Answer all questions

The mean is the sum of all values divided by the number of values.

Rounding to three significant figures:

The median is the middle value when the data is ordered. Since there are 13 values, the median is the $\frac{13+1}{2} = 7^{\text{th}}$ value.
The ordered data is: 6.2, 6.7, 6.8, 8.1, 8.1, 8.5, 8.6, 9.0, 9.9, 10.1, 11.0, 11.8, 22.8

The midrange is the average of the minimum and maximum values.
Minimum value = 6.2
Maximum value = 22.8

b) Comment on how useful each of these is as a measure of central tendency for these data, giving a brief reason of each of your answers.

• Mean:

  • Usefulness: Less useful.
  • Reason: The mean is significantly affected by the outlier (22.8 tonnes), which pulls the average towards the higher end and does not accurately represent the typical emission value for most countries.

• Median:

  • Usefulness: Most useful.
  • Reason: The median is robust to outliers. It represents the middle value, so it is not skewed by the unusually high emission of 22.8 tonnes, providing a better indication of the central tendency for the majority of the data.

• Midrange:

  • Usefulness: Least useful.
  • Reason: The midrange is calculated using only the minimum and maximum values, making it extremely sensitive to outliers (like 22.8 tonnes). It does not reflect the distribution of the bulk of the data at all.

: Time Taken to Type Documents

The frequency distribution for the time taken to type 250 documents is:

a) Calculate the modal time taken.

The modal class is the class with the highest frequency. Here, it is $20 < t \le 30$ with a frequency of 81.
From this, we have:
Lower boundary of modal class ($L$) = 20
Frequency of modal class ($f_m$) = 81
Frequency of class before modal class ($f_1$) = 54
Frequency of class after modal class ($f_2$) = 57
Class width ($w$) = 10

Rounding to one decimal place:

b) Calculate the median time taken.

The median position is $\frac{N}{2} = \frac{250}{2} = 125^{\text{th}}$ value.

From the cumulative frequency column, the 125th value falls into the class $20 < t \le 30$ (since its cumulative frequency is 163, which is the first to exceed 125).
From this, we have:
Lower boundary of median class ($L$) = 20
Total frequency ($N$) = 250
Cumulative frequency of class before median class ($C_f$) = 82
Frequency of median class ($f_m$) = 81
Class width ($w$) = 10

Rounding to one decimal place:

: Biased Coin Test

A coin is tossed 10 times ($n=10$).
The coin is considered biased if there are 9 or 10 heads, or 9 or 10 tails.
Let $X$ be the number of heads.
The rejection region (R) for the null hypothesis is $X \le 1$ (0 or 1 head, meaning 9 or 10 tails) or $X \ge 9$ (9 or 10 heads).
Null Hypothesis ($H_0$): The coin is fair, $p = 0.5$.
Alternative Hypothesis ($H_1$): The coin is biased, $p \ne 0.5$.

a) Show that the probability of making a Type I error in this test is approximately 0.0215.

A Type I error occurs when we reject $H_0$ when $H_0$ is true.
Under $H_0$, the probability of a head is $p=0.5$. The number of heads $X$ follows a binomial distribution $B(10, 0.5)$.

$P(\text{Type I error}) = P(X \in R | p=0.5) = P(X=0 \text{ or } X=1 \text{ or } X=9 \text{ or } X=10 | p=0.5)$
Using the binomial probability formula $P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$:

Rounding to four decimal places:

b) Find the significance of the test.

The significance level of a hypothesis test is the probability of making a Type I error.

From part a), the probability of a Type I error is 0.021484375.

c) Find the probability of making a Type II error in this test when the probability of a head is actually 0.7.

A Type II error occurs when we fail to reject $H_0$ when $H_1$ is actually true.
Here, $H_1$ is true with $p=0.7$. The number of heads $X$ follows a binomial distribution $B(10, 0.7)$.
The acceptance region (fail to reject $H_0$) is $2 \le X \le 8$.

It is easier to calculate $1 - P(\text{Reject } H_0 | p=0.7)$.
The rejection region is $X=0, 1, 9, 10$.
Calculate individual probabilities for $B(10, 0.7)$:

Rounding to four decimal places: