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Statistics, SPEC2 2024 VCAA 6

A machine fills bottles with olive oil. The volume of olive oil dispensed into each bottle may be assumed to be normally distributed with mean \(\mu\) millilitres (mL) and standard deviation \(\sigma=\)4.2 mL. When the machine is working properly \(\mu=1000\).

The volume dispensed is monitored regularly by taking a random sample of nine bottles and finding the mean volume dispensed.

The machine will be paused and adjusted if the mean volume of olive oil in the nine bottles is significantly less than 1000 mL at the 5% level of significance.

When checked, a random sample of nine bottles gave a mean volume of 997.5 mL .

A one-sided statistical test is to be performed.

  1. Write down suitable null and alternative hypotheses \(H_0\) and \(H_1\) for the test.   (1 mark)

    1. Find the \(p\) value for this test correct to three decimal places.  (1 mark)

    2. Using the \(p\) value found in part b.i, state with a reason whether the machine should be paused.  (1 mark)

  3. Assuming that the mean volume dispensed by the machine each time is in fact 997 mL and not 1000 mL, find the probability of a type \(\text{II}\) error for the test using nine bottles at the 5% level of significance. Assume that the population standard deviation is 4.2 mL, and give your answer correct to two decimal places.  (2 marks)

  4. Let \(\overline{X}\) denote the sample mean of a random sample of nine bottles. As a quality-control measure, the machine will be paused if  \(\overline{X}<a\)  or if  \(\overline{X}>b\),  where  \(\operatorname{Pr}(\overline{X}<a)=0.01\)  and  \(\operatorname{Pr}(\overline{X}>b)=0.01\).
  5. Assume \(\mu=1000\) mL and \(\sigma=4.2\) mL.
  6. Find the values of \(a\) and \(b\) correct to one decimal place.  (1 mark)

A new machine is purchased, and it is observed that the volume dispensed by the new machine in 50 randomly chosen bottles provided a sample mean of 1005 mL and a sample standard deviation of 4 mL .

  1. Find a 95% confidence interval for the population mean volume dispensed by the new machine, giving values correct to one decimal place. You may assume a population standard deviation of 4 mL .  (1 mark)

  2. Forty samples, each consisting of 50 randomly chosen bottles, are taken, and a 95% confidence interval is calculated for each sample.
  3. In how many of these confidence intervals would the population mean volume dispensed by the machine be expected to lie?  (1 mark)

  4. What minimum size sample should be used so that, with 95% confidence, the sample mean is within 1 mL of the population mean volume dispensed by the new machine? Assume a population standard deviation of 4 mL.  (1 mark)

Show Answers Only

a.    \(H_0: \mu=1000, \ H_1: \mu<1000\)

b.i.  \(p \text{–value}=0.037\)

b.ii. \(p<0.05 \Rightarrow \text{Pause the machine}\)

c.   \(0.31\)

d.   \(a=996.7, b=1003.3\)

e.   \((1003.9,1006.1)\)

f.   \(38\)

g.  \(\text{Mean sample size}=62.\)

Show Worked Solution

a.    \(H_0: \mu=1000\)

\(H_1: \mu<1000\)
 

b.i.  \(\text {Sample mean:}\  \mu_{\overline{X}}=1000\)

\(\dfrac{\sigma_{\overline{X}}}{\sqrt{n}}=\dfrac{4.2}{3}=1.4\)

\(p \text{–value}=\operatorname{Pr}\left(z<\dfrac{997.5-1000}{1.4}\right)=0.037\)
 

b.ii. \(p<0.05 \Rightarrow \text{Pause the machine}\)
 

c.   \(\text{Let \(\overline{X}\)}=\text{sample mean}\)

\(\text{Null hypothesis rejected when}\ \overline{X}<c\)

\(\operatorname{Pr}\left(z<\dfrac{c-1000}{1.4}\right)=0.05\)

\(\text{Solve for \(c\):}\)

\(\dfrac{c-1000}{1.4}=-1.6449 \ \Rightarrow \ c=997.697\)

\(\text{Probability of a type II error:}\)

\(\operatorname{Pr}(\overline{X}>997.697 \mid \mu=997)=\operatorname{Pr}\left(z<\dfrac{0.697}{1.4}\right) \approx 0.31\)

♦ Mean mark (c) 50%.

