Unknown variance (t-distribution)

4 The error, \(X ^ { \circ } \mathrm { C }\), made in measuring a patient's temperature at a local doctors' surgery may be modelled by a normal distribution with mean \(\mu\) and standard deviation \(\sigma\). The errors, \(x ^ { \circ } \mathrm { C }\), made in measuring the temperature of each of a random sample of 10 patients are summarised below. $$\sum x = 0.35 \quad \text { and } \quad \sum ( x - \bar { x } ) ^ { 2 } = 0.12705$$ Construct a \(99 \%\) confidence interval for \(\mu\), giving the limits to three decimal places.
(5 marks)

1 At the start of the 2012 season, the ages of the members of the Warwickshire Acorns Cricket Club could be modelled by a normal random variable, \(X\) years, with mean \(\mu\) and standard deviation \(\sigma\). The ages, \(x\) years, of a random sample of 15 such members are summarised below. $$\sum x = 546 \quad \text { and } \quad \sum ( x - \bar { x } ) ^ { 2 } = 1407.6$$

1. Construct a \(98 \%\) confidence interval for \(\mu\), giving the limits to one decimal place.
   (6 marks)
2. At the start of the 2005 season, the mean age of the members was 40.0 years. Use your confidence interval constructed in part (a) to indicate, with a reason, whether the mean age had changed.

1 Gemma, a biologist, studies guillemots, which are a species of seabird. She has found that the weight of an adult guillemot may be modelled by a normal distribution with mean \(\mu\) grams. During 2012, she measured the weight, \(x\) grams, of each of a random sample of 9 adult guillemots and obtained the following results. $$\sum x = 8532 \quad \text { and } \quad \sum ( x - \bar { x } ) ^ { 2 } = 38538$$

1. Construct a 98\% confidence interval for \(\mu\) based on these data.
2. The corresponding confidence interval for \(\mu\) obtained by Gemma based on a random sample of 9 adult guillemots measured during 2011 was \(( 927,1063 )\), correct to the nearest gram.
   1. Find the mean weight of guillemots in this sample.
   2. Studies of some other species of seabird have suggested that their mean weights were less in 2012 than in 2011. Comment on whether Gemma's two confidence intervals provide evidence that the mean weight of guillemots was less in 2012 than in 2011.
      (2 marks)

6. A supervisor wishes to cheek the typing speed of a new typist. On 10 randomly selected occasions, the supervisor records the time taken for the new typist to type 100 words. The results, in seconds, are given below. $$110,125,130,126,128,127,118,120,122,125$$ The supervisor assumes that the time taken to type 100 words is normally distributed.

1. Calculate a 95\% confidence interval for
   1. the mean,
   2. the variance
      of the population of times taken by this typist to type 100 words. The supervisor requires the average time needed to type 100 words to be no more than 130 seconds and the standard deviation to be no more than 4 seconds.
2. Comment on whether or not the supervisor should be concerned about the speed of the new typist.

7. A doctor wishes to study the level of blood glucose in males. The level of blood glucose is normally distributed. The doctor measured the blood glucose of 10 randomly selected male students from a school. The results, in mmol/litre, are given below. $$\begin{array} { l l l l l l l l l l } 4.7 & 3.6 & 3.8 & 4.7 & 4.1 & 2.2 & 3.6 & 4.0 & 4.4 & 5.0 \end{array}$$

1. Calculate a \(95 \%\) confidence interval for the mean.
2. Calculate a 95\% confidence interval for the variance. A blood glucose reading of more than 7 mmol/litre is counted as high.
3. Use appropriate confidence limits from parts (a) and (b) to find the highest estimate of the proportion of male students in the school with a high blood glucose level. \section*{END}

1. 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

$$\hat { \mu } = 560 \quad s ^ { 2 } = 25.2$$ Calculate a 95\% confidence interval for,

1. the mean weight of jam,
2. the variance of the weight of jam. A weight of more than 565 g is regarded as too high and suggests the machine is not working properly.
3. Use appropriate confidence limits from parts (a) and (b) to find the highest estimate of the proportion of jars that weigh too much.

