STAT 120

Bastola

**Two Competing Hypotheses**:

**Null Hypothesis \((H_0)\)**: Assumes no effect or no difference.**Alternative Hypothesis \((H_a)\)**: Suggests a specific effect or difference exists.

**Specifying Hypotheses**: We frame these hypotheses in terms of population parameters (like mean, proportion, etc.).**Example**:- \(H_0\): The mean weight of fish = 3 pounds.
- \(H_a\): The mean weight of fish ≠ 3 pounds.

Our goal is to determine which hypothesis is more likely given our sample data.

Source: click here

`10:00`

Suppose you did this 10 times and guessed correctly 3 times. Is this evidence that you have ESP abilities?

There are five cards with five different symbols. If there is no such thing as ESP, what proportion p of guesses should be correct?

- \(\mathrm{p}=0\)
- \(\mathrm{p}=1/ 4\)
- \(\mathrm{p}=1 / 5\)
- \(\mathrm{p}=1 / 2\)

Let \(\hat{p}\) denote the sample proportion of correct guesses. Which of the statistics below would give the strongest evidence for ESP?

- \(\hat{\mathrm{p}}=0\)
- \(\hat{\mathrm{p}}=1 / 5\)
- \(\hat{\mathrm{p}}=1 / 2\)
- \(\hat{\mathrm{p}}=3 / 4\)

**Goal**: Use sample data to assess population claims.**Null Hypothesis \((H_0)\)**: No effect or difference.**Alternative Hypothesis \((H_a)\)**: Evidence-seeking claim.- Tests reveal if observed stats suggest genuine difference or result from random chance.
- Aim: Refute \(H_0\) with enough evidence.

- As we’ve learned, statistics vary from sample to sample
- Even if the “population/true” proportion is \(\mathrm{p}=1/5\), not every sample proportion will be exactly \(1/5\)

How do we determine when a sample proportion is far enough above 1/5 to provide evidence of ESP?

For the ESP experiment:

- \(\mathrm{H}_{0}: \mathrm{p}=1 / 5\)
- \(\mathrm{H}_{\mathrm{a}}: \mathrm{p}>1 / 5\)

- \(\mathrm{H}_{0}\) usually includes \(=\)
- \(\mathrm{H}_{\mathrm{a}}\) usually includes \(>,<\), or \(\neq\)
- The direction in \(\mathrm{H}_{\mathrm{a}}\) depends on the question being asked, not based on what the data shows!

How to test if observed difference between groups is real or just due to random chance?

Students were given words to memorize, then randomly assigned to take either a 90 min nap, or a caffeine pill. \(2 \frac{1}{2}\) hours later, they were tested on their recall ability.

- Explanatory variable: sleep or caffeine
- Response variable: number of words recalled

Research Question: Is sleep or caffeine better for memory?

Mednick, Cai, Kanady, and Drummond (2008). “Comparing the benefits of caffeine, naps and placebo on verbal, motor and perceptual memory,” Behavioral Brain Research, 193, 79-86.

What is the parameter of interest in the sleep versus caffeine experiment?

- Proportion
- Difference in proportions
- Mean
- Difference in means
- Correlation

The correct answer is 4.

Difference in mean responses \(\mu_1 - \mu_2\), where \(\mu_1\) and \(\mu_2\) are the mean words recalled in the two different conditions

Let \(\mu_{\mathrm{s}}\) and \(\mu_{\mathrm{c}}\) be the mean number of words recalled after sleeping and after caffeine.

- Is there a difference in average word recall between sleep and caffeine?

What are the null and alternative hypothesis?

- \(\mathrm{H}_{0}: \mu_{\mathrm{s}} \neq \mu_{\mathrm{c}}, \mathrm{H}_{\mathrm{a}}: \mu_{\mathrm{s}}=\mu_{\mathrm{c}}\)
- \(\mathrm{H}_{0}: \mu_{\mathrm{s}}=\mu_{\mathrm{c}}, \mathrm{H}_{\mathrm{a}}: \mu_{\mathrm{s}} \neq \mu_{\mathrm{c}}\)
- \(\mathrm{H}_{0}: \mu_{\mathrm{s}} \neq \mu_{\mathrm{c}}, \mathrm{H}_{\mathrm{a}}: \mu_{\mathrm{s}}>\mu_{\mathrm{c}}\)
- \(\mathrm{H}_{0}: \mu_{\mathrm{s}}=\mu_{\mathrm{c}}, \mathrm{H}_{\mathrm{a}}: \mu_{\mathrm{s}}>\mu_{\mathrm{c}}\)
- \(\mathrm{H}_{0}: \mu_{\mathrm{s}}=\mu_{\mathrm{c}}, \mathrm{H}_{\mathrm{a}}: \mu_{\mathrm{s}}<\mu_{\mathrm{c}}\)

How can we be sure if an observed difference between two groups is real, or if it’s just something that could happen by random chance?

- Finish the rest of the class activity and submit to moodle when done.

`20:00`