Calculate CI from summary stats

7 An analyst routinely examines bottles of hair shampoo in order to check that the average percentage of a particular chemical which the shampoo contains does not exceed the value of \(1.0 \%\) specified by the manufacturer. The percentages of the chemical in a random sample of 12 bottles of the shampoo are as follows. \(\begin{array} { l l l l l l l l l l l } 1.087 & 1.171 & 1.047 & 0.846 & 0.909 & 1.052 & 1.042 & 0.893 & 1.021 & 1.085 & 1.096 \end{array} 0.931\) The analyst uses software to draw a Normal probability plot for these data, and to carry out a Normality test as shown below. \includegraphics[max width=\textwidth, alt={}, center]{c692fb20-436f-4bc1-89bd-10fdba41ceba-08_524_1539_694_264}

1. The analyst is going to carry out a hypothesis test to check whether the average percentage exceeds 1.0\%. Explain which test the analyst should use, referring to each of the following.
   Carry out the test at the 5\% significance level.

The mean Intelligence Quotient (IQ) of a random sample of 15 pupils at School A is 109. The mean IQ of a random sample of 20 pupils at School B is 112. You may assume that the IQs for the populations from which these samples are taken are normally distributed, and that both distributions have standard deviation 15. Find a 90% confidence interval for \(\mu_B - \mu_A\), where \(\mu_A\) and \(\mu_B\) are the population mean IQs. [6]

The heights of a certain species of deer are known to have standard deviation \(0.35\) m. A zoologist takes a random sample of \(150\) of these deer and finds that the mean height of the deer in the sample is \(1.42\) m.

1. Calculate a \(96\%\) confidence interval for the population mean height. [3]
2. Bubay says that \(96\%\) of deer of this species are likely to have heights that are within this confidence interval. Explain briefly whether Bubay is correct. [1]

The time taken for a particular type of paint to dry was measured for a sample of 150 randomly chosen points on a wall. The sample mean was 192.4 minutes and an unbiased estimate of the population variance was 43.6 minutes\(^2\). Find a 98\% confidence interval for the mean drying time. [3]

The time taken, \(T\) minutes, for a special anti-rust paint to dry was measured for a random sample of 120 painted pieces of metal. The sample mean was 51.2 minutes and an unbiased estimate of the population variance was 37.4 minutes\(^2\). Determine a 99% confidence interval for the mean drying time. [3]

Leaves from a certain type of tree have lengths that are distributed with standard deviation 3 cm. A random sample of 6 of these leaves is taken and the mean length of this sample is found to be 8 cm.

1. Calculate a 95\% confidence interval for the population mean length. [3]
2. Write down the probability that the whole 95\% confidence interval will lie below the population mean. [1]

The continuous random variable \(U\) has a normal distribution with unknown mean \(\mu\) and known variance 1. A random sample of four observations of \(U\) gave the values \(3.9, 2.1, 4.6\) and \(1.4\).

1. Calculate a \(90\%\) confidence interval for \(\mu\). [3]
2. The probability that the sum of four random observations of \(U\) is less than 11 is denoted by \(p\). For each of the end points of the confidence interval in part (i) calculate the corresponding value of \(p\). [5]

A teacher gives each student in his class a list of 30 numbers. All the numbers have been generated at random by a computer from a normal distribution with a fixed mean and variance. The teacher tells the class that the variance of the distribution is 25 and asks each of them to calculate a 95\% confidence interval based on their list of numbers. The sum of the numbers given to one student is 1419.

1. Find the confidence interval that should be obtained by this student. [5]

Assuming that all the students calculate their confidence intervals correctly,

1. state the proportion of the students you would expect to have a confidence interval that includes the true mean of the distribution, [1]
2. explain why the probability of any one student's confidence interval including the true mean is not 0.95 [1]

Brickland and Goodbrick are two manufacturers of bricks. The lengths of the bricks produced by each manufacturer can be assumed to be normally distributed. A random sample of 20 bricks is taken from Brickland and the length, \(x\) mm, of each brick is recorded. The mean of this sample is 207.1 mm and the variance is 3.2 mm².

1. Calculate the 98\% confidence interval for the mean length of brick from Brickland. [4]

A random sample of 10 bricks is selected from those manufactured by Goodbrick. The length of each brick, \(y\) mm, is recorded. The results are summarised as follows. \(\sum y = 2046.2\) \(\sum y^2 = 418785.4\) The variances of the length of brick for each manufacturer are assumed to be the same.

