22 Day 22

Announcements

That’s all folks

• This is the last new content we’ll cover

• From here on, it’s review, review, review

• Wednesday we’ll do a refresh of some of the stuff we probably completely forgot

• Thanksgiving break is next week

Some of you will get the TEVAL soon (we administer them out semi-randomly)

Review

Null Hypothesis $H_0$

The statement we are holding as known and established information

• i.e., The average body weight of an adult cat is $10$ lbs.

$$H_0:\mu=10$$

Alternate Hypothesis $H_a$ or $H_1$

The statement we are testing to determine the accuracy of

• I believe that the cats I interact with regularly have a different average body weight than the population

$$H_a:\mu \neq 10$$

Test Statistic $t^*$

A value calculated as part of the hypothesis testing process. We place it into a $t$-table (or $z$-table depending) to get a $p$-value.

$$t^* = \frac{\bar{x} - \mu_0}{{s}/{\sqrt{n}}}$$

• I weighed $4$ of my friends cats and my own cat and found that their average body weight was $8$ pounds, with a standard deviation of $2.49$

$$t^* = \frac{8 - 10}{{2.49}/{\sqrt{5}}}$$

$$t^*=-1.796039$$

A reminder of our key study participant:

Significance level $\alpha$

The percentage probability we incur Type 1 Error in our hypothesis testing process

• I want to test my cat weight hypothesis at $\alpha=0.05$

P-value

The final statistic calculated in a hypothesis test, used to determine if we reject or fail to reject the null hypothesis

$$2*P(T>t^*)=0.15$$

$$0.15>\alpha \quad \text{Fail to Reject} \ H_0$$

Statistically Significant

We refer to a result as statistically significant if we tested it against a null hypothesis and proceeded to reject the null hypothesis

• There is insufficient evidence to suggest that the body weight of the cats that interact with regularly have a statistically significant difference in average body weight from the population

Hypothesis Tests for Difference Between Two Means (Independent)

• We’ve covered hypothesis testing for a single population parameter

  • (e.g., population mean $\mu$)

• Let’s look at testing a claim about the difference between two population means

$$\mu_1-\mu_2$$

• We need two independent random samples from two distinct populations

  • Independence implies that $X$ and $Y$ have no effect on one another

• As with everything we do in this class, we need to confirm our sample can be assumed as approximately normal

$$n>30$$

• We want to see if population means $\mu_1$ and $\mu_2$ are equal:

$$H_0:\mu_1=\mu_2$$

• There are three possibile alternate hypotheses:

  • Left-tailed: $H_1:\mu_1<\mu_2$

  • Right-tailed: $H_1:\mu_1>\mu_2$

  • Two-tailed: $H_1:\mu_1\neq\mu_2$

We need a test statistic, $t^*$:

$$t^*=\frac{(\bar{x}_1-\bar{x}_2)-(\mu_1-\mu_2)}{\sqrt{({s_1^2}/{n_1})+({s_2^2}/{n_2})}}$$

Under $H_0$, $\mu_1=\mu_2$, so $\mu_1-\mu_2=0$

$$t=\frac{(\bar{x}_1-\bar{x}_2)-0}{\sqrt{({s_1^2}/{n_1})+({s_2^2}/{n_2})}}$$

• $\mu_1$, $\mu_2$ are population means (under the assumption that $H_0$ is true)

• $\bar{x}_1$, $\bar{x}_2$ are sample means

• $s_1$, $s_2$ are sample standard deviations

• $n_1$, $n_2$ are sample sizes

• The test statistic measures how large the sample mean difference $(\bar{x}_1-\bar{x}_2)$ differs from the hypothesized value $\mu_1-\mu_2$ in $H_0$

• The test statistic comes from a Student’s $t$ distribution with degrees of freedom:

  $$df=\min(n_1-1,n_2-1)$$

  • (i.e., the smaller of $n_1-1$ and $n_2-1$).

For the P-value calculation:

• Left-tailed: $H_1:\mu_1<\mu_2$

  $$\text{P-value}=P(T<t)$$

• Right-tailed: $H_1:\mu_1>\mu_2$

  $$\text{P-value}=P(T>t)$$

• Two-tailed: $H_1:\mu_1\neq\mu_2$

  $$\text{P-value}=2\cdot P(T<-|t|) \text{ OR } 2\cdot P(T>|t|)$$

The steps for this hypothesis test are:

1. State the null and alternate hypotheses

2. Choose a significance level $\alpha$

3. Compute the test statistic:

   $$t=\frac{(\bar{x}_1-\bar{x}_2)}{\sqrt{({s_1^2}/{n_1})+({s_2^2}/{n_2})}}$$

4. Compute the P-value of the test statistic $t$

• Left-tailed: $\text{P-value}=P(T<t)$

• Right-tailed: $\text{P-value}=P(T>t)$

• Two-tailed: $\text{P-value}=2\cdot P(T<-|t|)$ or $2\cdot P(T>|t|)$

Note: The degrees of freedom of the $t$ distribution is: $$df=\min(n_1-1,n_2-1)$$

5. Determine whether to reject $H_0$:
   • Reject $H_0$ if $\text{P-value} \leq \alpha$.
6. State a conclusion

Example 1

The National Assessment Educational Progress tested a sample of students who had used a computer in their mathematics classes, and another sample of students who had not used a computer. The sample mean score for students using a computer was 309, with a sample standard deviation of 29. For students not using a computer, the sample mean was 303, with a sample standard deviation of 32. Assume there were 60 students in the computer sample and 40 students in the sample that hadn’t used a computer.

