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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 20 MC

The lifespan of a certain electronic component is normally distributed with a mean of \(\mu\) hours and a standard deviation of \(\sigma\) hours.

Given that a 99% confidence interval, based on a random sample of 100 such components, is (10 500, 15 500), the value of \(\sigma\) is closest to

  1. 9710
  2. 10 750
  3. 12 750
  4. 15 190
  5. 19 390
Show Answers Only

\(A\)

Show Worked Solution

\(\sigma(\bar{X}) = \dfrac{\sigma}{\sqrt{100}} = \dfrac{\sigma}{10} \)

\(\bar{x} = \dfrac{10\ 500+15\ 500}{2} = 13\ 000 \)
 

\(\text{Find}\ z\ \text{for 99% CI:}\)

\(\text{Pr}(Z<z) = 0.995\ \ \Rightarrow \ z=2.57583 \)
 

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

\(13\ 000+2.57583 \times \dfrac{\sigma}{10} = 15\ 500 \)

\(\sigma \approx 9710\)

\(\Rightarrow A\)

Statistics, SPEC1 2023 VCAA 6

Josie travels from home to work in the city. She drives a car to a train station, waits, and then rides on a train to the city. The time, \(X_c \) minutes, taken to drive to the station is normally distributed with a mean of 20 minutes \( (\mu_c=20) \) and standard deviation of 6 minutes \((\sigma_c=6) \). The waiting time, \( X_w \) minutes, for a train is normally distributed with a mean of 8 minutes \( (\mu_w=8) \) and standard deviation of \( \sqrt{3} \) minutes \( (\sigma_w=\sqrt{3}) \). The time, \( X_t \) minutes, taken to ride on a train to the city is also normally distributed with a mean of 12 minutes \( (\mu_t=12) \) and standard deviation of 5 minutes \( (\sigma_t=5) \). The three times are independent of each other.

  1. Find the mean and standard deviation of the total time, in minutes, it takes for Josie to travel from home to the city.   (2 marks)
  1. Josie's waiting time for a train on each work day is independent of her waiting time for a train on any other work day. The probability that, for 12 randomly chosen work days, Josie's average waiting time is between 7 minutes 45 seconds and 8 minutes 30 seconds is equivalent to \( \text{Pr}(a<Z<b)\), where \(Z \sim \text{N}(0,1)\) and \(a\) and \(b\) are real numbers.
  2. Find the values of \(a\) and \(b\).   (2 marks)
Show Answers Only

a.    \(E[X_c+X_w+X_t]=40\)

\(\sigma[X_c+X_w+X_t]=8\)

b.    \(a=-0.5, \ b=1\)

Show Worked Solution

a.    \(X_c ∼ N(20, 6^2), X_w ∼ N(8, (\sqrt{3})^2), X_t ∼ N(12, 5^2) \)

\(E[X_c+X_w+X_t]\) \(=E[X_c]+E[X_w]+E[X_t]\)  
  \(=20+8+12\)  
  \(=40\)  

 

\(\text{Var}[X_c+X_w+X_t]\) \(=\text{Var}[X_c]+\text{Var}[X_w]+\text{Var}[X_t]\)  
  \(=36+3+25\)  
  \(=64\)  
\(\sigma[X_c+X_w+X_t]\) \(=8\)  

 
b.    \(E(\bar X)=\mu_w=8,\ \ \sigma(\bar X)=\dfrac{\sigma_w}{\sqrt{12}}=\dfrac{\sqrt3}{\sqrt{12}}=\dfrac{1}{2} \)

\(\text{Pr}(7.75<\bar X<8.5)\) \(=\text{Pr} \Bigg{(} \dfrac{7.75-8}{\frac{1}{2}}<Z<\dfrac{8.5-8}{\frac{1}{2}}\Bigg{)} \)  
  \(=\text{Pr} \Bigg{(} \dfrac{-0.25}{\frac{1}{2}}<Z<\dfrac{0.5}{\frac{1}{2}}\Bigg{)} \)  
  \(=\text{Pr}(-0.5<Z<1)\)  

 
\(\therefore a=-0.5, \ b=1\)

Statistics, SPEC2-NHT 2019 VCAA 6

A paint company claims that the mean time taken for its paint to dry when motor vehicles are repaired is 3.55 hours, with a standard deviation of 0.66 hours.

Assume that the drying time for the paint follows a normal distribution and that the claimed standard deviation value is accurate.

  1. Let the random variable  `barX`  represent the mean time taken for the paint to dry for a random sample of 36 motor vehicles.

