[Lecture3.1 PDF]
[Lecture3.2 PDF]
[Lecture3.3 PDF]
[Lecture3.4 PDF]
Basic Formula:
- One-sample $t$ test
- Test for a population mean (unknown SD); sample size $n$. That is, known sample mean and SD but unknown population mean (needs to be tested) and SD.
- Same $H_0$ and $H_A$, same calculation as $z$ test, except:
- Assume population distribution roughly normal, unknown mean and SD.
- Approximate unknown SD of population by sample SD, with $n-1$ degree of freedom (i.e. denominator).
- Two independent samples
- Comparing the means
- Known parameters: $\mu$, $n$ and $\sigma$ (SD) of the independent samples.
- $H_0: \mu_1=\mu_2\Rightarrow$ the difference between the two sample means is expected to be $\mu_1-\mu_2$.
- The SE of the difference is $$SE_1=\frac{\sigma_1}{\sqrt{n_1}},\ SE_2=\frac{\sigma_2}{\sqrt{n_2}}$$ $$\Rightarrow SE=\sqrt{SE_1^2+SE_2^2}$$ \item The above result derives from $$SE(X-Y)=\sqrt{Var(X-Y)}$$ $$=\sqrt{1^2\cdot Var(X)+(-1)^2\cdot Var(Y)}=\sqrt{SE^2(X)+SE^2(Y)}$$
- $z$ statistics is $$z=\frac{\mu_1-\mu_2}{SE}$$
- Comparing the percents
- Known parameters: $p$ and $n$ of the independent samples
- $H_0: p_1=p_2$
- It should use the same percents estimation for both samples (i.e. pooled estimate), since the P-valuse is computed under the null $H_0: p_1=p_2$.
- Pooled estimate of $p$: $$\hat{p}=\frac{n_1\cdot p_1+n_2\cdot p_2}{n_1+n_2}$$
- SE of the sample proportion, for the two samples: $$SE_1=\sqrt{\frac{\hat{p}\cdot(1-\hat{p})}{n_1}}, SE_2=\sqrt{\frac{\hat{p}\cdot(1-\hat{p})}{n_2}}$$
- Similar to the sample mean, the SE of the difference between the sample percents is approximately $$SE=\sqrt{SE_1^2+SE_2^2}$$
- $z$ statistic is $$z=\frac{p_1-p_2}{SE}$$
- Comparing the means
EXERCISE SET 3
If a problem asks for an approximation, please use the methods described in the video lecture segments. Unless the problem says otherwise, please give answers correct to one decimal place according to those methods. Some of the problems below are about simple random samples. If the population size is not given, you can assume that the correction factor for standard errors is close enough to 1 that it does not need to be computed. Please use the 5% cutoff for P-values unless otherwise instructed in the problem.
PROBLEM 1
A statistical test is performed, and its P-value turns out to be about 3%. Which of the following must be true? Pick ALL that are correct.
a. The null hypothesis is true.
b. There is about a 3% chance that the null hypothesis is true.
c. The alternative hypothesis is true.
d. There is about a 97% chance that the alternative hypothesis is true.
e. If the null hypothesis were true, there would be about a 3% chance of getting data that were like those that were observed in the sample or even further in the direction of the alternative.
f. The P-value of about 3% was computed assuming that the null hypothesis was true.
Solution
(e) and (f) are correct. Firstly, 3% is computed under assuming $H_0$ is true. So (f) is correct. Then, P-value means: assuming the null is true, the chance of getting data like the data in the sample or even more like the alternative. So (e) is correct (definition of P-value). Note that (a) and (c) could be wrong because there are two types of error, Type I and Type II error. There is no such thing as "the chance that the null / alternative is true". So (b) and (d) are incorrect.
PROBLEM 2
The distribution of cholesterol levels of the residents of a state closely follows the normal curve. Investigators want to test whether the mean cholesterol level of the residents is 200 mg/dL or lower. A simple random sample of 12 residents has a mean cholesterol level of 185 mg/dL, with an SD of 20 mg/dL (computed as the ordinary SD of a list of 12 numbers, with 12 in the denominator).
Let $m$ be the mean cholesterol level of the residents of the state, measured in mg/dL. In Problems 2A-2E, perform a $t$ test of the hypotheses $$\text{Null}: m = 200$$ $$\text{Alternative}: m < 200$$ 2A You are using data from a simple random sample to test the given hypotheses. Which of the following sets of assumptions is further required to justify the use of a $t$ distribution to compute the P-value?
a. The distribution of the cholesterol levels of the residents of the state is close to normal.
b. The distribution of the cholesterol levels of the residents of the state is close to normal, with an unknown mean.
c. The distribution of the cholesterol levels of the residents of the state is close to normal, with an unknown mean and an unknown SD.
