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Statistics, EXT1 EQ-Bank 6 MC

The lifetime of a certain brand of batteries has a mean lifetime of 20 hours and a standard deviation of 2 hours. A random sample of 40 batteries is selected.

The probability that the mean lifetime of this sample of 40 batteries exceeds 19.5 hours is closest to

  1. 0.0571
  2. 0.5987
  3. 0.8944
  4. 0.9429
Show Answers Only

\(D\)

Show Worked Solution

\(\text{Sample size is > 30} \ \ \Rightarrow \ \ \text{CLT applies}\)

\(\overline{X} \sim N\left( \mu, \dfrac{\sigma^2}{n}\right) \sim N\left( 20, \dfrac{4}{40}\right)\)

\(\sigma_{\bar{X}} = \dfrac{1}{\sqrt{10}} \)
 

\(\text{By the central limit theorem:}\)

\(Z=\dfrac{\overline{X}-20}{\frac{1}{\sqrt{10}}}\sim N(0,1)\)

\(Z=\dfrac{19.5-20}{\frac{1}{\sqrt{10}}}=-1.58 \ \text{(2 d.p.)}\)
 

\(\text{Using Normal Distribution Table of Values:}\)

\(\Pr(\overline{X}>19.5)=\Pr(Z>-1.58)\)

\(\phantom{\Pr(\overline{X}>19.5)}=0.9429\)

\(\Rightarrow D\)

Statistics, SPEC2 2022 VCAA 6

A company produces soft drinks in aluminium cans.

The company sources empty cans from an external supplier, who claims that the mass of aluminium in each can is normally distributed with a mean of 15 grams and a standard deviation of 0.25 grams.

A random sample of 64 empty cans was taken and the mean mass of the sample was found to be 14.94 grams.

Uncertain about the supplier's claim, the company will conduct a one-tailed test at the 5% level of significance. Assume that the standard deviation for the test is 0.25 grams.

  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 random sample of 64 empty cans support the supplier's claim at the 5% level of significance for a one-tailed test? Justify your answer.   (1 mark)

  4. What is the smallest value of the mean mass of the sample of 64 empty cans for \(H_0\) not to be rejected? Give your answer correct to two decimal places.   (1 mark)

The equipment used to package the soft drink weighs each can after the can is filled. It is known from past experience that the masses of cans filled with the soft drink produced by the company are normally distributed with a mean of 406 grams and a standard deviation of 5 grams.

  1. What is the probability that the masses of two randomly selected cans of soft drink differ by no more than 3 grams? Give your answer correct to three decimal places.   (2 marks)

Show Answers Only

a.   \(H_0: \mu=15, \quad H_1: \mu<15\)

b.   \(p=0.027\)

c.    \(\text{Since \(\ p<0.05\), claim is not supported.}\)

d.   \(a=14.95\)

e.  \(\text{Pr}(-3<D<3)=0.329\)

Show Worked Solution

a.    \(H_0: \mu=15, \quad H_1: \mu<15\)
 

b.    \(\mu=15, \ \ \sigma=0.25\)

\(\bar{x}=14.94, \ \sigma_{\bar{x}}=\dfrac{0.25}{\sqrt{64}}=0.03125\)

\(\text{By CAS:}\)

\(p=\text{Pr}\left(\bar{X}<14.94 \mid \mu=15\right)=0.027\ \text {(3 d.p.)}\)
 

c.    \(\text{Since \(\ p<0.05\), claim is not supported.}\)

\(\text{Evidence is against \(H_0\)  at the \(5 \%\) level.}\)
 

d.    \(\text{Pr}\left(\bar{X}<a \mid \mu=15\right)>0.05\)

\(\text{Pr}\left(Z<\dfrac{a-15}{0.03125}\right)>0.05\)

\(\text{By CAS:}\ \ a=14.95\ \text{(2 d.p.)}\)
 

e.    \(\text{Let}\ \ M=\ \text{mass of one can}\)

\(M \sim N\left(406,5^2\right)\)

\(E\left(M_1\right)=E\left(M_2\right)=\mu=406\)