     

    Write down the mean and standard deviation of  `barX`.   (2 marks)

At a car crash repair centre, it was found that the mean time taken for the paint company's paint to dry on randomly selected vehicles was 3.85 hours. The management of this crash repair centre was not happy and believed that the claim regarding the mean time taken for the paint to dry was too low. To test the paint company's claim, a statistical test was carried out.

  1. Write down suitable null and alternative hypotheses `H_0` and `H_1` respectively to test whether the mean time taken for the paint to dry is longer than claimed.   (1 mark)

  2. Write down an expression for the  `p`  value of the statistical test and evaluate it correct to three decimal places.   (2 marks)

  3. Using a 1% level of significance, state with a reason whether the crash repair centre is justified in believing that the paint company's claim of mean time taken for its paint to dry of 3.55 hours is too low.  (1 mark)

  4. At the 1% level of significance, find the set of sample mean values that would support the conclusion that the mean time taken for the paint to dry exceeded 3.55 hours. Give your answer in hours, correct to three decimal places.  (2 marks)

  5. If the true time taken for the paint to dry is 3.83 hours, find the probability that the paint company's claim is not rejected at the 1% level of significance, assuming the standard deviation for the paint to dry is still 0.66 hours. Give your answer correct to two decimal places.  (1 mark)

Show Answers Only
  1. `0.11`
  2. `H_0 : \ mu = 3.55`
    `H_1 : \ mu > 3.55`
  3. `0.003 \ text{(to 3 decimal places)}`
  4. `text(S) text(ince) \ \ p < 0.01 , H_0 \ text(should be rejected at the 1% level.)`
  5. `barX > 3.806`
  6. `0.41`
Show Worked Solution
a.   `E(barX)` `= 3.55`
`sigma(barX)` `= (sigma)/(sqrtn)`
  `= (0.66)/(sqrt36)`
  `= 0.11`

 

b.   `H_0 : \ mu = 3.55`

`H_1 : \ mu > 3.55`

 

c.   `p` `= text(Pr) (barX > 3.85)`
  `= text(Pr) (z > (3.85-3.55)/(0.11))`
  `= text(Pr) (z >2.326)`
  `= 0.003 \ text{(to 3 decimal places)}`

 

d.   `text(S) text(ince)\ \ p < 0.01 , H_0 \ text(should be rejected at the 1% level.)`

`text(i.e. repair centre is justified that the mean time 3.55 hours is too low.)`

 

e.   `text(If) \ \ mu = 3.55`

`(barX-mu)/(sigma)` `> 2.3263`
`barX` `> 2.3263 xx 0.11 + 3.55`
`barX` `> 3.806`

 

f.   `text(Pr) (barX< 3.806 | mu = 3.83)` `= text(Pr) (z < (3.806-3.83)/(0.11))`
  `= text(Pr) (z < -0.21818)`
  `= 0.41`

Statistics, SPEC2 2019 NHT 19 MC

Bags of peanuts are packed by a machine. The masses of the bags are normally distributed with a standard deviation of three grams.

The minimum size of a sample required to ensure that the manufacturer can be 98% confident that the sample mean is within one gram of the population mean is

  1.  37
  2.  38
  3.  48
  4.  49
  5.  60
Show Answers Only

`D`

Show Worked Solution

`M\ ~\ N(mu, 3^2)`

`text(Confidence interval 98%) => \ z = 2.326\ \ \ text{(by CAS)}`
 


 

`mu ± z sigma/sqrtn\ \ text(is within 1 gram of)\ mu:`

`text(Solve:)\ \ 2.326 xx 3/sqrtn <=1`

`n >= 48.7`

`n = 49`

`=>\ D`

Statistics, SPEC2 2019 VCAA 6

A company produces packets of noodles. It is known from past experience that the mass of a packet of noodles produced by one of the company's machines is normally distributed with a mean of 375 grams and a standard deviation of 15 grams.

To check the operation of the machine after some repairs, the company's quality control employees select two independent random samples of 50 packets and calculate the mean mass of the 50 packets for each random sample.

  1. Assume that the machine is working properly. Find the probability that at least one random sample will have a mean mass between 370 grams and 375 grams. Give your answer correct to three decimal places.   (2 marks)

  2. Assume that the machine is working properly. Find the probability that the means of the two random samples differ by less than 2 grams. Give your answer correct to three decimal places.   (3 marks)

To test whether the machine is working properly after the repairs and is still producing packets with a mean mass of 375 grams, the two random samples are combined and the mean mass of the 100 packets is found to be 372 grams. Assume that the standard deviation of the mass of the packets produced is still 15 grams. A two-tailed test at the 5% level of significance is to be carried out.

  1. Write down suitable hypotheses `H_0` and `H_1` for this test.   (1 mark)

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

  3. Does the mean mass of the sample of 100 packets suggest that the machine is working properly at the 5% level of significance for a two-tailed test? Justify your answer.   (1 mark)

  4. What is the smallest value of the mean mass of the sample of 100 packets for `H_0` to be not rejected? Give your answer correct to one decimal place.   (1 mark)

Show Answers Only
  1. `0.741`
  2. `0.495`
  3. `H_0: mu = 375`
    `H_1: mu != 375`
  4. `0.046`
  5. `text(See Worked Solutions)`
  6. `372.1`
Show Worked Solution

a.   `barX\ ~\ N(375, (15/sqrt50)^2)`

`text(Pr)(370 <= barX <= 375)`

`= 0.490788…`
 

`text(Pr)(text(not in range)) ~~ 1-0.490788 ~~ 0.509212`

`text(Pr)(text(Samples within range) >= 1)`

`= 1-text(Pr)(text(0 samples in range))`

`~~ 1-(0.509212)^2`

`~~ 0.741`
 

b.   `text(Let)\ \ Y = barX_1-barX_2`

`Y\ ~\ N (0, 2 xx (15^2)/50)`

`text(Pr)(−2 < Y < 2)\ \ (text(by CAS))`

`= text(norm cdf)\ (−2,2,0, sqrt(2 xx (15^2)/50))`

`~~ 0.495`
 

c.   `H_0: \ mu = 375`

`H_1: \ mu != 375`
 

d.   `text(By CAS:)\ \ mu_0 = 375, \ barx = 372, \ sigma = 15, \ n = 100`

`p ~~ 0.046`
 

e.   `p < 0.05 \ => \ text(reject)\ H_0`

`=>\ text(The machine is NOT working properly.)`
 

f.   `text(Pr)(barx <= a) = 0.025\ \ (text(2 sided test))`

`a_text(min)` `= text(inv norm)\ (0.025, 375, 15/sqrt100)`
  `= 372.06…`
  `=372.1`

Statistics, SPEC2-NHT 2018 VCAA 7

According to medical records, the blood pressure of the general population of males aged 35 to 45 years is normally distributed with a mean of 128 and a standard deviation of 14. Researchers suggested that male teachers had higher blood pressures than the general population of males.

To investigate this, a random sample of 49 male teachers from this age group was obtained and found to have a mean blood pressure of 133.

  1. State two hypotheses and perform a statistical test at the 5% level to determine if male teachers belonging to the 35 to 45 years age group have higher blood pressures than the general population of males. Clearly state your conclusion with a reason.   (3 marks)

  2. Find a 90% confidence interval for the mean blood pressure of all male teachers aged 35 to 45 years using a standard deviation of 14. Give your answers correct to the nearest integer.   (1 mark)

    [wa_lines lines="3" style="lined"
Show Answers Only
  1. `text(See Worked Solutions)`
  2. `(130, 136)`
Show Worked Solution

a.   `B\ ~\ N (128, 14^2)`

`H_0: mu = 128`

`H_1: mu > 128`

`bar B\ ~\ N (128, 14^2/49)`

`p` `= text(Pr) (bar B > 133 | mu=128)`
  `= text(Pr)(Z>(133-128)/(14/sqrt49))`
  `= text(Pr)(Z>5/2)`
  `~~ 0.00621`

 
`p < 0.05 => text(Reject)\ H_0\ text(at the 5% significance level):`

`text{evidence supports the contention that male teachers 35 to 45}`

`text(have higher blood pressure than the general male population.)`

 

b.  `text(If)\ \ text(Pr)(- z < Z < z) = 0.9`

`text(Pr)(Z < z) = 0.95\ \ =>\ \ z ~~ 1.64485`

`:.\ text(90% CI): (133-(z xx 14)/sqrt 49, 133 + (z xx 14)/sqrt 49)`

`~~(130, 136)`

Statistics, SPEC2 2018 VCAA 6

The heights of mature water buffaloes in northern Australia are known to be normally distributed with a standard deviation of 15 cm. It is claimed that the mean height of the water buffaloes is 150 cm.

To decide whether the claim about the mean height is true, rangers selected a random sample of 50 mature water buffaloes. The mean height of this sample was found to be 145 cm.

A one-tailed statistical test is to be carried out to see if the sample mean height of 145 cm differs significantly from the claimed population mean of 150 cm.

Let `bar X` denote the mean height of a random sample of 50 mature water buffaloes.

