Estimated variance confidence interval

3 The times, in minutes, taken by people to complete a walk are normally distributed with mean \(\mu\). The times, \(t\) minutes, for a random sample of 80 people were summarised as follows. $$\Sigma t = 7220 \quad \Sigma t ^ { 2 } = 656060$$

1. Calculate a \(97 \%\) confidence interval for \(\mu\).
2. Explain whether it was necessary to use the Central Limit theorem in part (i).

7 The diameter, in cm, of pistons made in a certain factory is denoted by \(X\), where \(X\) is normally distributed with mean \(\mu\) and variance \(\sigma ^ { 2 }\). The diameters of a random sample of 100 pistons were measured, with the following results. $$n = 100 \quad \Sigma x = 208.7 \quad \Sigma x ^ { 2 } = 435.57$$

1. Calculate unbiased estimates of \(\mu\) and \(\sigma ^ { 2 }\). The pistons are designed to fit into cylinders. The internal diameter, in cm , of the cylinders is denoted by \(Y\), where \(Y\) has an independent normal distribution with mean 2.12 and variance 0.000144 . A piston will not fit into a cylinder if \(Y - X < 0.01\).
2. Using your answers to part (i), find the probability that a randomly chosen piston will not fit into a randomly chosen cylinder.

4 The probability density function of a random variable \(X\) is given by $$\mathrm { f } ( x ) = \begin{cases} k x & 0 \leqslant x \leqslant a \\ k ( 2 a - x ) & a < x \leqslant 2 a \\ 0 & \text { otherwise } \end{cases}$$ where \(a\) and \(k\) are positive constants.

1. Sketch \(\mathrm { f } ( x )\). Hence explain why \(\mathrm { E } ( X ) = a\).
2. Show that \(k = \frac { 1 } { a ^ { 2 } }\).
3. Find \(\operatorname { Var } ( X )\) in terms of \(a\). In order to estimate the value of \(a\), a random sample of size 50 is taken from the distribution. It is found that the sample mean and standard deviation are \(\bar { x } = 1.92\) and \(s = 0.8352\).
4. Construct a symmetrical \(95 \%\) confidence interval for \(a\). Give one reason why the answer is only approximate.
5. A non-statistician states that the probability that \(a\) lies in the interval found in part (iv) is 0.95 . Comment on this statement. \section*{END OF QUESTION PAPER} \section*{OCR \(^ { \text {® } }\)}

9 The continuous random variable \(C\) has the distribution \(\mathrm { N } \left( \mu , \sigma ^ { 2 } \right)\). The sum of a random sample of 16 observations of \(C\) is 224.0 .

1. Find an unbiased estimate of \(\mu\).
2. It is given that an unbiased estimate of \(\sigma ^ { 2 }\) is 0.24. Find the value of \(\Sigma c ^ { 2 }\). \(D\) is the sum of 10 independent observations of \(C\).
3. Explain whether \(D\) has a normal distribution. The continuous random variable \(F\) is normally distributed with mean 15.0, and it is known that \(\mathrm { P } ( F < 13.2 ) = 0.115\).
4. Use the unbiased estimates of \(\mu\) and \(\sigma ^ { 2 }\) to find \(\mathrm { P } ( D + F > 157.0 )\). \section*{OCR} \section*{Oxford Cambridge and RSA}

5 The mass, \(X\), in grams of a particular type of apple is modelled using a normal distribution. A random sample of 12 apples is collected and the summarised results are $$\sum x = 1038 \quad \text { and } \quad \sum x ^ { 2 } = 90100$$ 5

1. A 99\% confidence interval for the population mean of the masses of the apples is constructed using the random sample. Show that the confidence interval is \(( 81.7,91.3 )\) with values correct to three significant figures.
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2. Padraig claims that the population mean mass of the apples is 85 grams. He carries out a hypothesis test at the \(1 \%\) level of significance using the random sample of 12 apples. The hypotheses are $$\begin{aligned} & \mathrm { H } _ { 0 } : \mu = 85 \\ & \mathrm { H } _ { 1 } : \mu \neq 85 \end{aligned}$$ State, with a reason, whether the null hypothesis is accepted or rejected.
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3. Interpret, in context, the conclusion to the hypothesis test in part (b).