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1. Stating hypotheses

Warm-up

Three quick questions from earlier lessons.

Discover

A courtroom presumes innocence until the evidence is convincing. In a significance test, which hypothesis plays the role of "innocent"?

The null hypothesis H₀
The alternative Hₐ

Key idea

State the question as two hypotheses about a parameter.

H₀ (null): the parameter equals a specific value (no effect). Hₐ (alternative): what we look for evidence of:

<

>

(one-sided) or

\ne

(two-sided). Hypotheses are about parameters (

p

\mu

), never statistics.

Worked example

A factory says bulbs last 1,000 hours on average; a buyer suspects less.

H_0: \mu = 1000

Status quo.

H_a: \mu < 1000

The suspicion.

One-sided test

Direction given.

Your turn, guided

Your turn: in "H₀: p = 0.3, Hₐ: p ≠ 0.3", what is the null value $p_0$ ?

p_0 = 0.3

The value in H₀.

Practice ★

A company claims that 80% of customers are satisfied. A researcher suspects fewer are. Which is the null hypothesis?
1. H₀: p = 0.80
2. H₀: p < 0.80
3. H₀: p̂ = 0.80

The mean wait at a clinic used to be 81 minutes. A manager wants to know whether it has changed, in either direction. Which alternative?
1. Hₐ: μ ≠ 81
2. Hₐ: μ > 81
3. Hₐ: x̄ ≠ 81

Watch out!

Practice ★★

Why are hypotheses written about parameters such as p, not statistics such as p̂ (sample of 445)?
1. Because we know p̂ exactly; the question is about the unknown population value
2. Because statistics are harder to compute
3. There is no difference between p and p̂

Exit ticket

Three questions to check you have got it.

A company claims that 80% of customers are satisfied. A researcher suspects fewer are. Which is the null hypothesis?
1. H₀: p = 0.80
2. H₀: p < 0.80
3. H₀: p̂ = 0.80
A company claims that 60% of customers are satisfied. A researcher suspects fewer are. Which is the null hypothesis?
1. H₀: p = 0.60
2. H₀: p < 0.60
3. H₀: p̂ = 0.60
A company claims that 62% of customers are satisfied. A researcher suspects fewer are. Which is the null hypothesis?
1. H₀: p = 0.62
2. H₀: p < 0.62
3. H₀: p̂ = 0.62

Summary card

Hypotheses

H₀: parameter = value. Hₐ: <, >, or ≠.

Watch out: parameters, never statistics.

2. The one-proportion z test and P-values

Warm-up

Three quick questions from earlier lessons.

Discover

A coin lands heads 60 times in 100 tosses. If it were fair, $\hat{p}$ would have SD 0.05. How many SDs above 0.5 is 0.6?

Key idea

How surprising is the data if H₀ is true?

One-proportion z test:

z = \dfrac{\hat{p} - p_0}{\sqrt{p_0(1 - p_0)/n}}

(the SE uses the null value

p_0

). The P-value is the probability, assuming H₀ is true, of a result at least as extreme as the one observed, in the direction(s) of Hₐ.

Worked example

H₀: $p = 0.25$ , $n = 300$ , $\hat{p} = 0.30$ .

SE = 0.025

√(0.25 × 0.75/300).

z = 2

(0.30 − 0.25)/0.025.

P-value

\approx 0.023

Upper tail for Hₐ: p > 0.25.

Your turn, guided

Your turn: SE = 0.04, $\hat{p} - p_0 = 0.06$ . Find $z$ .

z = 1.5

0.06/0.04.

Practice ★

Test H₀: p = 0.5 with a random sample of 100 in which p̂ = 52/100. Find the z statistic.
Test H₀: p = 0.5 with a random sample of 100 in which p̂ = 58/100. Find the z statistic.

Test H₀: p = 0.2 with n = 400 and p̂ = 28/100. Find the z statistic.

Watch out!

Practice ★★

A test gives a P-value of 0.05. What does it mean?
1. If H₀ were true, a result at least as extreme as ours would happen with probability 0.05
2. There is a 0.05 probability that H₀ is true
3. The result is due to chance with probability 0.05

Boss problem. A coin is tossed 100 times and lands heads 52 times. Test H₀: p = 0.5 against Hₐ: p > 0.5.

Exit ticket

Three questions to check you have got it.

Test H₀: p = 0.5 with a random sample of 100 in which p̂ = 52/100. Find the z statistic.
Test H₀: p = 0.5 with a random sample of 100 in which p̂ = 58/100. Find the z statistic.
Test H₀: p = 0.5 with a random sample of 100 in which p̂ = 56/100. Find the z statistic.

Summary card

z test for p

$z = (\hat{p} - p_0)/\sqrt{p_0(1 - p_0)/n}$ ; P-value assumes H₀.

Watch out: use $p_0$ in the SE.

3. Conclusions, errors and power

Warm-up

Three quick questions from earlier lessons.

Discover

At α = 0.05 a test gives P-value 0.30. Does this prove H₀ is true?

No: we just lack evidence against it
Yes

Key idea

Decide, and know what can go wrong.