\(\text {Let}\ \ D=M_1-M_2\)

\(E(D)=406-406=0\)

\(\text{Var}(D)=1^2 \times \text{Var}\left(M_1\right)+(-1)^2 \times  \text{Var}\left(M_2\right)=50\)

\(\sigma_D=\sqrt{50}\)

\(D \sim N\left(0,(\sqrt{50})^2\right)\)

\(\text{By CAS: Pr\((-3<D<3)=0.329\) (3 d.p.) }\)

♦ Mean mark (e) 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 6

The maximum load of a lift in a chocolate company's office building is 1000 kg. The masses of the employees who use the lift are normally distributed with a mean of 75 kg and a standard deviation of 8 kg. On a particular morning there are `n` employees about to use the lift.

  1. What is the maximum possible value of `n` for there to be less than a 1% chance of the lift exceeding the maximum load?   (2 marks)

Clare, who is one of the employees, likes to have a hot drink after she exits the lift. The time taken for the drink machine to dispense a hot drink is normally distributed with a mean of 2 minutes and a standard deviation of 0.5 minutes. Times taken to dispense successive hot drinks are independent.

  1. Clare has a meeting at 9.00 am and at 8.52 am she is fourth in the queue for a hot drink. Assume that the waiting time between hot drinks dispensed is negligible and that it takes Clare 0.5 minutes to get from the drink machine to the meeting room.
  2. What is the probability, correct to four decimal places, that Clare will get to her meeting on time?   (2 marks)

Clare is a statistician for the chocolate company. The number of chocolate bars sold daily is normally distributed with a mean of 60 000 and a standard deviation of 5000. To increase sales, the company decides to run an advertising campaign. After the campaign, the mean daily sales from 14 randomly selected days was found to be 63 500.

Clare has been asked to investigate whether the advertising campaign was effective, so she decides to perform a one-sided statistical test at the 1% level of significance.

  1.   i. Write down suitable null and alternative hypotheses for this test.   (1 mark)

  2.  ii. Determine the `p` value, correct for decimal places, for this test.   (1 mark)

  3. iii. Giving a reason, state whether there is any evidence for the success of the advertising campaign.   (1 mark)

  4. Find the range of values for the mean daily sales of another 14 randomly selected days that would lead to the null hypothesis being rejected when tested at the 1% level of significance. Give your answer correct to the nearest integer.   (1 mark)

  5. The advertising campaign has been successful to the extent that the mean daily sales is now 63 000.
  6. A statistical test is applied at the 5% level of significance.
  7. Find the probability that the null hypothesis would be incorrectly accepted, based on the sales of another 14 randomly selected days and assuming a standard deviation of 5000. Give your answer correct to three decimal places.   (2 marks)

Show Answers Only
  1. `12`
  2. `0.3085`
  3.   i. `H_0: \ mu = 60 \ 000, \ H_1: \ mu > 60 \ 000`
  4.  ii.  `0.0044`
  5. iii.  `text{Successful as} \ p text{-value is below 0.01.}`
  6. `n ≥ 63\ 109`
  7. `0.274`
Show Worked Solution

a.    `text{Method 1}`

♦♦ Mean mark 22%.

`text{Let} \ \ M_i\ ~\ N (75, 8^2) \ \text{for} \ \ i = 1, 2, 3, … , n`

`W_n = W_1 + W_2 + … + W_n`

`E (W_n) = E(M_1 + … + M_n) = 75n`

`text(s.d.) (W_n) = text{s.d.} (M_1 + … M_n) = 8 sqrtn`

`W_n\ ~\ N (75n, 8^2 n)`

`text{Using} \ Z\ ~\ N (0, 1)`

`text(Pr) (W_n > 1000) = 0.01`

`text(Pr) (Z > {1000-75 n}/{8 sqrtn}) = 0.01`
 
`n = 12.5`

`:. \ text{Largest} \ n = 12`
 

`text{Method 2}`

`text{By trial and error}`
 
`text(Pr) (M_1 + … + M_11 > 1000) ≈ 0`

`text(Pr) (M_1 + … + M_12 > 1000) ≈ 0.0002`

`text(Pr) (M_1 + … + M_13 > 1000) ≈ 0.193`

`:. \ text{Largest} \ n = 12`

 

b.   `T_i\ ~\ N (2, 0.5^2) \ \ text{for} \ \ i = 1, 2, …`

Mean mark part (b) 51%.