4. A random sample of 15 strawberries is taken from a large field and the weight \(x\) grams of each strawberry is recorded. The results are summarised below. $$\sum x = 291 \quad \sum x ^ { 2 } = 5968$$ Assume that the weights of strawberries are normally distributed. Calculate a 95\% confidence interval for

1. 1. the mean of the weights of the strawberries in the field,
   2. the variance of the weights of the strawberries in the field. Strawberries weighing more than 23 g are considered to be less tasty.
2. Use appropriate confidence limits from part (a) to find the highest estimate of the proportion of strawberries that are considered to be less tasty.

5. A large company has designed an aptitude test for new recruits. The score, \(S\), for an individual taking the test, has a normal distribution with mean \(\mu\) and standard deviation \(\sigma\). In order to estimate \(\mu\) and \(\sigma\), a random sample of 15 new recruits were given the test and their scores, \(x\), are summarised as $$\sum x = 880 \quad \sum x ^ { 2 } = 54892$$

1. Calculate a 95\% confidence interval for
   1. \(\mu\),
   2. \(\sigma\). The company wants to ensure that no more than \(80 \%\) of new recruits pass the test.
2. Using values from your confidence intervals in part (a), estimate the lowest pass mark they should set.

3. A large number of chicks were fed a special diet for 10 days. A random sample of 9 of these chicks is taken and the weight gained, \(x\) grams, by each chick is recorded. The results are summarised below. $$\sum x = 181 \quad \sum x ^ { 2 } = 3913$$ You may assume that the weights gained by the chicks are normally distributed.
Calculate a 95\% confidence interval for

1. 1. the mean of the weights gained by the chicks,
   2. the variance of the weights gained by the chicks. A chick which gains less than \(16 g\) has to be given extra feed.
2. Using appropriate confidence limits from part (a), find the lowest estimate of the proportion of chicks that need extra feed.

1. Fred is a new employee in a delicatessen. He is asked to cut cheese into 100 g blocks. A random sample of 8 of these blocks of cheese is selected. The weight, in grams, of each block of cheese is given below

$$94 , \quad 106 , \quad 115 , \quad 98 , \quad 111 , \quad 104 , \quad 113 , \quad 102$$

1. Calculate a \(90 \%\) confidence interval for the standard deviation of the weights of the blocks of cheese cut by Fred. Given that the weights of the blocks of cheese are independent,
2. state what further assumption is necessary for this confidence interval to be valid. The delicatessen manager expects the standard deviation of the weights of the blocks of cheese cut by an employee to be less than 5 g. Any employee who does not achieve this target is given training.
3. Use your answer from part (a) to comment on Fred's results. A second employee, Olga, has just been given training. Olga is asked to cut cheese into 100 g blocks. A random sample of 20 of these blocks of cheese is selected. The weight of each block of cheese, \(x\) grams, is recorded and the results are summarised below. $$\bar { x } = 102.6 \quad s ^ { 2 } = 19.4$$ Given that the assumption in part (b) is also valid in this case,
4. test, at a \(10 \%\) level of significance, whether or not the mean weight of the blocks of cheese cut by Olga after her training is 100 g . State your hypotheses clearly.
   (6)

1. Jeremiah currently uses a Fruity model of juicer. He agrees to trial a new model of juicer, Zesty. The amounts of juice extracted, \(x \mathrm { ml }\), from each of 9 randomly selected oranges, using the Zesty are summarised as

$$\sum x = 468 \quad \sum x ^ { 2 } = 24560$$ Given that the amounts of juice extracted follow a normal distribution,

1. calculate a 95\% confidence interval for
   1. the mean amount of juice extracted from an orange using the Zesty,
   2. the standard deviation of the amount of juice extracted from an orange using the Zesty. Jeremiah knows that, for his Fruity, the mean amount of juice extracted from an orange is 38 ml and the standard deviation of juice extracted from an orange is 5 ml . He decides that he will replace his Fruity with a Zesty if both
      • the mean for the Zesty is more than \(20 \%\) higher than the mean for his Fruity and
2. the standard deviation for the Zesty is less than 5.5 ml .
3. Using your answers to part (a), explain whether or not Jeremiah should replace his Fruity with the Zesty.
   

