Unit 4 · Topic 4.5 Beta

Carrying Out a Test for a Population Mean or Population Mean Difference

Carry out a one-sample (or paired) t test by computing t = (x-bar − μ_0)/(s/√n) with n − 1 df, finding the p-value in the direction of H_a , and comparing it with α .

Practice 3: Analyze DataPractice 4: Interpret Results

Part 1 · Hook

Why this matters

The delivery sample came back: 45 deliveries, mean 32.4 minutes, standard deviation 6.8. That is 2.4 minutes over the advertised 30. Is that a real slowdown or the ordinary wobble of a random sample? One number, the t statistic, measures the gap in standard errors, and the t distribution turns it into a p-value.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. For $H_0: \mu = 30$ and $H_a: \mu > 30$ , a p-value of 0.03 at $\alpha = 0.05$ means

reject $H_0$ : there is convincing evidence that $\mu > 30$
accept $H_0$ : $\mu = 30$
fail to reject $H_0$
there is a 3% chance that $H_0$ is true

Show the answer

$0.03 < 0.05$ , so reject $H_0$ ; the evidence supports $H_a$ .

Correct: reject $H_0$ : there is convincing evidence that $\mu > 30$ :
accept $H_0$ : $\mu = 30$ :
fail to reject $H_0$ :
there is a 3% chance that $H_0$ is true:

2. A standardized test statistic has the form

(statistic minus null value) divided by the standard error
statistic divided by the sample size
null value minus statistic
standard error divided by the statistic

Show the answer

It counts how many standard errors the statistic is from the null value.

Correct: (statistic minus null value) divided by the standard error:
statistic divided by the sample size:
null value minus statistic:
standard error divided by the statistic:

3. For paired data, the hypotheses $H_0: \mu_d = 0$ , $H_a: \mu_d > 0$ are tested with

a one-sample t test on the differences
two separate one-sample tests
a one-proportion z test
a z test using $\sigma = 1$

Show the answer

A paired t test is a one-sample t test on the differences.

Correct: a one-sample t test on the differences:
two separate one-sample tests:
a one-proportion z test:
a z test using $\sigma = 1$ :

Part 4 · See it

See it first

The t distribution with 44 degrees of freedom, centered at 0. A dashed line marks the observed t statistic 2.37, and the small area under the curve to its right is shaded: the one-sided p-value, 0.0112. — The t distribution with 44 degrees of freedom, the null model for the delivery test. The observed $t = 2.37$ is marked, and the shaded area to its right, 0.011, is the p-value. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

Assume $H_0$ is true: the true mean is $\mu_0$ sample means vary around $\mu_0$ with standard error $\frac{s}{\sqrt{n}}$
Measure the gap in standard errors $t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}}$ , with $\text{df} = n - 1$
Find the area beyond $t$ in the direction of $H_a$ that area is the p-value (double the tail for $\ne$ )
Compare the p-value with $\alpha$ reject $H_0$ if $p < \alpha$ , fail to reject otherwise, and state the evidence for $H_a$ in context

Part 6 · Key ideas

Key ideas

t statistic: $t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}}$ (paired: $t$ $= \frac{\bar{x}_d - 0}{s_d/\sqrt{n}}$ ), $\text{df} = n - 1$ .
p-value: right tail for $>$ , left tail for $<$ , twice the smaller tail for $\ne$ . Table B brackets it; technology gives it exactly.
Conclusion: "Because the p-value of 0.0112 is less than 0.05, we reject $H_0$ . There is convincing evidence that the true mean ... is greater than ..."
A large p-value is "not convincing evidence", never "accept $H_0$ " or "the mean is $\mu_0$ ".

Part 7 · Misconception

A common mistake

The wrong idea: A p-value of 0.0112 means there is a 1.12% chance that the null hypothesis is true.

What actually happens: The p-value is computed assuming $H_0$ is true. It is the probability of a sample mean at least as extreme as the one observed, by chance alone, when the true mean is $\mu_0$ .