`text{Wait time} \ (T) = T_1 + T_2 + T_3 + T_4`

`E(T) = 4 xx 2 = 8`

`text{s.d.}(T) = text{s.d.}(T_1 + T_2 + T_3 + T_4) = sqrt4 xx 0.5 = 1`
 
`T\ ~\ N (8, 1)`

`text(Pr) (T < 7.5) = 0.3085`

`text{By CAS: normCdf} (0, 7.5, 8, 1)`

 

c.i.  `H_0: \ mu = 60 \ 000`

`H_1: \ mu > 60 \ 000`
 

c.ii.  `text(Pr) (barX > 63\ 500 | mu = 60 \ 000) = 0.004407`

  `:. \ p \ text{value} = 0.0044`

 `text{By CAS: norm Cdf} (63\ 500, oo, 60\ 000, 5000/sqrt14)`
 

c.iii.  `text{S} text{ince the} \ p text{-value is below 0.01, there is strong evidence}`

   `text{the advertising was effective (against the null hypothesis).}`

 

d.   `text(Pr) (barX > n | mu = 60 \ 000) < 0.01`

♦♦♦ Mean mark part (d) 16%.

`n ≥ 63\ 109`

`text{By CAS: inv Norm} (0.99, 60 \ 000, 5000/sqrt14)`

 

e.    `text{Similar to part (d):}`

♦♦♦ Mean mark part (e) 10%.

`text(Pr) (barX > n | mu = 60 \  000) < 0.05`

`n ≤ 62198`

`text{By CAS: invNorm} \ (0.95, 60 \ 000, 5000/sqrt14)`
 
`text{If} \ \ mu = 63\ 000, text{find probability null hypothesis incorrectly accepted:}`

`text(Pr) (barX < 62\ 198 | mu = 63\ 000) = 0.274`

`text{By CAS: normCdf} (0, 62\ 198, 63\ 000, 5000/sqrt14)`


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, 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, 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 20 MC

The random number function of a calculator is designed to generate random numbers that are uniformly distributed from 0 to 1. When working properly, a calculator generates random numbers from a population where  `mu = 0.5`  and  `sigma = 0.2887`

When checking the random number function of a particular calculator, a sample of 100 random numbers was generated and was found to have a mean of  `barx = 0.4725`.

Assuming  `H_0: mu = 0.5`  and  `H_1: mu < 0.5`, and  `sigma = 0.2887`, the  `p`  value for a one-sided test is

  1. 0.0953
  2. 0.1704
  3. 0.4621
  4. 0.8296
  5. 0.9525
Show Answers Only

`B`

Show Worked Solution

`text(Standard Deviation)\ (barx) = 0.2887/sqrt100 = 0.02887`

`p` `= text(Pr)(barx <= 0.4725 | mu = 0.5)`
  `~~ 0.1704\ \ \ text{(by CAS)}`

 
`=>B`

Statistics, SPEC2 2016 VCAA 20 MC

The lifetime of a certain brand of batteries is normally distributed with a mean lifetime of 20 hours and a standard deviation of two hours. A random sample of 25 batteries is selected.

The probability that the mean lifetime of this sample of 25 batteries exceeds 19.3 hours is

  1. 0.0401
  2. 0.1368
  3. 0.6103
  4. 0.8632
  5. 0.9599
Show Answers Only

`E`

Show Worked Solution

`X\ ~\ N (20, 2^2)`

`bar X\ ~\ N (20, (2^2)/25)`

`text(Pr) (bar X > 19.3) ~~ 0.9599`

`=>  E`

Statistics, SPEC2-NHT 2017 VCAA 6

A bank claims that the amount it lends for housing is normally distributed with a mean of $400 000 and a standard deviation of $30 000.

A consumer organisation believes that the average loan amount is higher than the bank claims.

To check this, the consumer organisation examines a random sample of 25 loans and finds the sample mean to be $412 000.

  1. Write down the two hypotheses that would be used to undertake a one-sided test.   (1 mark)

  2. Write down an expression for the `p` value for this test and evaluate it to four decimal places.   (2 marks)

  3. State with a reason whether the bank’s claim should be rejected at the 5% level of significance.   (1 mark)

  4. What is the largest value of the sample mean that could be observed before the bank’s claim was rejected at the 5% level of significance? Give your answer correct to the nearest 10 dollars.   (1 mark)

  5. If the average loan made by the bank is actually $415 000 and not $400 000 as originally claimed, what is the probability that a random selection of 25 loans has a sample mean that is at most $410 000? Give your answer correct to three decimal places.   (2 marks)

Show Answers Only
  1. `H_0: mu = 400\ 000`

     

    `H_1: mu > 400\ 000`

  2. `p ~~ 0.0228`
  3. `text(reject)`
  4. `x_max ~~ 409\ 860`
  5. `~~ 0.202`
Show Worked Solution

a.   `H_0: \ mu = 400\ 000`

`H_1: \ mu > 400\ 000`

 

b.   `L\ ~\ N (400\ 000, 30\ 000^2)`

`bar L\ ~\ N (400\ 000, (30\ 000^2)/25)`

`p = text(Pr)(bar L > $412\ 000)`
 

`:. p ~~ 0.0228`

 

c.   `text(The bank’s claim)\ (h_0: mu = 400\ 000)`

`text(should be rejected at the 5% level of)`

`text(significance as)\ \ p ~~ 0.0228 < 0.05.`

 

d.   `p = text(Pr)(bar L > x) = 0.05`

`=> text(Pr)(bar L < x) = 0.95`

`=> x ~~ 409\ 869.12`

`:. x_max ~~ $409\ 870\ \ text{(nearest $10)}`

 

e.   `text(New distribution):\ \ L_2\ ~\ N (415\ 000, 30\ 000^2)`

`bar L_2\ ~\ N (415\ 000, (30\ 000^2)/25)`

`text(Pr)(bar L_2 <= 410\ 000) ~~ 0.202`

Statistics, SPEC2 2017 VCAA 6

A dairy factory produces milk in bottles with a nominal volume of 2 L per bottle. To ensure most bottles contain at least the nominal volume, the machine that fills the bottles dispenses volumes that are normally distributed with a mean of 2005 mL and a standard deviation of 6 mL.

  1. Find the percentage of bottles that contain at least the nominal volume of milk, correct to one decimal place.   (1 marks)

Bottles of milk are packed in crates of 10 bottles, where the nominal total volume per crate is 20 L.

  1. Show that the total volume of milk contained in each crate varies with a mean of 20 050 mL and a standard deviation of  `6sqrt10`  mL.   (2 marks)

  2. Find the percentage, correct to one decimal place, of crates that contain at least the nominal volume of 20 L.   (1 mark)

  3. Regulations require at least 99.9% of crates to contain at least the nominal volume of 20 L.
  4. Assuming the mean volume dispensed by the machine remains 2005 mL, find the maximum allowable standard deviation of the bottle-filling machine needed to achieve this outcome. Give your answer in millilitres, correct to one decimal place.   (3 marks)

  5. A nearby dairy factory claims the milk dispensed into its 2 L bottles varies normally with a mean of 2005 mL and a standard deviation of 2 mL.
  6. When authorities visit the nearby dairy factory and check a random sample of 10 bottles of milk, they find the mean volume to be 2004 mL.
  7. Assuming that the standard deviation of 2 mL is correct, carry out a one-sided statistical test and determine, stating a reason, whether the nearby dairy’s claim should be accepted at the 5% level of significance.   (2 marks)

Show Answers Only
  1. `79.8text(%)`
  2. `text(See Worked Solutions)`
  3. `99.6text(%)`
  4. `text(max) (σ_x) ~~ 5.1`
  5. `text(The claim should be accepted at a 5% significance level.)`
Show Worked Solution

a.   `V_B~N(2005, 6^2)`

`text(Pr)(V_B > 2000)` `~~ 0.797672\ \ \ text{(by CAS)}`
  `~~ 79.8text(%)`

 

b.   `text(Let)\ \ V_C=\ text(Volume of a crate)`

`mu_(V_C)` `= 10 xx mu_(V_B)=20\ 050`  
     
  `σ_(V_C)^2` `= σ_(V_B)^2 xx 10`
    `= 360`
  `:. σ_(V_C)` `= 6sqrt(10)`

 

c.  `V_C~N(20\ 050, (6sqrt10)^2)`

`text(Pr)(V_C > 20\ 000)` `~~ 0.995796`
  `~~ 99.6text(%)`

 

d.   `text(Let)\ \ V_A =\ text(New distribution)`

`V_A~N(20\ 050, (σ_x  sqrt10)^2)`

`text(Pr)(V_A > 20\ 000)` `>= 0.999`  
`text(Pr)(Z=a)` `=0.999`  
`:.a` `=-3.0902`  

 

`(20\ 000-20\ 050)/(σ_x sqrt10)` `=-3.0902`
`σ_x` `=5.116…`

 
`:.\ text(max) (σ_x) ~~ 5.1`

 

e.   `H_0: mu=2005`

`H_1: mu<2005`

`D~N(2005, 2^2)\ \ =>\ \ barD~N(2005, (2^2)/10)`

`p` `= text(Pr)(barD < 2004)`
  `~~ 0.056923`

  

`:.\ text(S)text(ince)\ \ p>0.05,\ text(the claim should be accepted)`

`text(at a 5% significance level.)`

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-NHT 2018 VCAA 6

A coffee machine dispenses coffee concentrate and hot water into a 200 mL cup to produce a long black coffee. The volume of coffee concentrate dispensed varies normally with a mean of 40 mL and a standard deviation of 1.6 mL.

Independent of the volume of coffee concentrate, the volume of water dispensed varies normally with a mean of 150 mL and a standard deviation of 6.3 mL.

  1. State the mean and the standard deviation, in millilitres, of the total volume of liquid dispensed to make a long black coffee.   (2 marks)

  2. Find the probability that a long black coffee dispensed by the machine overflows a 200 mL cup. Give your answer correct to three decimal places.   (1 mark)

  3. Suppose that the standard deviation of the volume of water dispensed by the machine can be adjusted, but that the mean volume of water dispensed and the standard deviation of the volume of coffee concentrate dispensed cannot be adjusted.
  4. Find the standard deviation of the volume of water dispensed that is needed for there to be only a 1% chance of a long black coffee overflowing a 200 mL cup. Give your answer in millilitres, correct to two decimal places.   (2 marks)

Show Answers Only
  1. `mu_V = 40 + 150 = 190\ text(mL)`

     

    `sigma_V = sqrt (1.6^2 + 6.3^2) = 6.5\ text(mL)`

  2. `0.062`
  3. `3.99\ text(mL)`
Show Worked Solution

a.   `C~N (40, 1.6^2)`

`W~N (150, 6.3^2)`

`V = C + W`

`mu_V = 40 + 150 = 190\ text(mL)`

`{:sigma^2:}_V` `= 1.6^2 + 6.3^2`  
`:.sigma_V ` `= sqrt (1.6^2 + 6.3^2)`  
  `= 6.5\ text(mL)`  

 

b.   `V~N (190, 6.5^2)`

`text(Pr)(V > 200) ~~ 0.062\ \ text{(by CAS)}`

 

c.  `C~N (40, 1.6^2)`

`W_2~N (150, sigma_w^2)`

`V_2 = C + W_2`

`V_2~N (190, 1.6^2 + sigma_w^2)`

`Z~N (0, 1)`
 

`text(Pr)(V_2 > 200) = 0.01`

`text(Pr)(Z > a) = 0.01\ \ =>\ \ a=2.326…`
  

`text(Find)\ \ sigma_w:`

`2.326…` `= (200-190)/sqrt(1.6^2 + sigma_w^2)`
`:. sigma_w` `~~ 3.99\ text(mL)\ \ \ text{(by CAS)}`

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 19 MC

The petrol consumption of a particular model of car is normally distributed with a mean of 12 L/100 km and a standard deviation of 2 L/100 km.

