Detailed Answer

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.

Detailed Answer

a) 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” \( H_0 \).

Detailed Answer

\(H_0 \colon \mu_A = \mu_B\) 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.

Detailed Answer

\(H_0 \colon \mu_1 = \mu_2\) 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.

Detailed Answer

a) 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.

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:

• \( P(\text{weight} \leq 3) = 0.1587 \)
• \( P(3 < \text{weight} \leq 3.4) = 0.3413 \)
• \( P(3.4 < \text{weight} \leq 3.8) = 0.3413 \)
• \( P(\text{weight} > 3.8) = 0.1587 \)

Then, we need to multiply each probability to 60 and round it to the nearest integer:

Detailed Answer

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.

Detailed Answer

a) 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:

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.

Detailed Answer

a) 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 \(H_0\).

Detailed Answer

a) 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.

Detailed Answer

a) 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 \(H_0\).

Detailed Answer

a) (i) \( \mu = 20 \)

a) (ii) \( \mu < 20 \)

b) Using our GDC's z-Test function, we find:

\[ z = -1.79, \, p = 0.0368 \]

c) Since the found \( p \)-value is smaller than 0.05, we can conclude that the mean length of pencils is less than 20 cm.

Detailed Answer

a) The null and alternative hypotheses are:

\[ H_0: m = 10, \quad H_1: m < 10 \]

b) A Type I error occurs when \( H_0 \) is rejected even though it is true. If \( D \) is the distance they run in 1 hour, then:

\[ Pr(D < 9 | m = 10) = Pr(D \leq 8 | m = 10) = \text{poissonCdf}(10, 0, 8) = 0.333 \]

c) A Type II error occurs when the null hypothesis is accepted while it is false. The calculation is as follows:

\[ Pr(D \geq 9 | m = 9.5) = \text{poissonCdf}(9.5, 9, 100) = 0.608 \]

Detailed Answer

a) We can input the given lengths into our GDC and use One-variable statistics to find the values. The mean \(\mu\) is straightforward to find as it is the average. We are also asked to estimate the variance, which can be calculated by squaring the standard deviation provided by the calculator:

\[ \sigma^2 = 0.904^2 = 0.817 \]

b) Since the population variance is unknown, we need to use the t-distribution. We know \(\alpha = 0.05\) and \(v = 9\):

\[ \text{Error} = t_{v, \frac{\alpha}{2}} \cdot \frac{s}{\sqrt{n}} \]

\[ = \text{invT}(0.975, 9) \cdot \frac{\sqrt{0.817}}{\sqrt{10}} = 0.647 \]

Thus, the required confidence interval is:

\[ 20.22 - 0.647 < \mu < 20.22 + 0.647 \]

\[ 19.573 < \mu < 20.867 \]

c) (i) We can continue using the same spreadsheet as in the previous part and conduct a t-test for the data with our GDC. We find \(t = 4.27\) and \(p\text{-value} = 0.0021\).

c) (ii) Our \(p\)-value is smaller than 0.05, so we reject \(H_0\), indicating that the initial assumption that their lengths are 19 cm was incorrect.

Detailed Answer

a) First, we need to calculate the mean of the provided data: \( \frac{11 + 10 + 9.5 + 10.5 + 10.2}{5} = 10.24 \)

We can then use the z-values. We know that \(1 - \alpha = 0.95\), so \(\alpha = 0.05\) and \(\frac{\alpha}{2} = 0.025\). We now calculate the margin of error:

\[ \text{Error} = Z_{0.025} \cdot \frac{\sigma}{\sqrt{n}} \]

\[ = 1.96 \cdot \frac{1.5}{\sqrt{5}} \approx 1.31 \]

So, the 95% confidence interval is:

\[ 10.24 - 1.31 < \mu < 10.24 + 1.31 \]

\[ 8.93 < \mu < 11.6\]

b)

\[ Z_{\frac{\alpha'}{2}} \cdot \frac{n}{\sqrt{\sigma}} = \frac{1}{3} \cdot Z_{\frac{\alpha}{2}} \cdot \frac{n}{\sqrt{\sigma}} \]

\[ Z_{\frac{\alpha'}{2}} = \frac{1}{3} \cdot Z_{\frac{\alpha}{2}} \]

\[ Z_{\frac{\alpha'}{2}} = \frac{1}{3} \cdot 1.96 \approx 0.65 \]

\[ \frac{\alpha'}{2} = \text{normCDF}(0.65, 100, 0, 1) \approx 0.26 \]

\[ \alpha' \approx 0.52 \]

\[ 1 - \alpha' \approx 0.48 \]

So, the confidence level of the interval is approximately 48%.