  1. State suitable hypotheses `H_0` and `H_1` for the statistical test.   (1 mark)

  2. Find the standard deviation of `bar X`.   (1 mark)

  3. Write down an expression for the `p` value of the statistical test and evaluate your answer to four decimal places.   (2 marks)

  4. State with a reason whether `H_0` should be rejected at the 5% level of significance.   (1 mark)

  5. What is the smallest value of the sample mean height that could be observed for `H_0` to be not rejected? Give your answer in centimetres, correct to two decimal places.   (1 mark)

  6. If the true mean height of all mature water buffaloes in northern Australia is in fact 145 cm, what is the probability that `H_0` will be accepted at the 5% level of significance? Give your answer correct to two decimal places.   (1 mark)

  7. Using the observed sample mean of 145 cm, find a 99% confidence interval for the mean height of all mature water buffaloes in northern Australia. Express the values in your confidence interval in centimetres, correct to one decimal place.   (1 mark)

Show Answers Only
  1.  `H_0: mu = 150; qquad H_1: mu < 150`
  2. `3/sqrt 2`
  3. `p = text(Pr)(bar X < 145); qquad p~~ 0.0092`
  4. `text(Yes, he should be rejected as)`
    `p~~ 0.0092 < 0.05`
  5. `bar X_min = 146.52`
  6. `0.24`
  7. `(139.5, 150.5)`
Show Worked Solution
a.    `H_0: mu =150`
  `H_1: mu < 150`

 

b.    `sigma_bar X` `= (sigma_X)/sqrt n`
    `= 15/sqrt 50`
    `= (3sqrt 2)/2`

 

c.   `p = text(Pr)(bar X < 145), qquad bar X\ ~\ N (150, 9/2)`

`p ~~ 0.0092`

 

d.   `text(Yes, he should be rejected as)`

`p~~ 0.0092 < 0.05`

 

e.  `text(Pr)(bar X < x) = 0.05`

♦ Mean mark part (e) 48%.

`x ~~ 146.511`

`text(NOT rejected:) quad bar X_min = 146.52`

 

f.   `bar X_2\ ~\ N (145, 9/2)`

♦♦♦ Mean mark part (f) 11%.

`text(Pr)(bar X_2 > x)` `= text(Pr)(bar X_2 > 146.51074)`  
  `~~ 0.24`  

 

g.   `(145-(Z_99 xx 3)/sqrt 2, 145 + (Z_99 xx 3)/sqrt 2)`

`text(Pr)(Z < Z_99) = 0.995, \ Z\ ~\ N(0, 1)`

`=>Z_99 ~~ 2.57583`

`:. 99%\ text(C.I:)\ (139.5, 150.5)`

Statistics, SPEC2-NHT 2017 VCAA 20 MC

The mass of suspended matter in the air in a particular locality is normally distributed with a mean of `mu` micrograms per cubic metre and a standard deviation of  `sigma = 8`  micrograms per cubic metre. The mean of 100 randomly selected air samples was found to be 40 micrograms per cubic metre.

Based on this, a 90% confidence interval for `mu`, correct to two decimal places, is

A.   `(38.68, 41.32)`

B.   `(26.84, 53.16)`

C.   `(38.43, 41.57)`

D.   `(24.32, 55.68)`

E.   `(37.93, 42.06)`

Show Answers Only

`A`

Show Worked Solution

`M\ ~\ N (mu, 8^2)`

`bar M\ ~\ N (mu, 8^2/100)`
 

`text(Find 90%  CI),\ bar m =40:`

`=> (40 – Z_0.9 xx 8/sqrt 100,\ 40 + Z_0.9 xx 8/sqrt 100)`
 

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

`Pr(Z < Z_0.9) = 0.95\ \ => \ \ Z_0.9 ~~ 1.64485\ \ \ text{(by CAS)}`
 

`:.­text(CI)\ => (40 – 1.64485 xx 8/sqrt 100,\ 40 + 1.64485 xx 8/sqrt 100)`

`~~ (38.68, 41.32)`

`=>   A`

Statistics, SPEC2 2017 VCAA 19 MC

A confidence interval is to be used to estimate the population mean  `mu`  based on a sample mean  `barx`.

To decrease the width of a confidence interval by 75%, the sample size must be multiplied by a factor of

  1.    2
  2.    4
  3.    9
  4.  16
  5.  25
Show Answers Only

`D`

Show Worked Solution

`text(Original width) = 2m = 2 xx z σ/sqrtn`

♦ Mean mark 44%.

`text(Decreased width)\ = (2zσ)/sqrtn_2`

`text(If decreased width =)\ 1/4\ text(original width:)`

`(2zσ)/sqrtn_2` `= 1/4 ((2zσ)/sqrtn)`
`(2zσ)/sqrtn_2` `= (2zσ)/(4sqrtn)`
`1/(sqrt(n_2))` `= 1/(4sqrtn)`
`sqrt(n_2)` `= 4sqrtn`
`n_2` `= 16n`

  
`=> D`