d.    \(\text{Sample mean has mean}=1000, \dfrac{\sigma}{\sqrt{n}}=\dfrac{4.2}{\sqrt{9}}=1.4\)

\(\text{Solve (by CAS):}\)

\(\operatorname{Pr}\left(z<\dfrac{a-1000}{1.4}\right)=0.01 \ \Rightarrow \ a=996.7\)

\(\operatorname{Pr}\left(z>\dfrac{b-1000}{1.4}\right)=0.01 \ \Rightarrow \ b=1003.3\)
 

e.    \(\text{95% C.I.}\ (\overline{x}=1005):\)

\(\left(1005-1.96 \times \dfrac{4}{\sqrt{50}}, 1005+1.96 \times \dfrac{4}{\sqrt{50}}\right)=(1003.9,1006.1)\)
 

f.    \(\text{Expect \(95 \%\) of C.I.’s to contain population mean}\)

\(\text{Number of C.I.’s}=0.95 \times 40=38\)
 

g.    \(\text{CI. extends  \(1.96 \times \dfrac{4}{\sqrt{n}}\)  each side of the mean.}\)

\(\text{Solve for \(n\):}\ 1.96 \times \dfrac{4}{\sqrt{n}}<1\ \Rightarrow \ n \approx 61.5\)

\(\therefore \text{Mean sample size}=62.\)

♦ Mean mark (f) 51%.
♦ Mean mark (g) 46%.

Statistics, SPEC2 2023 VCAA 6

A forest ranger wishes to investigate the mass of adult male koalas in a Victorian forest. A random sample of 20 such koalas has a sample mean of 11.39 kg.

It is known that the mass of adult male koalas in the forest is normally distributed with a standard deviation of 1 kg.

  1. Find a 95% confidence interval for the population mean (the mean mass of all adult male koalas in the forest). Give your values correct to two decimal places.  (1 mark)

  2. Sixty such random samples are taken and their confidence intervals are calculated.
  3. In how many of these confidence intervals would the actual mean mass of all adult male koalas in the forest be expected to lie?  (1 mark)

The ranger wants to decrease the width of the 95% confidence interval by 60% to get a better estimate of the population mean.

  1. How many adult male koalas should be sampled to achieve this?  (1 mark)

It is thought that the mean mass of adult male koalas in the forest is 12 kg. The ranger thinks that the true mean mass is less than this and decides to apply a one-tailed statistical test. A random sample of 40 adult male koalas is taken and the sample mean is found to be 11.6 kg.

  1. Write down the null hypothesis, \(H_0\), and the alternative hypothesis, \(H_1\), for the test.   (1 mark)

The ranger decides to apply the one-tailed test at the 1% level of significance and assumes the mass of adult male koalas in the forest is normally distributed with a mean of 12 kg and a standard deviation of 1 kg.

  1.  i. Find the \(p\) value for the test correct to four decimal places.  (1 mark)

  2. ii. Draw a conclusion about the null hypothesis in part d. from the \(p\) value found above, giving a reason for your conclusion.  (1 mark)

  3. What is the critical sample mean (the smallest sample mean for \(H_0\) not to be rejected) in this test? Give your answer in kilograms correct to three decimal places.  (1 mark)

Suppose that the true mean mass of adult male koalas in the forest is 11.4 kg, and the standard deviation is 1 kg. The level of significance of the test is still 1%.

  1. What is the probability, correct to three decimal places, of the ranger making a type \(\text{II}\) error in the statistical test?  (1 mark)

Show Answers Only

a.    \((10.95,11.83)\)

b.    \(57\)

c.    \(n=125\)

d.    \(H_0: \mu=12, \quad H_1: \mu<12\)

e.i.  \(p=0.0057\)

e.ii. \(\text{Since } p<0.01 \text { : reject } H_0 \text {, favour } H_1\)

f.    \(\text {Critical sample mean } \bar{x} \approx 11.632\)

g.    \(\text{Pr}(\bar{x} \geqslant 11.63217 \mid \mu=11.4) \approx 0.071\)

Show Worked Solution

a.    \(\sigma_{\text{pop}}=1\)

\(\text{Sample:}\ \ n=20,\ \ \bar{x}=11.39,\ \ \sigma_{\text {sample }}=\dfrac{1}{\sqrt{20}}\)

\(\text{Find 95% C.I. (by CAS):}\)