The probability that the average petrol consumption of 16 such cars exceeds 13 L/100 km is closest to

A.   0.0104

B.   0.0193

C.   0.0228

D.   0.3085

E.   0.3648

Show Answers Only

`C`

Show Worked Solution

`P\ ~\ N (12, 2^2) \ => \ bar P\ ~\ N (12, 2^2/16)`

`Pr(bar P > 13) ~~ 0.0228\ \ \ text{(by CAS)}`

`=>   C`

Statistics, SPEC1-NHT 2017 VCAA 9

The random variables `X` and `Y` are independent with  `mu_X = 4,\ text(Var)(X) = 36`  and  `mu_Y = 3,\ text(Var)(Y) = 25`.

  1. The random variable `Z` is such that  `Z = 2X + 3Y`.
  2.  i. Find `E(Z)`.   (1 mark)

  3. ii. Find the standard deviation of `Z`.   (1 mark)

  1. Researchers have reason to believe that the mean of `X` has decreased. They collect a random sample of 64 observations of `X` and find that the sample mean is  `bar X = 3.8`
  2.  i. State the null hypothesis and the alternative hypothesis that should be used to test that the mean has decreased.   (1 mark)

  3. ii. Calculate the mean and standard deviation for a distribution of sample means, `bar X`, for samples of 64 observations.   (1 mark)

Show Answers Only
  1.  i. `17`
  2. ii. `3 sqrt 41`
  3.  i. `H_0: mu = 4,\ \ H_1: mu < 4`
  4. ii. `mu_bar X = 4,\ \ sigma_bar X = 3/4`
Show Worked Solution
a.i.    `E(Z)` `= 2E(X) + 3E(Y)`
    `= 2(4) + 3(3)`
    `= 8 + 9`
    `= 17`

 

a.ii.    `text(Var)(Z)` `= 2^2 text(var)(X) + 3^2 text(var)(Y)`
    `= 4(36) + 9(25)`
    `= (4xx9xx4) + 9(25)`
    `= 9(16 + 25)`

 

`:. sigma_Z` `= sqrt(9 (41))`
  `= 3 sqrt 41`

 

b.i.    `H_0: \ mu = 4`
  `H_1: \ mu < 4`

 

b.ii.    `bar X~N(mu, (sigma^2)/n)`
  `bar X~N(4, 36/64)`

 
`E(barX) = 4`

`sigma_bar X` `= sqrt(36/64)`
  `= 3/4`

Statistics, SPEC2 2017 VCAA 20 MC

In a one-sided statistical test at the 5% level of significance, it would be concluded that

  1. `H_0` should not be rejected if  `p = 0.04`
  2. `H_0` should be rejected if  `p = 0.06`
  3. `H_0` should be rejected if  `p = 0.03`
  4. `H_0` should not be rejected if  `p != 0.05`
  5. `H_0` should not be rejected if  `p = 0.01`
Show Answers Only

`C`

Show Worked Solution

`p < 0.05:\ text(Reject)\ H_0`

`p > 0.05:\ text(Do not reject)\ H_0`

`=> C`

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 2018 VCAA 19 MC

The gestation period of cats is normally distributed with mean  `mu = 66`  days and variance  `sigma^2 = 16/9`.

The probability that a sample of five cats chosen at random has an average gestation period greater than 65 days is closest to

  1. 0.5000
  2. 0.7131
  3. 0.7734
  4. 0.8958
  5. 0.9532
Show Answers Only

`E`

Show Worked Solution

`G~N (66 , 16/9)`

`bar G~N (66, 16/(9 xx 5))`

`text(Pr)(bar G > 65) ~~ 0.9532`

`=>  E`