If P-value

\le \alpha

, reject H₀ (convincing evidence for Hₐ); otherwise fail to reject H₀. Type I error: rejecting a true H₀ (probability α). Type II error: failing to reject a false H₀. Power = probability of correctly rejecting a false H₀; it grows with

n

and with α.

Worked example

H₀: a new teaching method is no better.

Type I: adopting a method that is not better

Reject a true H₀.

Type II: keeping the old method when the new one is better

Miss a real improvement.

Higher power, fewer Type II errors

More students in the study.

Your turn, guided

Your turn: α = 0.01. What is the probability of a Type I error when H₀ is true?

0.01

It equals α.

Practice ★

At significance level α = 0.05, a test gives P-value 0.04. What do you conclude?
1. Reject H₀: there is convincing evidence for Hₐ
2. Accept H₀
3. Hₐ is proven true

A test of a new drug (sample of 388) has H₀: the drug has no effect. What is a Type I error?
1. Concluding the drug works when it really has no effect
2. Concluding there is no effect when the drug really works
3. Making a mistake in the data

Watch out!

Practice ★★

How can a researcher increase the power of a test that now uses 82 subjects?
1. Use a larger sample
2. Use a smaller α
3. Use fewer subjects

Exit ticket

Three questions to check you have got it.

At significance level α = 0.05, a test gives P-value 0.04. What do you conclude?
1. Reject H₀: there is convincing evidence for Hₐ
2. Accept H₀
3. Hₐ is proven true
At significance level α = 0.05, a test gives P-value 0.03. What do you conclude?
1. Reject H₀: there is convincing evidence for Hₐ
2. Accept H₀
3. Hₐ is proven true
At significance level α = 0.05, a test gives P-value 0.02. What do you conclude?
1. Reject H₀: there is convincing evidence for Hₐ
2. Accept H₀
3. Hₐ is proven true

Summary card

Conclusions and errors

P ≤ α: reject H₀. Otherwise: fail to reject.
Type I: reject true H₀. Type II: keep false H₀.
Power rises with $n$ .

Watch out: never "accept" or "prove" H₀.

4. The one-sample t test for a mean

Warm-up

Three quick questions from earlier lessons.

Discover

H₀: μ = 50. A sample of 16 has x̄ = 53 and s = 8. How many standard errors is 53 above 50?

1.5
0.375

Key idea

The one-sample t test.

t = \dfrac{\bar{x} - \mu_0}{s/\sqrt{n}}

with

n - 1

degrees of freedom. For a two-sided Hₐ, the P-value counts both tails (double the one-tail area).

Worked example

H₀: μ = 100, Hₐ: μ ≠ 100; $n = 25$ , $\bar{x} = 104$ , $s = 10$ .

SE = 2

10/5.

t = 2

, df

= 24

4/2.

P-value

\approx 2 \times 0.028 = 0.057

Both tails.

Your turn, guided

Your turn: one-tail area 0.04, two-sided test. The P-value.

0.08

2 × 0.04.

Practice ★

Test H₀: μ = 39. A random sample of 9 has x̄ = 43 and s = 9. Find the t statistic.
Test H₀: μ = 24. A random sample of 64 has x̄ = 29 and s = 18. Find the t statistic.

For a test statistic, the area in one tail beyond it is 0.04. Find the P-value of the two-sided test.

Watch out!

Practice ★★

A t test gives P-value 0.20. What can you conclude at α = 0.05?
1. Fail to reject H₀: not enough evidence for Hₐ
2. H₀ is true
3. Reject H₀

Exit ticket

Three questions to check you have got it.

Test H₀: μ = 39. A random sample of 9 has x̄ = 43 and s = 9. Find the t statistic.
Test H₀: μ = 24. A random sample of 64 has x̄ = 29 and s = 18. Find the t statistic.
Test H₀: μ = 71. A random sample of 9 has x̄ = 76 and s = 8. Find the t statistic.

Summary card

t test for μ

$t = (\bar{x} - \mu_0)/(s/\sqrt{n})$ , df $= n - 1$ ; two-sided: double the tail.

Watch out: failing to reject is not proving H₀.

5. Comparing two groups: two proportions and paired data

Warm-up

Three quick questions from earlier lessons.

Discover

To test whether two groups have the same success rate, group A had 30 of 100 and group B 50 of 100. Under H₀ (same rate), what is the best estimate of that common rate?

0.4
It cannot be estimated

Key idea

Two independent groups, or one group measured twice.

Two proportions: pooled

\hat{p} = \dfrac{x_1 + x_2}{n_1 + n_2}

z = \dfrac{\hat{p}_1 - \hat{p}_2}{\sqrt{\hat{p}(1 - \hat{p})(1/n_1 + 1/n_2)}}

. Paired data (same subjects twice): a one-sample t test on the differences.

Worked example

$x_1 = 30$ , $n_1 = 100$ ; $x_2 = 50$ , $n_2 = 100$ .

\hat{p} = 0.4

Pool.

SE \approx 0.069

√(0.4 × 0.6 × 0.02).

z \approx -2.89

(0.3 − 0.5)/0.069.