1. [(b)] Find a 90\% confidence interval for the value by which the mean length of brick made by Brickland exceeds the mean length of brick made by Goodbrick. [8]

(Total 12 marks)

A machine fills jars with jam. The weight of jam in each jar is normally distributed. To check the machine is working properly the contents of a random sample of 15 jars are weighed in grams. Unbiased estimates of the mean and variance are obtained as $$\mu = 560 \quad s^2 = 25.2$$ Calculate a 95\% confidence interval for,

1. the mean weight of jam, [4]
2. the variance of the weight of jam. [5]

A weight of more than 565g is regarded as too high and suggests the machine is not working properly.

1. Use appropriate confidence limits from parts (a) and (b) to find the highest estimate of the proportion of jars that weigh too much. [5]

A nutritionist studied the levels of cholesterol, \(X\) mg/cm³, of male students at a large college. She assumed that \(X\) was distributed \(\text{N}(\mu, \sigma^2)\) and examined a random sample of 25 male students. Using this sample she obtained unbiased estimates of \(\mu\) and \(\sigma^2\) as $$\hat{\mu} = 1.68, \quad \hat{\sigma}^2 = 1.79.$$

1. Find a 95% confidence interval for \(\mu\). [4]
2. Obtain a 95% confidence interval for \(\sigma^2\). [5]

A cholesterol reading of more than 2.5 mg/cm³ is regarded as high.

1. Use appropriate confidence limits from parts \((a)\) and \((b)\) to find the lowest estimate of the proportion of male students in the college with high cholesterol. [4]

Murni is investigating the annual salary of people from a particular town. She takes a random sample of 200 people from the town and records their annual salary. The mean annual salary is £28 500 and the standard deviation is £5100 Calculate a 97% confidence interval for the population mean of annual salaries for the people who live in the town, giving your values to the nearest pound. [3 marks]

The continuous random variable \(X\) has the distribution \(\text{N}(\mu, 30)\). The mean of a random sample of 8 observations of \(X\) is 53.1. Determine a 95\% confidence interval for \(\mu\). You should give the end points of the interval correct to 4 significant figures. [4]

Rugby players sometimes use protein powder to aid muscle increase. The monthly weight gains of rugby players taking protein powder may be modelled by a normal distribution having a standard deviation of 40 g and a mean which may depend on the type of protein powder they consume. A rugby team coach gives the same amount of protein powder over a trial month to 22 randomly selected players. Protein powder \(A\) was used by 12 players, randomly selected, and their mean weight gain was 900 g. Protein powder \(B\) was used by the other 10 players and their mean weight gain was 870 g. Let \(\mu_A\) and \(\mu_B\) be the mean monthly weight gains, in grams, of the populations of rugby players who use protein powder \(A\) and protein powder \(B\) respectively.

1. Calculate a 98\% confidence interval for \(\mu_A - \mu_B\). [4]
2. In the given context, what can you conclude from your answer to part (a)? Give a reason for your answer. [2]
3. Find the confidence level of the largest confidence interval that would lead the coach to favour protein powder \(A\) over protein powder \(B\). [4]
4. State one non-statistical assumption you have made in order to reach these conclusions. [1]

During practice sessions, a basketball coach makes his players run several 'line drills'.

1. He records the times taken, in seconds, by one of his players to run the first 'line drill' on a random sample of 8 practice sessions. The results are shown below. 29.4 \quad 31.1 \quad 28.9 \quad 30.0 \quad 29.9 \quad 30.4 \quad 29.7 \quad 30.2 Assuming that these data come from a normal distribution with mean \(\mu\) and variance 0.6, calculate a 95\% confidence interval for \(\mu\). [5]
2. State the two ways in which the method used to calculate the confidence interval in part (a) would change if the variance were unknown. [2]
3. During a practice session, a player recorded a mean time of 35.6 seconds for 'line drills'.
   1. Give a reason why this player may not be the same as the player in part (a).
   2. Give a reason why this player could be the same as the player in part (a). [2]

Alan's journey time to work can be modelled by a normal distribution with standard deviation 6 minutes. Alan measures the journey time to work for a random sample of 5 journeys. The mean of the 5 journey times is 36 minutes.

1. Construct a 95\% confidence interval for Alan's mean journey time to work, giving your values to one decimal place. [2 marks]
2. Alan claims that his mean journey time to work is 30 minutes. State, with a reason, whether or not the confidence interval found in part (a) supports Alan's claim. [1 mark]
3. Suppose that the standard deviation is not known but a sample standard deviation is found from Alan's sample and calculated to be 6. Explain how the working in part (a) would change. [1 mark]