Confidence intervals for variance using chi-squared

1. A diabetic patient records her blood glucose readings in \(\mathrm { mmol } / \mathrm { l }\) at random times of day over several days. Her readings are given below.

$$\begin{array} { l l l l l l l } 5.3 & 5.7 & 8.4 & 8.7 & 6.3 & 8.0 & 7.2 \end{array}$$ Assuming that the blood glucose readings are normally distributed calculate

1. an unbiased estimate for the variance \(\sigma ^ { 2 }\) of the blood glucose readings,
2. a \(90 \%\) confidence interval for the variance \(\sigma ^ { 2 }\) of blood glucose readings.
3. State whether or not the confidence interval supports the assertion that \(\sigma = 0.9\). Give a reason for your answer.

4. A random sample of 15 tomatoes is taken and the weight \(x\) grams of each tomato is found. The results are summarised by \(\sum x = 208\) and \(\sum x ^ { 2 } = 2962\).

1. Assuming that the weights of the tomatoes are normally distributed, calculate the \(90 \%\) confidence interval for the variance \(\sigma ^ { 2 }\) of the weights of the tomatoes.
2. State with a reason whether or not the confidence interval supports the assertion \(\sigma ^ { 2 } = 3\).

4. The length \(X \mathrm {~mm}\) of a spring made by a machine is normally distributed \(\mathrm { N } \left( \mu , \sigma ^ { 2 } \right)\). A random sample of 20 springs is selected and their lengths measured in mm . Using this sample the unbiased estimates of \(\mu\) and \(\sigma ^ { 2 }\) are $$\bar { x } = 100.6 , \quad s ^ { 2 } = 1.5 .$$ Stating your hypotheses clearly test, at the \(10 \%\) level of significance,

1. whether or not the variance of the lengths of springs is different from 0.9 ,
2. whether or not the mean length of the springs is greater than 100 mm .

5. A machine is filling bottles of milk. A random sample of 16 bottles was taken and the volume of milk in each bottle was measured and recorded. The volume of milk in a bottle is normally distributed and the unbiased estimate of the variance, \(s ^ { 2 }\), of the volume of milk in a bottle is 0.003

1. Find a 95\% confidence interval for the variance of the population of volumes of milk from which the sample was taken. The machine should fill bottles so that the standard deviation of the volumes is equal to 0.07
2. Comment on this with reference to your 95\% confidence interval.

1. Two independent random samples \(X _ { 1 } , X _ { 2 } , \ldots , X _ { 7 }\) and \(Y _ { 1 } , Y _ { 2 } , Y _ { 3 } , Y _ { 4 }\) were taken from different normal populations with a common standard deviation \(\sigma\). The following sample statistics were calculated.

$$s _ { x } = 14.67 \quad s _ { y } = 12.07$$ Find the \(99 \%\) confidence interval for \(\sigma ^ { 2 }\) based on these two samples.

1. A machine produces components whose lengths are normally distributed with mean 102.3 mm and standard deviation 2.8 mm . After the machine had been serviced, a random sample of 20 components were tested to see if the mean and standard deviation had changed. The lengths, \(x \mathrm {~mm}\), of each of these 20 components are summarised as

$$\sum x = 2072 \quad \sum x ^ { 2 } = 214856$$

1. Stating your hypotheses clearly, test, at the \(5 \%\) level of significance, whether or not there is evidence of a change in standard deviation.
2. Stating your hypotheses clearly, test, at the \(5 \%\) level of significance, whether or not the mean length of the components has changed from the original value of 102.3 mm using
   1. a normal distribution,
   2. a \(t\) distribution.
3. Comment on the mean length of components produced after the service in the light of the tests from part (a) and part (b). Give a reason for your answer.