Your turn, guided

Your turn: 20 of 50 and 30 of 50. The pooled proportion.

0.5

50/100.

Practice ★

To test H₀: p₁ = p₂, group 1 had 21 successes out of 82 and group 2 had 22 out of 102. Find the pooled proportion.
To test H₀: p₁ = p₂, group 1 had 51 successes out of 85 and group 2 had 51 out of 105. Find the pooled proportion.

24 students take a test before and after a course. Which procedure compares the scores?
1. A paired t test on the 24 differences (after − before)
2. A two-sample t test treating the groups as independent
3. A two-proportion z test

Watch out!

Practice ★★

Testing H₀: p₁ = p₂ with n₁ = n₂ = 100, the pooled proportion is 3/10. Find the standard error of p̂₁ − p̂₂.

Exit ticket

Three questions to check you have got it.

To test H₀: p₁ = p₂, group 1 had 21 successes out of 82 and group 2 had 22 out of 102. Find the pooled proportion.
To test H₀: p₁ = p₂, group 1 had 51 successes out of 85 and group 2 had 51 out of 105. Find the pooled proportion.
To test H₀: p₁ = p₂, group 1 had 32 successes out of 93 and group 2 had 53 out of 113. Find the pooled proportion.

Summary card

Comparing two groups

Two proportions: pooled $\hat{p}$ in the SE.
Paired: test the differences.

Watch out: pool counts, not proportions.

6. Chi-square tests

Warm-up

Three quick questions from earlier lessons.

Discover

A die is rolled 60 times. If it is fair, how many times do you expect each face?

Key idea

Compare observed counts with expected counts.

\chi^2 = \sum \dfrac{(O - E)^2}{E}

. Goodness of fit:

E = n \times p_i

, df

=

(categories

- 1

). Two-way table (independence or homogeneity):

E = \dfrac{\text{row total} \times \text{column total}}{\text{grand total}}

, df

= (r - 1)(c - 1)

. Large

\chi^2

(small P-value) is evidence against H₀. Condition: all expected counts at least 5.

Worked example

Die: observed 14 sixes, expected 10.

Contribution

= 1.6

(14 − 10)²/10.

= 5

Six faces.

\chi^2

is the sum

Add all six contributions.

Your turn, guided

Your turn: a 3 × 4 two-way table. Degrees of freedom.

6

(3 − 1)(4 − 1).

Practice ★

A die is rolled 114 times to test whether it is fair. What is the expected count for each face?

In a chi-square test, one category has observed count 39 and expected count 21. Find its contribution to χ².
In a chi-square test, one category has observed count 16 and expected count 27. Find its contribution to χ².

Watch out!

Practice ★★

In a two-way table with grand total 158, a row total is 55 and a column total is 41. Find the expected count in that cell under independence.
In a two-way table with grand total 134, a row total is 53 and a column total is 40. Find the expected count in that cell under independence.

Exit ticket

Three questions to check you have got it.

In a chi-square test, one category has observed count 39 and expected count 21. Find its contribution to χ².
In a chi-square test, one category has observed count 16 and expected count 27. Find its contribution to χ².
In a chi-square test, one category has observed count 20 and expected count 16. Find its contribution to χ².

Summary card

Chi-square tests

$\chi^2 = \sum (O - E)^2/E$ ; GOF: $E = np$ ; two-way: $E =$ row × column / total.

Watch out: divide by $E$ ; use counts, not percentages.

Answers: Significance tests

1. Stating hypotheses

H₀: p = 0.80

Hₐ: μ ≠ 81

Because we know p̂ exactly; the question is about the unknown population value

H₀: p = 0.80
H₀: p = 0.60
H₀: p = 0.62

2. The one-proportion z test and P-values

$\frac{2}{5}$
$\frac{8}{5}$

If H₀ were true, a result at least as extreme as ours would happen with probability 0.05

$\frac{3}{5}$

$\frac{2}{5}$
$\frac{8}{5}$
$\frac{6}{5}$

3. Conclusions, errors and power

Reject H₀: there is convincing evidence for Hₐ

Concluding the drug works when it really has no effect

Use a larger sample

Reject H₀: there is convincing evidence for Hₐ
Reject H₀: there is convincing evidence for Hₐ
Reject H₀: there is convincing evidence for Hₐ

4. The one-sample t test for a mean

$\frac{4}{3}$
$\frac{20}{9}$

$\frac{2}{25}$

Fail to reject H₀: not enough evidence for Hₐ

$\frac{4}{3}$
$\frac{20}{9}$
$\frac{15}{8}$

5. Comparing two groups: two proportions and paired data

$\frac{43}{184}$
$\frac{51}{95}$

A paired t test on the 24 differences (after − before)

$\frac{\sqrt{42}}{100}$

$\frac{43}{184}$
$\frac{51}{95}$
$\frac{85}{206}$

6. Chi-square tests

$19$

$\frac{108}{7}$
$\frac{121}{27}$

$\frac{2255}{158}$
$\frac{1060}{67}$

$\frac{108}{7}$
$\frac{121}{27}$
$1$