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Table

Testing the delivery time claim

A company advertises a mean delivery time of 30 minutes in one city. A manager suspects the true mean is longer and takes a random sample of 45 of last month's thousands of deliveries (invented data). She tests $H_0: \mu = 30$ against $H_a: \mu > 30$ , where $\mu$ is the true mean delivery time of all the company's deliveries in the city, at $\alpha = 0.05$ .

Summary of the 45 delivery times
Sample size	Sample mean	Sample standard deviation
45	32.4 minutes	6.8 minutes

1. Write the expression for the test statistic. Type it as on the exam, for example (a - b)/(s/sqrt(n)).

Type the expression as on the exam, with the values substituted.

Show the answer

$t$ $= \frac{\bar{x} - \mu_0}{s/\sqrt{n}}$ $= \frac{32.4 - 30}{6.8/\sqrt{45}}$ $= 2.37$ , with $\text{df} = 44$ .

Answer: $\frac{32.4 - 30}{\frac{6.8}{\sqrt{45}}}$

2. What is the p-value of the test?

Type a number. A table answer and a calculator answer are both accepted.

Show the answer

With $t = 2.37$ and $\text{df} = 44$ , the p-value is the area to the right: $P(t > 2.37) = 0.011$ . Technology: a one-sample t test gives $t = 2.37$ , $p = 0.0112$ . Table B (df 40 row) puts it between 0.01 and 0.02.

Answer: 0.0112

3. State the conclusion of the test at $\alpha = 0.05$ .

Write the conclusion in context.

Show the answer

Compare the p-value with alpha, then state the evidence for $H_a$ in context, naming the parameter.

A full-credit answer: Because the p-value of 0.0112 is less than 0.05, we reject the null hypothesis. There is convincing evidence that the true mean delivery time of all the company's deliveries in the city is greater than 30 minutes.

4. Interpret the p-value of 0.0112 in context.

Write the conclusion in context.

Show the answer

A p-value is a probability computed assuming the null hypothesis is true: the chance of a sample result at least as extreme as the one observed, in the direction of $H_a$ .

A full-credit answer: Assuming the true mean delivery time of all the company's deliveries in the city is 30 minutes, there is a 0.0112 probability of getting a sample mean of 32.4 minutes or more by chance in a random sample of 45 deliveries.

5. For the sleep study ( $H_0: \mu = 8$ , $H_a: \mu \ne 8$ , $p = 0.080$ ), state the conclusion at $\alpha = 0.05$ .

Write the conclusion in context.

Show the answer

0.080 is greater than 0.05, so fail to reject $H_0$ . The data are not convincing evidence of a difference, which is not the same as showing the mean is 8.

A full-credit answer: Because the p-value of 0.080 is greater than 0.05, we fail to reject the null hypothesis. There is not convincing evidence that the true mean nightly sleep of students at the university is different from 8 hours.

6. A t test gives $p = 0.27$ at $\alpha = 0.05$ . Which statement is correct?

The data do not give convincing evidence against the null hypothesis
The data show that the null hypothesis is true
There is a 27% chance that the null hypothesis is true
The sample mean equals the hypothesized mean

Show the answer

$0.27 > 0.05$ : fail to reject $H_0$ . That is "not convincing evidence", never "accept $H_0$ ".

Correct: The data do not give convincing evidence against the null hypothesis: A large p-value means results like these are common when $H_0$ is true, so there is not convincing evidence for $H_a$ .
The data show that the null hypothesis is true: Failing to reject is not proof of $H_0$ ; the true mean could differ by an amount this sample could not detect.
There is a 27% chance that the null hypothesis is true: A p-value is computed assuming $H_0$ is true; it is not the probability of $H_0$ .
The sample mean equals the hypothesized mean: The sample mean almost surely differs from $\mu_0$ ; the difference is just not convincing.

Part 9 · Summary

Summary

Carry out a one-sample (or paired) t test by computing

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

with

n - 1

df, finding the p-value in the direction of

H_a

, and comparing it with

\alpha

. Conclude in context with evidence language about the true mean, and interpret the p-value as a probability computed assuming

H_0

Part 10 · Up next

What comes next

Part 11 · Connections

Connections

Question set for this topic