\))

• \(SE = \frac{s}{\sqrt{n}}= \frac{0.069}{\sqrt{25}}= 0.0138\)

• For \(df= n-1=25-1 = 24\), and 90% confidence level, \(t_{0.05}= 1.71\)

qt(0.05,24, lower.tail = FALSE)
qt(0.95,24)

• \(M.E= t_{\frac{\alpha}{2}} \times S.E= 1.71 \times 0.0138\)

• \(CI: \bar{x} \pm t_{\frac{\alpha}{2}} \times S.E = 0.287 \pm 1.71 \times 0.0138= (0.2634,0.3106)\)

Set up the null and alternative hypotheses. (Similar to that for proportion)

What are the hypotheses for testing for the mean of college student heights different from 66.5 inches?

1. \(H_0: \mu = 66.5\) \(H_a: \mu \ne 66.5\)

2. \(H_0: \mu = 66.5\) \(H_a: \mu > 66.5\)

3. \(H_0: \mu = 66.5\) \(H_a: \mu < 66.5\)

4. \(H_0: \mu \ne 66.5\) \(H_a: \mu > 66.5\)

• With \(\alpha\) =0.05, \(df\)=9, the critical t-value is \(t_{0.05}\)=1.833113

• The rejection region is

  \(R.R.=\{𝑇≤−𝑡_{0.05}\}=\{𝑇≤− 1.833\}\) (for left-sided}

• Our t-test statistic is −2.14, is in the rejection region.

• So we have the same decision: We have strong evidence to reject \(𝐻_0\) and support \(𝐻_𝑎\). That is, the water of that lake is within safety level.

• With \(\alpha\)=0.05, \(df\)=14, the critical t-value is \(t_{0.05}\)=1.7613

• The rejection region is

  \(R.R.=\{𝑇\ge 𝑡_{0.05}\}=\{𝑇\ge 1.7613\}\) (for right-sided}

• Our t-test statistic is 3.458, is in the rejection region.

• So we have the same decision: We have strong evidence to reject \(𝐻_0\) and support \(𝐻_𝑎\). That is, the water in the lakes is non-acidic(PH>6).

• With \(\alpha=\frac{0.05}{2}\) =0.025, \(df\)=99, the critical t-value is \(t_{\frac{0.05}{2}}= t_{0.025}\)=1.984

• The rejection region (for two-sided} is

  \(R.R.=\{|𝑇| \ge −𝑡_{0.025}\}=\{|𝑇|≤ -1.984 \hspace{0.2cm} or \hspace{0.2cm}𝑇\ge 1.984\}\)

Our t-test statistic is 2.3734, is in the rejection region, so we have the same decision: We have strong evidence to reject \(𝐻_0\) and support \(𝐻_𝑎\). That is, the data provides strong evidence that the average running time in 2017 is different from the 2006 average.

Note: Rejection region depends on the significance level and the degree of freedom, the smaller n allows larger R.R.

• Compare \(t_{0.025}=1.984\), (\(df\)=99), with \(z_{0.025}\)=1.96.

• Compare \(t_{0.05}\)=1.833 (\(df\)=9) ,\(t_{0.05}\)=1.761, (\(df\)=14), with \(z_{0.05}=1.645\).

For the previous example, for two-sided hypothesis testing

Hypotheses

\(𝐻_0: \mu = 93.29, 𝐻_𝑎: \mu \ne 93.29\)

\(n\)=100, \(\bar{x}=97.32\), \(s\)=16.98, \(\mu_0=93.29\) (null value)

\(S.E = 1.698\), \(t_{0.025}\)=1.984 (with \(df\)=99) (for \(\alpha\)=0.05).

\(\text{The 95% CI is}\hspace{0.2cm} 97.32 \pm 1.984 \times 1.698= (93.95117, 100.6888)\).

\(\mu_0=93.29\) is not in the CI, so we reject \(𝐻_0\) to substantiate \(𝐻_𝑎\).

1. We can also construct CI using the null value as the center, then check the if the observed statistic is in the CI (not reject \(𝐻_0\)) or not (reject \(𝐻_0\))

2. For One-sided test, we can do the same, but we must be very careful with the confidence level and significance level, and the direction of interval.

We do not recommend using CI for one-sided tests.

• If \(\sigma\) is unknown, use 𝑡-distribution with \(S.E =\frac{s}{\sqrt{n}}\)

• Conditions:

  • Independence of observations (often verified by random sample, and if sampling w/o replacement, n < 10% of population).
  • No extreme skew.

• Point Estimate: \(\bar{x}\)

• Confidence Interval: \(\bar{x} \pm 𝑡_\frac{\alpha}{2}×𝑆.𝐸\), \((\color{blue}{𝑡_\frac{\alpha}{2}}\) depends on \(df\)=𝑛−1).

• Hypothesis testing: \(T = \frac{\text{sample mean- null value}}{S.E.}\)

• P-value depends on \(df\)=𝑛−1.