\((10.95,11.83)\)
 

b.    \(\text{95% C.I. for 60 samples calculated}\)

\(\text{Number expected }(\mu \text{ within C.I.)}=0.95 \times 60=57\)
 

c.    \(\text {C.I.}=\left(11.39-1.96 \times \dfrac{1}{\sqrt{20}}, 11.39+1.96 \times \dfrac{1}{\sqrt{20}}\right)\)

\(\Rightarrow \text { Interval }=2 \times 1.96 \times \dfrac{1}{\sqrt{20}}\)

\(\text{Interval reduced by } 60\%\)

\(\Rightarrow \text{ New interval }=0.40 \times 2 \times 1.96 \times \dfrac{1}{\sqrt{20}}\)

\(\text{Solve for } n\) :

\(2 \times 1.96 \times \dfrac{1}{\sqrt{n}}=0.40 \times 2 \times 1.96 \times \dfrac{1}{\sqrt{20}}\)

\(\Rightarrow n=125\)
 

♦♦♦ Mean mark (c) 28%.

d.    \(H_0: \mu=12,\ \ H_1: \mu<12\)
 

e.i.  \(E(\bar{X})=\mu=12\)

 \(\bar{x}=11.6, \ \sigma(\bar{X})=\dfrac{1}{\sqrt{40}}\)

 \(p=\operatorname{Pr}(\bar{X} \leqslant 11.6)=0.0057\)

 

e.ii. \(\text{Since}\ \  p<0.01 \text {: reject } H_0 \text {, favour } H_1\)
 

f.    \(\text{Pr}(\bar{X} \leqslant a \mid \mu=12) \geqslant 0.01\) 

\(\text{Find } a \text{ (by CAS):}\)

\(\text{inv Norm}\left(0.01,12, \dfrac{1}{\sqrt{40}}\right) \ \Rightarrow \ a \geqslant 11.63217\)

\(\text {Critical sample mean}\ \ \bar{x} \approx 11.632\)
 

g.    \(\mu=11.4,\ \ \sigma_{\text{pop}}=1,\ \  n=40\)

\(\text{Pr}(\bar{X} \geqslant 11.63217 \mid \mu=11.4) \approx 0.071\)

♦♦ Mean mark (g) 39%.

Statistics, SPEC2 2021 VCAA 18 MC

A scientist investigates the distribution of the masses of fish in a particular river. A 95% confidence interval for the mean mass of a fish, in grams, calculated from a random sample of 100 fish is (70.2, 75.8).

The sample mean divided by the population standard deviation is closest to

  1.   1.3
  2.   2.6
  3.   5.1
  4. 10.2
  5. 13.0
Show Answers Only

`C`

Show Worked Solution

`barx = (70.2 + 75.8)/2 = 73`

`1.96 xx s/sqrt100` `= 73 – 70.2`
`s` `= (2.8 xx 10)/1.96`
  `= 14.29`

 
`:. barx/s = 73/14.29 ~~ 5.1`

`=>\ C`

Statistics, SPEC1 2021 VCAA 3

A company produces a particular type of light globe called Shiny. The company claims that the lifetime of these globes is normally distributed with a mean of 200 weeks and it is known that the standard deviation of the lifetime of Shiny globes is 10 weeks. Customers have complained, saying Shiny globes were lasting less than the claimed 200 weeks. It was decided to investigate the complaints. A random sample of 36 Shiny globes was tested and it was found that the mean lifetime of the sample was 195 weeks.

Use  `text(Pr)(-1.96 < Z < 1.96) = 0.95`  and  `text(Pr)(-3 < Z < 3) = 0.9973`  to answer the following questions.

  1. Write down the null and alternative hypotheses for the one-tailed test that was conducted to investigate the complaints.   (1 mark)

    1. Determine the `p` value, correct to three places decimal places, for the test.   (2 marks)

    2. What should the company be told if the test was carried out at the 1% level of significance?   (1 mark)

  3. The company decided to produce a new type of light globe called Globeplus.
    Find the approximate 95% confidence interval for the mean lifetime of the new globes if a random sample of 25 Globeplus globes is tested and the sample mean is found to be 250 weeks. Assume that the standard deviation of the population is 10 weeks. Give your answer correct to two decimal places.   (1 mark)

Show Answers Only
  1. `H_0 : mu = 200`
    `H_1 : mu < 200`
    1. `0.001\ \ (text(to 3 d.p.))`
    2. `(246.08, 253.92)`
  3. `text(Reject the null hypothesis.)`
Show Worked Solution

a.   `H_0 : mu = 200`

`H_1 : mu < 200`

 

b.i.   `E(barX) = mu = 200`

♦ Mean mark part (b)(i) 48%.