2. The time, \(t\) hours, that a typist can sit before incurring back pain is modelled by \(\mathrm { N } \left( \mu , \sigma ^ { 2 } \right)\). A random sample of 30 typists gave unbiased estimates for \(\mu\) and \(\sigma ^ { 2 }\) as shown below. $$\hat { \mu } = 2.5 \quad s ^ { 2 } = 0.36$$

1. Find a 95\% confidence interval for \(\sigma ^ { 2 }\)
2. State with a reason whether or not the confidence interval supports the assertion that \(\sigma ^ { 2 } = 0.495\)

1. A researcher is investigating the accuracy of IQ tests. One company offers IQ tests that it claims will give any individual's IQ with a standard deviation of 5

The researcher takes these tests 9 times with the following results $$123 , \quad 118 , \quad 127 , \quad 120 , \quad 134 , \quad 120 , \quad 118 , \quad 135 , \quad 121$$

1. Find the sample mean, \(\bar { x }\), and the sample variance, \(s ^ { 2 }\), of these scores.
   (2) Given that any individual's IQ scores on these tests are independent and have a normal distribution,
2. use the hypotheses $$\mathrm { H } _ { 0 } : \sigma ^ { 2 } = 25 \text { against } \mathrm { H } _ { 1 } : \sigma ^ { 2 } > 25$$ to test the company's claim at the \(5 \%\) significance level.
   (4) Gurdip works for the company and has taken these IQ tests 12 times. Gurdip claims that the sample variance of these 12 scores is \(s ^ { 2 } = 8.17\)
3. Use this value of \(s ^ { 2 }\) to calculate a \(95 \%\) confidence interval for the variance of Gurdip's IQ test scores.
   [0pt] [You may use \(\mathrm { P } \left( \chi _ { 11 } ^ { 2 } > 3.816 \right) = 0.975\) and \(\mathrm { P } \left( \chi _ { 11 } ^ { 2 } > 21.920 \right) = 0.025\) ]
4. Assuming that \(\sigma ^ { 2 } = 25\), comment on Gurdip's claim.

7. The times taken to travel to school by sixth form students are normally distributed. A head teacher records the times taken to travel to school, in minutes, of a random sample of 10 sixth form students from her school. Based on this sample, the \(95 \%\) confidence interval for the mean time taken to travel to school for sixth form students from her school is
[0pt] [28.5, 48.7] Calculate a 99\% confidence interval for the variance of the time taken to travel to school for sixth form students from her school.
(9)

6 A new employee, Kim, joins an existing employee, Jiang, to work in the quality control department of a company producing steel rods.
Each day a random sample of rods is taken, their lengths measured and a \(95 \%\) confidence interval for the mean length of the rods, in metres, is calculated. It is assumed that the lengths of the rods produced are normally distributed. Kim took a random sample of 25 rods and used the \(t\) distribution to obtain a \(95 \%\) confidence interval of \(( 1.193,1.367 )\) for the mean length of the rods. Jiang commented that this interval was a little wider than usual and explained that they usually assume that the standard deviation does not change and can be taken as 0.175 metres.

1. Test, at the \(10 \%\) level of significance, whether or not Kim's sample suggests that the standard deviation is different from 0.175 metres. State your hypotheses clearly. Using Kim's sample and the normal distribution with a standard deviation of 0.175 metres, (b) find a 95\% confidence interval for the mean length of the rods.

1. 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 \(\hat { \mu }\) and \(\hat { \sigma } ^ { 2 }\)

A \(95 \%\) confidence interval for \(\mu\) was found to be \(( 1.128,2.232 )\)

1. Show that \(\hat { \sigma } ^ { 2 } = 1.79\) (correct to 3 significant figures)
2. Obtain a \(95 \%\) confidence interval for \(\sigma ^ { 2 }\)

5. A machine is filling bottles of milk. A random sample of 16 bottles was taken and the volume of milk in each bottle was measured and recorded. The volume of milk in a bottle is normally distributed and the unbiased estimate of the variance, \(s ^ { 2 }\), of the volume of milk in a bottle is 0.003

1. Find a 95\% confidence interval for the variance of the population of volumes of milk from which the sample was taken. The machine should fill bottles so that the standard deviation of the volumes is equal to 0.07
2. Comment on this with reference to your 95\% confidence interval.