Question 7
The heights of all students at a school were measured. A mean height of $157.0$ cm was calculated from this data.
A random sample of $35$ students from this school was selected. The mean height of this sample was $159.7$ cm with a standard deviation of $8.7$ cm.
The smallest confidence level that could be used to produce a confidence interval that contains $\mu$, based on this sample, is
Not quite. $85\%$ is the smallest level on offer, but its interval misses $\mu$. With $z\approx1.440$, $E\approx1.440\times\frac{8.7}{\sqrt{35}}\approx2.12$, so the interval is about $(157.58,\ 161.82)$. It starts above $157.0$.
Not quite. The level you need is at least about $93.4\%$, and $90\%$ is below that. With $z\approx1.645$, $E\approx2.42$, so the interval is about $(157.28,\ 162.12)$ and still starts above $157.0$. Any level under $93.4\%$ misses $\mu$, so round up to the next option.
That’s it. Step through the hints below to see why.
Not quite. The $99\%$ interval, about $(155.91,\ 163.49)$, does contain $157.0$. So does the $95\%$ interval, about $(156.82,\ 162.58)$, and the question asks for the smallest level that works.
Height in centimetres. Blue: the interval at your confidence level. Grey: the intervals at the four levels in the options, with yours in black. Orange: the population mean $\mu=157.0$.
The margin of error is at least the $2.7$ gap between $\bar{x}$ and $\mu$, so the interval reaches $157.0$ and contains $\mu$.
The margin of error is less than the $2.7$ gap between $\bar{x}$ and $\mu$, so the interval starts above $157.0$ and misses $\mu$. Raise the level to widen it.
$157.0$ is the mean of every student at the school, so it is $\mu$. The interval is $\bar{x}\pm E$, so it contains $\mu$ when the margin of error $E$ is at least the gap between $\bar{x}$ and $\mu$.
$$\bar{x}-\mu=159.7-157.0=2.7$$
$$E\ge2.7$$
Work backwards from the margin of error to $z$.
$$z\times\frac{8.7}{\sqrt{35}}\ge2.7$$
$$z\ge\frac{2.7\sqrt{35}}{8.7}\approx1.836$$
The confidence level is the area under the standard normal curve between $-z$ and $z$. Find it with normal cdf on your calculator.
$$P(-1.836<Z<1.836)\approx0.934$$
You need a level of at least $93.4\%$. $85\%$ and $90\%$ are too low, and $95\%$ is the smallest option above it.
The interval $\bar{x}\pm z\frac{s}{\sqrt{n}}$ contains $\mu=157.0$ when its margin of error is at least $159.7-157.0=2.7$.
$$z\times\frac{8.7}{\sqrt{35}}\ge2.7\ \Rightarrow\ z\ge1.836$$
$$P(-1.836<Z<1.836)\approx0.934$$
The level must be at least $93.4\%$, so the smallest option is $95\%$. Check: $159.7\pm1.96\times\frac{8.7}{\sqrt{35}}\approx(156.82,\ 162.58)$, which contains $157.0$.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording is from the 2020 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2020, licensed under CC BY 4.0, and has been adapted. Tangent Tuition is not affiliated with the QCAA.
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