Notes
Why $t$ instead of $z$
We rarely know $\sigma$, so we estimate it with $s$. The standardized statistic then follows a $t$ distribution with $df = n - 1$. $t$ distributions are symmetric and bell-shaped but have heavier tails than the normal; as $df$ grows they approach the normal.
Conditions for means
- Random: random sample or random assignment.
- 10%: $n \le 10\%$ of the population when sampling without replacement.
- Normal/large sample: $n \ge 30$, or the population is approximately normal, or a graph of the sample shows no strong skew or outliers.
One-sample $t$ procedures
- Interval: $\displaystyle \bar x \pm t^*\frac{s}{\sqrt n}$.
- Test: $\displaystyle t = \frac{\bar x - \mu_0}{s/\sqrt n}$, with $df = n - 1$.
- Paired data (two measurements on the same individual, like before and after): take the differences, then do one-sample $t$ on $\mu_{diff}$.
Two-sample $t$ procedures
For two independent groups:
- Interval: $\displaystyle (\bar x_1 - \bar x_2) \pm t^*\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}$.
- Test of $H_0: \mu_1 = \mu_2$: $\displaystyle t = \frac{(\bar x_1 - \bar x_2) - 0}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}$. Use technology for $df$ (or the conservative smaller $n - 1$).
Topics in this unit
Each topic has its own short notes and practice.
- 4.1Sampling Distributions for Sample Means
- 4.2Constructing a Confidence Interval for a Population Mean or Population Mean Difference
- 4.3Justifying a Claim Based on a Confidence Interval for a Population Mean or Population Mean Difference
- 4.4Setting Up a Test for a Population Mean or Population Mean Difference
- 4.5Carrying Out a Test for a Population Mean or Population Mean Difference
- 4.6Sampling Distributions for the Difference Between Two Sample Means
- 4.7Constructing a Confidence Interval for the Difference Between Two Population Means
- 4.8Justifying a Claim Based on a Confidence Interval for the Difference Between Two Population Means
- 4.9Setting Up a Test for the Difference Between Two Population Means
- 4.10Carrying Out a Test for the Difference Between Two Population Means
Practice
Estimating $\sigma$ with $s$ adds variability, so more area is in the tails.
$\displaystyle \frac{s}{\sqrt n} = \frac{0.8}{4} = 0.2$.
The same students are measured twice, so the data are paired.
A small sample with strong skew or outliers violates the normality condition.
$df = n - 1$.
The whole interval is positive, so $\mu_1 > \mu_2$ is supported. Intervals describe means, not individuals.
The one-sided upper-tail area beyond 2.4 with 24 df is about 0.012.
Higher confidence needs a larger $t^*$.
Free response
Write your answer first, then open the scoring guide.
Free-response question
A random sample of 36 phone batteries lasted a mean of 18.6 hours with standard deviation 2.4 hours. The maker claims a mean of 20 hours.
(a) Construct a 95% confidence interval for the true mean battery life. ($t^* \approx 2.030$ for $df = 35$.)
Show the answer and scoring guide
One-sample $t$ interval for $\mu$. Random sample, $n = 36 \ge 30$. $\displaystyle 18.6 \pm 2.030\left(\frac{2.4}{6}\right) = 18.6 \pm 0.812$, giving $(17.79, 19.41)$ hours.
Check off the points you earned:
(b) Interpret the interval.
Show the answer and scoring guide
We are 95% confident that the interval from 17.79 to 19.41 hours captures the true mean battery life of all batteries of this kind.
Check off the points you earned:
(c) Does the interval support the maker's claim? Explain.
Show the answer and scoring guide
No. 20 hours is not in the interval, so 20 is not a plausible value for the true mean; the data suggest the mean is lower.
Check off the points you earned: