Question 11
The manufacturer of a brand of LED light globe states that the lifetime of these globes is normally distributed with a mean of 12 000 hours and a standard deviation of 900 hours.
A consumer group tests a random sample of 45 of these globes.
Assuming the manufacturer’s statement is true, determine the probability that the mean lifetime of the sample is less than 11 650 hours.
[2 marks]The sample of 45 globes has a mean lifetime of 11 650 hours and a sample standard deviation of 950 hours.
Determine an approximate 95% confidence interval for the population mean lifetime of these globes.
[2 marks] c)State the margin of error of the interval in Question 11b).
[1 mark]The consumer group claims that the mean lifetime of these globes is less than the manufacturer states.
Use your result from Question 11b) to evaluate the reasonableness of the consumer group’s claim.
[2 marks]One globe, the mean of 45 globes, and an interval
First the spread of one globe shrinks to the spread of a sample mean, and the tail below 11 650 hours almost vanishes. Then the real sample builds a 95% interval, and you see which side of it 12 000 falls.
The tall curve is for the mean of 45 globes, so its spread is $\frac{900}{\sqrt{45}}$. The orange bar is the interval for the population mean, $\bar{x}\pm1.96\frac{s}{\sqrt{n}}$.
Pick the right spread, then build the interval
In part a), choose what goes in the $\sigma$ field of the normal distribution function and watch the tail below 11 650. Then switch to parts b) to d), pick the interval you would build, and see where 12 000 falls.
This is where most people start. 900 is the spread of one globe, not of the mean of 45 globes. It gives $P\approx0.3487$, far too big. Divide by $\sqrt{45}$.
That’s it. The mean of 45 globes has standard deviation $\frac{900}{\sqrt{45}}\approx134.16$, so $P(\bar{X}<11\,650)\approx0.0045$. Now switch to parts b) to d).
You divided by $n$ instead of $\sqrt{n}$. The spread of 20 is far too small, so $z=-17.5$ and $P\approx0$.
18 000 is the variance of $\bar{X}$, $\left(\frac{900}{\sqrt{45}}\right)^2$. The $\sigma$ field wants the standard deviation. Typing the variance gives $P\approx0.4922$.
950 is the consumer group’s sample standard deviation. Part a) assumes the manufacturer’s statement is true, so use 900. With 950 you get $\sigma_{\bar{X}}\approx141.618$ and $P\approx0.0067$.
That’s it: $(11\,372.4,\ 11\,927.6)$ hours, with margin of error $E\approx277.6$ hours for part c). 12 000 is above the upper limit, so the whole interval lies below 12 000 and the claim is reasonable.
This is where most people start. It uses the manufacturer’s 900. The interval is built from the consumer group’s data, so use $s=950$. You get $(11\,387.0,\ 11\,913.0)$, which loses Mark 3.
You left out $\sqrt{n}$. The interval $(9788,\ 13\,512)$ contains 12 000, so it leads to the wrong conclusion in d).
1.645 is the 90% value. For 95%, $z=1.96$, because 95% of the standard normal distribution lies between $-1.96$ and 1.96. With 1.645 you get $(11\,417.0,\ 11\,883.0)$.
The interval is centred on the sample mean, 11 650, not on 12 000. Centred on 12 000 you get $(11\,722.4,\ 12\,277.6)$.
First, in part a), find what goes in the $\sigma$ field.
Part a) is done. Now switch to parts b) to d) and build the interval.
The interval is done. Now go back to part a) and find what goes in the $\sigma$ field.
You have found both. Part a) uses the manufacturer’s 900, because it assumes the statement is true. The interval uses the consumer group’s 950.
Work through the solution one mark at a time
Try each step yourself before you reveal it. Keep unrounded values in your calculator until the last line.
Part a) says to assume the manufacturer is right, so divide $\sigma=900$ by $\sqrt{n}$ for the spread of the sample mean.
$$\begin{aligned}\sigma_{\bar{X}}&=\frac{900}{\sqrt{45}}\approx134.16\\\bar{X}&\sim N\left(12\,000,\ 134.16^2\right)\end{aligned}$$
Enter 900 ÷ √(45). The output is 134.164.
Marker: correctly determines the standard deviation of the distribution of the sample mean
Use the normal cumulative distribution function with the standard deviation of $\bar{X}$, not 900.