At 5% significance level, conduct a hypothesis test to determine whether the population mean scores differ in the between those students who use a computer and those who do not.

Step 1. State the null and alternative hypotheses

$$H_0: \mu_1 = \mu_2$$

$$H_A: \mu_1 \neq \mu_2 \quad (\rightarrow \text{two-tailed})$$

Step 2. The significance level is $\alpha=0.05$

Step 3. Compute the test statistic

$$\begin{array}{|c|c|c|c|} \hline \text{} & \text{Sample Mean} & \text{Sample Std. Dev.} & \text{Sample Size} \\ \hline \text{With Computer} & \bar{x}_1 = 309 & s_1 = 29 & n_1 = 60 \\ \hline \text{Without Computer} & \bar{x}_2 = 303 & s_2 = 32 & n_2 = 40 \\ \hline \end{array}$$

$$t = \frac{(\bar{x}_1 - \bar{x}_2) - (\mu_1 - \mu_2)}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}$$

$$t = \frac{(309 - 303) - (0)}{\sqrt{\frac{29^2}{60} + \frac{32^2}{40}}} \approx 0.953$$

Step 4. Compute the P-value

We use the t-table with $df=\min(n_1-1, n_2-1)=39$. Then:

$$P(T>0.953) \text{ is between } P(T>1.304)=0.10 \text{ and } P(T>0.681)=0.25$$

For the two-tailed test, the P-value:

$$\text{P-value} = 2 \cdot P(T>0.953) \text{ is between } 0.20 \text{ and } 0.50$$

Step 5. Determine whether to reject $H_0$

Since the P-value $> \alpha = 0.05$, we fail to reject $H_0$

Step 6. State a conclusion

There is not enough evidence to conclude that the mean scores differ between those students who use a computer and those who do not (i.e., the mean scores may be the same)

Hypothesis Tests for Difference Between Two Means (Paired)

• Next we turn our attention to a hypothesis test for paired (or matched) samples

• Example: Gas mileage before and after tune-up for automobiles

$$\begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline \text{Automobile} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline \text{After Tune-up} & 35.44 & 35.17 & 31.07 & 31.57 & 26.48 & 23.11 & 25.18 & 32.39 \\ \hline \text{Before Tune-up} & 33.76 & 34.30 & 29.55 & 30.90 & 24.92 & 21.78 & 24.30 & 31.25 \\ \hline \end{array}$$

• Both mileages before and after tune-up are obtained from the same automobile (i.e., the values are paired within the subject)

-Now, we are interested in testing the population mean difference for the matched pairs

-Our hypothesis test will involve two paired random samples from a single population

-The set of differences between the values in the matched pairs is considered as the sample data

-Required assumption: Each sample size is large (n > 30), or the differences in the matched pairs are normally distributed (at least approximately)

• The population mean difference for the matched pairs is denoted $\mu_d$ (unknown value)

• The sample mean of the differences is denoted $\bar{d}$

• The sample std. deviation of the differences is denoted $s_d$

$$\mu_d = \text{the mean mileage difference before and after tune-up}$$

$$\bar{d} = \frac{1.68 + 0.87 + \ldots + 1.14}{8} \approx 1.2063$$

$$s_d = \sqrt{\frac{(1.68 - 1.206)^2 + \ldots + (1.14 - 1.206)^2}{7}} \approx 0.3732$$

Step 1. State the null and alternate hypotheses. The null hypothesis is of the form

$$H_0: \mu_d = \mu_0$$

where $\mu_0$ is a prespecified value (e.g. $\mu_0 = 0$ is most common)

The alternate hypothesis:

• Left-tailed: $H_1: \mu_d < \mu_0$

• Right-tailed: $H_1: \mu_d > \mu_0$

• Two-tailed: $H_1: \mu_d \neq \mu_0$

Step 2. Choose a significance level $\alpha$

Step 3. Compute the test statistic:

$$t = \frac{\bar{d} - \mu_0}{s_d / \sqrt{n}}$$

which follows a Student’s $t$ distribution with $df = n - 1$

Step 4. Compute the P-value of the test statistic $t$

• Left-tailed: P-value = area under the Student’s $t$ distribution to the left of $t$, i.e., $P(T < t)$

• Right-tailed: P-value = area under the Student’s $t$ distribution to the right of $t$, i.e., $P(T > t)$

• Two-tailed: P-value = sum of the areas under the Student’s $t$ distribution to the left of $-|t|$ and right of $|t|$, i.e., $2 \cdot P(T < -|t|)$ or $2 \cdot P(T > |t|)$

Step 5. Determine whether to reject $H_0$:

• Reject $H_0$ if P-value $\leq \alpha$

Step 6. State a conclusion

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