2B The t distribution that should be used has ( ) degrees of freedom.
2C The value of the t statistic is closest to?
2D The P-value of the test is closest to?
2E The conclusion of the test is to reject the null hypothesis not reject the null hypothesis
Solution
2A) The sample is small, so unless the population distribution is bell-shaped, you might have trouble using a bell-shaped approximation to the probabilities for the sample mean. If the mean of the normally distributed population were already known, there would be no reason to perform the test. If the SD of the underlying normally distributed population (with unknown mean) were known, you would perform the $z$ test.
2B) The degree of freedom is sample size minus 1, that is, $12-1=11$.
2C) Sample SD is $$\sigma=20\times\sqrt{\frac{12}{11}}$$ and $$SE=\frac{\sigma}{12}$$ Thus the t-statistic is $$t=\frac{185-200}{SE}=-2.487469$$ R code:
sigma = 20 * sqrt(12 / 11) se = sigma / sqrt(12) t = (185 - 200) /se; t [1] -2.4874692D) Left-tailed $t$ test, 11 degree of freedom, P-value is $p=0.0150854$ R code:
pt(t, 11) [1] 0.01508542E) Because the P-value is less than 5%, so reject $H_0$.
PROBLEM 3
In a simple random sample of 1000 people taken from City A, 13% are senior citizens. In an independent simple random sample of 600 people taken from City B, 17% are senior citizens. Is the percent of senior citizens different in the two cities? Or is this just chance variation?
Answer in the steps described in Problems 3A-3C.
3A
Under the null hypothesis, the percent of senior citizens in each of the two cities is estimated to be ( )%.
3B
Under the null hypothesis, the estimated standard error of the difference between the percents of senior citizens in the two samples is closest to ( )%.
3C
The P-value of the test is closest to ( )%, so the null hypothesis is rejected.
Solution
This is two independent simple random samples, $$H_0: p_A=p_B$$ $$p_A\neq p_B$$ 3A) Pooled estimate of $p$ is $$\hat{p}=\frac{1000\times13%+600\times17%}{1000+600}=14.5%$$ 3B) $${SE}_{A}=\sqrt{\frac{\hat{p}\cdot(1-\hat{p})}{1000}}$$ $${SE}_{B}=\sqrt{\frac{\hat{p}\cdot(1-\hat{p})}{600}}$$ Thus, SE of the difference between the sample percents is approximately $$SE=\sqrt{({SE}_{A}^2+{SE}_{B}^2)}=1.818241%$$ R code:
p = 0.145; n1 = 1000; n2 = 600 se.a = sqrt(p * (1 - p) / n1) se.b = sqrt(p * (1 - p) / n2) se = sqrt(se.a^2 + se.b^2); se [1] 0.018182413C) The observed difference is 4%, so $$z=\frac{0.04-0}{SE}$$ Two-tailed $z$ test and the P-value is 2.781197% which is less than 5%, thus we reject $H_0$. R code:
z = (0.04 - 0) / se (1 - pnorm(z)) * 2 [1] 0.02781197PROBLEM 4
Last year, there were 30,000 students at a university; their GPA had a mean of 2.9 and an SD of 0.6. This year, in a simple random sample of 100 students taken from this university, the GPAs have a mean of 2.95 and an SD of 0.55. Has the mean GPA at the university gone up since last year, or is this just chance variation? Pick the correct calculation of the test statistic.
one-sample $z = (2.95 - 2.9)/0.06 = 0.833$; P large; conclude chance variation
one-sample $z = (2.95 - 2.9)/0.055 = 0.909$; P large; conclude chance variation
two-sample $z = (0.05 - 0)/\sqrt{0.055^2 + 0.0035^2} = 0.907$; P large; conclude chance variation
Solution
Firstly, the population is the students at the university of THIS YEAR. $$H_0: \mu=2.9$$ $$H_A: \mu > 2.9$$ The third choice must be wrong since the number of last year\rq s students was not a sample.Thus, this is one-sample $z$ test. $$n=100, \sigma=0.55$$ $$\Rightarrow SE=\frac{\sigma}{\sqrt{n}}=0.055, z=\frac{2.95-0}{SE}=0.909$$ R code:
n = 100; sigma = 0.55 se = sigma / sqrt(n) z = (2.95 - 2.9) /se; z [1] 0.9090909
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