TI-84 Plus CE: normalcdf in the DISTR menu, fields lower, upper, $\mu$, $\sigma$. Casio fx-CG50: STAT → DIST → NORM → Ncd, Data: Variable, fields Lower, Upper, $\sigma$, $\mu$.
Enter lower $-1\times10^{99}$, upper 11 650, $\mu=12\,000$ and $\sigma=900\div\sqrt{45}$. The output is 0.00454373.
Marker: correctly determines the probability
Use the 1-Sample $Z$ Interval on the calculator, with the sample values.
$$\begin{aligned}\sigma&=950,\quad \bar{x}=11\,650\\n&=45,\quad \text{C-Level}=0.95\end{aligned}$$
Casio fx-CG50: STAT → INTR → Z → 1-S. TI-84 Plus CE: STAT → TESTS → ZInterval. Choose Stats input and enter the values above.
The interval is built from the consumer group’s data, so use the sample mean and the sample standard deviation ($s=950$, not the manufacturer’s 900). With $n=45$ the sample is large, so $s$ stands in for $\sigma$.
Marker: correctly uses the sample values ($\bar{x}=11\,650$, $s=950$, $n=45$) and a 95% confidence level
Read off the interval.
Answers given to a different but reasonable degree of accuracy, such as 0.00454 in a) or $(11\,372,\ 11\,928)$ here, are accepted.
Marker: correctly determines the confidence interval
Halve the width of the interval from b).
$$E=\frac{11\,927.6-11\,372.4}{2}\approx277.6$$
The margin of error is the amount added to and subtracted from $\bar{x}$, so it is half the width of the interval.
$E=1.96\times\dfrac{950}{\sqrt{45}}\approx277.6$ is also accepted. Follow-through applies in c) and d) from an incorrect interval in b).
Marker: correctly states the margin of error
Compare the manufacturer’s 12 000 hours with the limits of the interval from b).
$$11\,372.4<11\,927.6<12\,000$$
12 000 hours is above the upper limit of the interval, so the interval lies entirely below 12 000 hours.
The claim is $\mu<12\,000$, so the side 12 000 falls on decides it. Being outside the interval is not enough.
Marker: correctly compares the manufacturer’s mean with the limits of the confidence interval
Turn the comparison into a verdict on the claim.
Every plausible value of $\mu$ (at 95% confidence) satisfies $\mu<12\,000$, which is exactly the claim. This is a conclusion from one sample, not a proof.
Marker: evaluates the reasonableness of the claim using a suitable comment
Each phrase of the question gave you something
If I had to guess where you might lose marks…
- Forgetting to divide by $\sqrt{n}$ in a). Using $\sigma=900$ for $\bar{X}$ gives $P\approx0.3487$. Dividing by $n$ instead ($\sigma=20$) gives $z=-17.5$ and $P\approx0$.
- Entering the variance as the standard deviation in a). Typing 18 000 into the $\sigma$ field gives $P\approx0.4922$.
- Mixing up the two standard deviations. Using $s=950$ in a) gives $\sigma_{\bar{X}}\approx141.618$ and $P\approx0.0067$. Using 900 in b) gives a half-width of 262.96 and the interval $(11\,387.0,\ 11\,913.0)$, which loses Mark 3.
- Building the interval incorrectly in b). Leaving out $\sqrt{n}$ gives $11\,650\pm1862=(9788,\ 13\,512)$, which contains 12 000 and so leads to the wrong conclusion in d). Using $z=1.645$ (the 90% value) gives $(11\,417.0,\ 11\,883.0)$. Centring the interval on 12 000 instead of $\bar{x}$ gives $(11\,722.4,\ 12\,277.6)$.
- Ignoring the direction of the claim in d). Writing “12 000 is outside the interval, so the claim is reasonable” leaves out why. The claim is $\mu<12\,000$, so your comment must say that 12 000 is above the upper limit (the whole interval is below 12 000). Had 12 000 been below the lower limit, the manufacturer would still be contradicted, but the claim would not be reasonable. Saying “the claim is proven” also overstates a 95% confidence result.
- Giving the full width as the margin of error in c). $11\,927.6-11\,372.4\approx555.1$ is twice the margin of error. In a), finding the upper tail $P\left(\bar{X}>11\,650\right)\approx0.9955$ instead of the lower tail is a similar reading error.
Only mark this done when you could do it without help. Reading the solution does not count.
A Tangent Tuition mock exam question. It is not a past QCAA question, and Tangent Tuition is not affiliated with the QCAA.
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