4. Gill, a member of the accounts department in a large company, is studying the expenses claims of company employees. She assumes that the claims, in \(\pounds\), follow a normal distribution with mean \(\mu\) and variance \(\sigma ^ { 2 }\). As a first stage in her investigation she took the following random sample of 10 claims. $$30.85,99.75,142.73,223.16,75.43,28.57,53.90,81.43,68.62,43.45 .$$

1. Find a 95\% confidence interval for \(\mu\). The chief accountant would like a \(95 \%\) confidence interval where the difference between the upper confidence limit and the lower confidence limit is less than 20 .
2. Show that \(\frac { \sigma ^ { 2 } } { n } < 26.03\) (to 2 decimal places), where \(n\) is the size of the sample required to achieve this. Gill decides to use her original sample of 10 to obtain a value for \(\sigma ^ { 2 }\) so that the chance of her value being an underestimate is 0.01 .
3. Find such a value for \(\sigma ^ { 2 }\).
4. Use this value for \(\sigma ^ { 2 }\) to estimate the size of sample the chief accountant requires.

4 Shellfish in the sea near nuclear power stations are regularly monitored for levels of radioactivity. On a particular occasion, the levels of caesium-137 (a radioactive isotope) in a random sample of 8 cockles, measured in becquerels per kilogram, were as follows. \(\begin{array} { l l l l l l l l } 2.36 & 2.97 & 2.69 & 3.00 & 2.51 & 2.45 & 2.21 & 2.63 \end{array}\) Software is used to produce a 95\% confidence interval for the level of caesium-137 in the cockles. The output from the software is shown in Fig. 4. The value for 'SE' has been deliberately omitted. T Estimate of a Mean
Confidence Level 0.95 Sample
Mean 2.6025
s 0.2793
□
0.2793 N □ 8 Result T Estimate of a Mean \begin{table}[h] \end{table}

1. State an assumption necessary for the use of the \(t\) distribution in the construction of this confidence interval.
2. State the confidence interval which the software gives in the form \(a < \mu < b\).
3. In the software output shown in Fig. 4, SE stands for standard error. Find the standard error in this case.
4. Show how the value of 0.2335 in the confidence interval was calculated.
5. State how, using this sample, a wider confidence interval could be produced.

5 Jane, who supplies fruit to a jam manufacturer, knows that the weight of fruit in boxes that she sends to the manufacturer can be modelled by a normal distribution with unknown mean, \(\mu\) grams, and unknown standard deviation, \(\sigma\) grams. Jane selects a random sample of 16 boxes and, using the \(t\)-distribution, calculates correctly that a \(98 \%\) confidence interval for \(\mu\) is \(( 70.65,80.35 )\).

1. 1. Show that the sample mean is 75.5 grams.
   2. Find the width of the confidence interval.
   3. Calculate an estimate of the standard error of the mean.
   4. Hence, or otherwise, show that an unbiased estimate of \(\sigma ^ { 2 }\) is 55.6 , correct to three significant figures.
2. Jane decides that the width of the \(98 \%\) confidence interval is too large. Construct a \(95 \%\) confidence interval for \(\mu\), based on her sample of 16 boxes.
3. Jane is informed that the manufacturer would prefer the confidence interval to have a width of at most 5 grams.
   1. Write down a confidence interval for \(\mu\), again based on Jane's sample of 16 boxes, which has a width of 5 grams.
   2. Determine the percentage confidence level for your interval in part (c)(i).

6 Bishen believes that the mean weight of boxes of black peppercorns is 45 grams. Abi, thinking that this is not the case, weighs, in grams, a random sample of 8 boxes of black peppercorns, with the following results. $$\begin{array} { l l l l l l l l } 44 & 44 & 43 & 46 & 42 & 40 & 43 & 46 \end{array}$$

1. 1. Construct a \(95 \%\) confidence interval for the mean weight of boxes of black peppercorns, stating any assumption that you make.
   2. Comment on Bishen's belief.
2. 1. Abi claims that the mean weight of boxes of black peppercorns is less than 45 grams. Test this claim at the \(5 \%\) level of significance.
   2. If Bishen's belief is true, state, with a reason, what type of error, if any, may have occurred when conclusions to the test in part (b)(i) were drawn.
      (2 marks)

6. A nutritionist studied the levels of cholesterol, \(X \mathrm { mg } / \mathrm { cm } ^ { 3 }\), of male students at a large college. She assumed that \(X\) was distributed \(\mathrm { N } \left( \mu , \sigma ^ { 2 } \right)\) 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 \(\mu\).
2. Obtain a \(95 \%\) confidence interval for \(\sigma ^ { 2 }\). A cholesterol reading of more than \(2.5 \mathrm { mg } / \mathrm { cm } ^ { 3 }\) is regarded as high.
3. 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.