a) 160cm is the assumed value. The alternative here is that this value is higher in reality.
b) The assumption is that the university is right with at least $50,000. The alternative is that this value is lower in reality.
c) The status quo is supposed to be 6 weeks. The alternative is that this is incorrect, and the average is different than 6.
d) The status quo is there is no difference between the classes. The alternative is that there is a difference.
e) The status quo / assumption is that the average salary of men and women is the same. The alternative here is that women have a lower average salary.
f) The assumption is the lightbulbs last the same amount. The alternative here is that lightbulbs A in fact last longer.
Note: the alternative hypothesis never has an equal sign, including in the inequalities.
Closea) We’re testing if the burgers weigh less than expected.
b) This is a one-sample, one-tailed \( t \)-test. The calculator gives the \( p \)-value of 0.226.
c) The \( p \)-value is higher than 0.01. The difference we observe is not improbable for a population that has a mean of 200g. We do not reject \( H_0 \). This is all we can conclude, as we cannot know for sure that the average is indeed 200g from this test. Be careful not to say we “accept” \( H_0 \).
\(H_0 \colon \mu_A = \mu_B\) and \(H_1 \colon \mu_B > \mu_A\)
This is a two-sample, one-tailed test with pooled variances. We carry out the calculations and find a \(p\)-value = 0.0125.
\(p\)-value < \(\alpha\) (0.05), so we have satisfying evidence against \(H_0\) and in favor of \(H_1\). It looks like, on average, people rate videogame B higher and that it is the better videogame.
Close\(H_0 \colon \mu_1 = \mu_2\) and \(H_1 \colon \mu_1 \neq \mu_2\)
This is a two-tailed, two-sample test with pooled variance. Running the test on a calculator, we find \(p\)-value = 0.0838.
\(p\)-value < \(\alpha\), so we can reject the null hypothesis. We have evidence that the fertilizer used makes a difference in plant growth.
Closea) If the 3,500 clicks are distributed evenly, then each day we expect \( \frac{3,500}{7} = 500 \) clicks.
b) For each category (number of computers), we calculate percentage × sample size.
| Number of computers | 0 | 1 | 2 | 3 | 4+ |
|---|---|---|---|---|---|
| Percentage of households | 180 * 10% = 18 | 180 * 15% = 27 | 180 * 40% = 72 | 180 * 20% = 36 | 180 * 15% = 27 |
c) For each weight category, we calculate the probability × the size of the sample. To get the probabilities, we can plug the values into the GDC:
Then, we need to multiply each probability to 60 and round it to the nearest integer:
| Weight | \(\text{weight} \leq 3\) | \(3 < \text{weight} \leq 3.4\) | \(3.4 < \text{weight} \leq 3.8\) | \(\text{weight} > 3.8\) |
|---|---|---|---|---|
| Number of newborns | \(0.1587 \times 60 \approx 10\) | \(0.3413 \times 60 \approx 20\) | \(0.3413 \times 60 \approx 20\) | \(0.1587 \times 60 \approx 10\) |
a) The table records 600 throws in total. We expect each face to be drawn \( \frac{600}{6} = 100 \) times each, since the probability of drawing each number should be the same in the case of a fair die.
b) A fair die means no face will have a higher chance to appear, and it is the assumption we’re testing against.
c) There are 6 possible values (categories), so the degrees of freedom are \(6 - 1 = 5\).
d) We run the test with the calculator and find the \(p\)-value of 0.2148.
e) \(p\)-value > \(\alpha = 5\%\), so there is no sufficient evidence to reject \( H_0 \), and the die appears balanced.
Closea) The assumption is the person is correct, and their time follows the expected distribution.
b) A day is 1440 minutes total. With the percentages of the first table, we calculate:
| Activity | Work | Chores | Hobbies | Socializing | Sleeping |
| Minutes spent | 1440 * 0.3 = 432 | 1440 * 0.1 = 144 | 1440 * 0.2 = 288 | 1440 * 0.1 = 144 | 1440 * 0.3 = 432 |
c) Given by a calculator.
d) The \(p\)-value (0.0553) is lower than 10% = 0.1, so we reject \(H_0\). The distribution is different from what the person expected.
e) At a significance level of 5%, we would not have rejected \(H_0\). In practice, this means the difference between the guess and the recorded data would need to be even bigger to allow us to conclude. This shows how our choices of sensitivity can affect our conclusion.
Closea) We expect the same amount of people at each concert, so we calculate:
\[ \frac{\text{total of spectators}}{\text{number of dates}} = \frac{20,361 + 19,946 + 19,799 + 20,194}{4} = 20,075 \]
b) There are 4 dates (categories), so the degrees of freedom are \(4 - 1 = 3\).
c) The calculator gives us the value of \(\chi^2 = 9.40\).
d) \(\chi^2\) is above the critical value of 7.81, so we reject \(H_0\).
Closea) The status quo is that preferences are independent of where you live. We’re trying to see if we have proof to the opposite.
b) The degrees of freedom is \((\text{number of columns} – 1) \times (\text{number of rows} – 1) = 2 \times 3 = 6\).
c) The calculator gives us the \(p\)-value of 0.192.
d) The \(p\)-value is higher than the significance level \(\alpha = 5\%\), so we cannot reject the null hypothesis. We don’t have sufficient evidence to suggest that people will prefer different vegetables based on where they live.
Closea) We are trying to test against assumed independence.
b) Degrees of freedom = \((\text{number of columns} - 1) \times (\text{number of rows} - 1)\) = \((4-1) \times (3-1) = 3 \times 2 = 6\)
c) The calculator gives us \(\chi^2 = 13.8\)
d) The \(\chi^2\) is above the critical value, so we reject \(H_0\).
Close| Height of plants with fertilizer 1 | Height of plants with fertilizer 2 |
|---|---|
| 68 | 62 |
| 59 | 55 |
| 64 | 73 |
| 38 | 66 |
| 49 | 63 |
| 34 | 73 |
| 62 | 49 |
| 57 | 69 |
| Number of computers | 0 | 1 | 2 | 3 | 4+ |
|---|---|---|---|---|---|
| Percentage of households | 10% | 15% | 40% | 20% | 15% |
| Result | Occurrences |
|---|---|
| 1 | 87 |
| 2 | 109 |
| 3 | 85 |
| 4 | 106 |
| 5 | 99 |
| 6 | 114 |
| Activity | Work | Chores | Hobbies | Socializing | Sleeping |
|---|---|---|---|---|---|
| Percentage of time spent | 30% | 10% | 20% | 10% | 30% |
| Activity | Work | Chores | Hobbies | Socializing | Sleeping |
|---|---|---|---|---|---|
| Minutes spent | 441 | 161 | 278 | 166 | 394 |
| Concert | Audience Members |
|---|---|
| Date 1 | 20,361 |
| Date 2 | 19,946 |
| Date 3 | 19,799 |
| Date 4 | 20,194 |
| City A | City B | City C | |
|---|---|---|---|
| Carrot | 20 | 20 | 25 |
| Zucchini | 15 | 10 | 15 |
| Green Beans | 10 | 20 | 15 |
| Cauliflower | 20 | 15 | 10 |
| Younger than 20 | Between 20 and 40 | Between 40 and 60 | Older than 60 | |
|---|---|---|---|---|
| Very active | 4 | 10 | 6 | 2 |
| Somewhat active | 8 | 15 | 8 | 11 |
| Not active | 5 | 4 | 12 | 15 |