`sigma(barX) = sigma/sqrtn = 10/sqrt36 = 5/3`

`p` `= text(Pr)(barX< 195 | mu = 200)`
  `= text(Pr)(z < (195-200)/(3/5))`
  `= text(Pr)(z < -3)`
  `= 1/2 (1-0.9973)`
  `= 0.00135`
  `= 0.001\ \ (text(to 3 d.p.))`

 

b.ii.   `text(1% level) => 0.01`

♦ Mean mark part (b)(ii) 49%.

  `text(S)text(ince)\ \ 0.001 < 0.01\ \ =>\ text(Strong evidence against)\ H_0`

  `:.\ text(Reject the null hypothesis.)`

 

c.   `sigma(barx) = sigma/sqrtn = 10/sqrt25 = 2`

`text(95% C.I.)` `= barx-1.96 xx 2, barx + 1.96 xx 2`
  `= (250-3.92, 250 + 3.92)`
  `= (246.08, 253.92)`

Statistics, SPEC1-NHT 2019 VCAA 3

The number of cars per day making a U-turn at a particular location is known to be normally distributed with a standard deviation of 17.5. In a sample of 25 randomly selected days, a total of 1450 cars were observed making the U-turn.

  1. Based on this sample, calculate an approximate 95% confidence interval for the number of cars making the U-turn each day. Use an integer multiple of the standard deviation in your calculations.   (3 marks)

  2. The average number of U-turns made at the location is actually 60 per day.

     

    Find an approximation, correct to two decimal places, for the probability that on 25 randomly selected days the average number of U-turns is less than 53.   (1 mark)

Show Answers Only
  1. `(51, 65)`
  2. `0 .02`
Show Worked Solution

a.      `barx = (1450)/(25) = 58`

`(σ)/(sqrtn) = (17.5)/(sqrt25) = (35)/(2xx5) = (7)/(2)`

`text(Limit:)\ ` `(58-2 xx (7)/(2) \ , \ 58 + 2 xx (7)/(2)) `
  `= (51, 65)`

 

b.     `μ = 60 \ , \ (σ)/(sqrtn) = (7)/(2)`
 
          `\ barX ∼ N (60, ((7)/(2))^2)`
 
 
         
 

`text(Pr) (barX < 53)` `= text(Pr) (z <–2)`
  `= (0.05)/(2)`
  `= 0.025`
  `≈ 0.02 \ \ text{(round down)}`

 
`text{(0.03 was also accepted as a correct answer)}`

Statistics, SPEC2 2019 VCAA 18 MC

The masses of a random sample of 36 track athletes have a mean of 65 kg. The standard deviation of the masses of all track athletes is known to be 4 kg.

A 98% confidence interval for the mean of the masses of all track athletes, correct to one decimal place, would be closest to

  1. (51.0, 79.0)
  2. (63.6, 66.4)
  3. (63.3, 66.7)
  4. (63.4, 66.6)
  5. (64.3, 65.7)
Show Answers Only

`=>D`

Show Worked Solution

`text(Solution 1)`

`mu = 65, \ sigma = 4, \ n = 36`

`text(Confidence level 98%:)`

`(63.4, 66.6)\ \ \ text{(by CAS)}`

 

`text(Solution 2)`

`text(Find)\ z\ text(such that)\ \ text(Pr)(z < 2) = 0.99`

`z ~~ 2.3263,\ \ \ text{(by CAS)}`
 

`:.\ text(Confidence Interval)`

`= (65 – 2.3263 xx 4/sqrt36, 65 + 2.3263 xx 4/sqrt36)`

`= (63.4, 66.6)`

`=>D`

Statistics, SPEC2 2016 VCAA 19 MC

A random sample of 100 bananas from a given area has a mean mass of 210 grams and a standard deviation of 16 grams.

Assuming the standard deviation obtained from the sample is a sufficiently accurate estimate of the population standard deviation, an approximate 95% confidence interval for the mean mass of bananas produced in this locality is given by

A.   `(178.7, 241.3)`

B.   `(206.9, 213.1)`

C.   `(209.2, 210.8)`

D.   `(205.2, 214.8)`

E.   `(194, 226)`

Show Answers Only

`B`

Show Worked Solution

`(210 – 1.96 xx 16/sqrt 100, 210 + 1.96 xx 16/sqrt 100)`

`~~ (206.9, 213.1)`

 
`=>  B`

Statistics, SPEC1 2016 VCAA 2

A farmer grows peaches, which are sold at a local market. The mass, in grams, of peaches produced on this farm is known to be normally distributed with a variance of 16. A bag of 25 peaches is found to have a total mass of 2625 g.

Based on this sample of 25 peaches, calculate an approximate 95% confidence interval for the mean mass of all peaches produced on this farm. Use an integer multiple of the standard deviation in your calculations.   (3 marks)

Show Answers Only

`(103.4, 106.6)`

Show Worked Solution
`bar x` `= 2625/25`  
  `= (2500 + 125)/25`  
  `= 100 + 5`  
  `= 105`  

 
`σ_(bar x) = 4/sqrt25 = 4/5`

♦ Mean mark 44%.
  

`:. 95 text(%  CI)` `= (105 – 2 xx 4/5, 105 + 2 xx 4/5)`
  `= (105 – 1.6, 105 + 1.6)`
  `= (103.4, 106.6)`

Statistics, SPEC1 2017 VCAA 4

The volume of soft drink dispensed by a machine into bottles varies normally with a mean of 298 mL and a standard deviation of 3 mL. The soft drink is sold in packs of four bottles.

Find the approximate probability that the mean volume of soft drink per bottle in a randomly selected four-bottle pack is less than 295 mL. Give your answer correct to three decimal places.   (3 marks)

Show Answers Only

`0.025`

Show Worked Solution

`X\ ~\ N(298, 3^2)`

♦ Mean mark 48%.
MARKER’S COMMENT: Many students incorrectly used the population std dev, not the sample std dev.

`barX\ ~\ N(298, (3^2)/4)`

`Z\ ~\ N(0,1)`

`text(Pr)(barX < 295)` `= text(Pr)(Z < (295 – 298)/(3/2))`
  `= text(Pr)(Z < −2)`

 


 

`text(Pr)(barX < 295)` `~~ 0.05/2`
  `= 0.025`

Statistics, SPEC2-NHT 2018 VCAA 18 MC

The heights of all six-year-old children in a given population are normally distributed. The mean height of a random sample of 144 six-year-old children from this population is found to be 115 cm.

If a 95% confidence interval for the mean height of all six-year-old children is calculated to be (113.8, 116.2) cm, the standard deviation used in this calculation is closest to

A.     1.20

B.     7.35

C.   15.09

D.   54.02

E.   88.13

Show Answers Only

`B`

Show Worked Solution
`M` `=(116.2 – 113.8)/2`
`(1.96 xx sigma)/sqrt 144` `= 1.2`
`sigma` `= (1.2 xx 12)/1.96`
  `= 7.35`

 
`=>  B`

Statistics, SPEC1-NHT 2018 VCAA 4

Throughout this question, use an integer multiple of standard deviations in calculations.

The standard deviation of all scores on a particular test is 21.0

  1. From the results of a random sample of `n` students, a 95% confidence interval for the mean score for all students was calculated to be  `(44.7, 51.7)`.

     

    Calculate the mean score and the size of this random sample.  (2 marks)

  2. Determine the size of another random sample for which the endpoints of the 95% confidence interval for the population mean of the particular test would be 1.0 either side of the sample mean.  (2 marks)

Show Answers Only
  1.  `bar x = 48.2; quad n = 144`
  2.  `1764`
Show Worked Solution

a.  `bar x = (51.7 + 44.7)/2 = 96.4/2 = 48.2`
 

`text{Difference: z-score –2 to 0 = 3.5}`

`2 xx 21/sqrt n` `= 7/2`
`7 sqrt n` `= 2 xx 2 xx 21`
`sqrt n` `= (2 xx 2 xx 7 xx 3)/7`
  `=12`
`:.n` `=144`

 

b.   `2 xx 21/sqrtn` `= 1`
  `sqrt n` `= 42`
  `n` `= 42^2`
    `= (40 + 2)^2`
    `= 1600 + 160 + 4`
    `= 1764`