Question 18
The mass of mangoes grown on a farm is normally distributed. A random sample of mangoes from the farm was used to calculate an approximate 95% confidence interval for the population mean mass of mangoes on the farm. This interval was $(414.5,\ 435.5)$ grams.
The sample mean and sample standard deviation of the same sample were used to estimate that approximately 95% of all mangoes on the farm have a mass between 362 grams and 488 grams.
The farm manager claims that if 64 more mangoes had been added to the original sample, and the sample mean and sample standard deviation were unchanged, the approximate 99% confidence interval for the population mean mass would not contain 415 grams.
Evaluate the reasonableness of this claim.
[6 marks]What the question doesn’t tell you
Have a go at the question cold first. Then open this to see the steps the question expects you to find for yourself.
- A confidence interval is symmetric about the sample mean. Its midpoint is $\bar{x}$ and its half-width is the margin of error $E$.
- The 362 g to 488 g statement is about single mangoes. It is an interval of values, $\bar{x}\pm1.96s$, with no $\sqrt{n}$. It is not a second confidence interval, and it is the only place you can get $s$ from.
- The original sample size $n$ is never given. You recover it by running the margin of error $E=z\dfrac{s}{\sqrt{n}}$ backwards, with $E=10.5$, $z=1.96$ and the $s$ you just found.
- “64 more mangoes had been added to the original sample” means the new sample size is $n+64$. You cannot find it until you know $n$.
- The $z$-values are not given. 1.96 (95%) and 2.58 (99%) are standard values and can be stated directly.
- “Unchanged” sample mean and standard deviation means the new interval is still centred at 425 with the same $s$. Only $z$ and $n$ change.
- “Evaluate the reasonableness” needs an explicit decision. Say whether 415 is inside the new interval and what that means for the claim.
One sample of mangoes, two very different intervals
The first interval is about single mangoes and gives you the standard deviation. The second is about the mean and hides the sample size. Then watch what 64 more mangoes do to a 99% interval.
The curve is for single mangoes, so its middle 95% is $\bar{x}\pm1.96s$. The bars are intervals for the mean, $\bar{x}\pm z\frac{s}{\sqrt{n}}$, so they shrink as the sample grows. The dashed bar is there for comparison only.
Find the hidden sample size, then test the claim
The standard deviation is already in from the interval of values. At 95%, change the sample size until your interval lands on the given blue one. Then switch to 99% and set the sample size the farm manager describes. Watch the curve for the sample mean narrow as the sample grows.
Your interval is wider than the given one, so this sample is too small. A bigger sample makes $\frac{s}{\sqrt{n}}$ smaller and narrows the interval.
Your interval is narrower than the given one, so this sample is too big. A smaller sample makes $\frac{s}{\sqrt{n}}$ bigger and widens the interval.
Your interval lands exactly on the given one. Here $\frac{63}{\sqrt{36}}=10.5$, so the original sample had 36 mangoes. Now switch to 99% and think about how many mangoes the new sample has.
At 99% with the same 36 mangoes, the interval is $(411.18,\ 438.82)$. Raising the confidence level on its own widens the interval, and 415 is still inside. But the new sample is not 36 mangoes: 64 more were added.
This is a new sample of 64, not 64 more. The interval is $(414.63,\ 435.37)$, which contains 415, so this reading leads to the wrong verdict.
With $36+64=100$ mangoes, the 99% interval is $(416.71,\ 433.29)$. 415 is 1.71 g below the lower limit, so the interval does not contain 415 and the claim is reasonable. The larger sample outweighs the higher confidence level.
This 99% interval contains 415. Is this the sample size the question describes? The new sample is the original sample plus 64 mangoes.
This 99% interval does not contain 415, but is this the sample size the question describes? The new sample is the original sample plus 64 mangoes.
First, at 95%, find the sample size that reproduces the given interval.
The original sample had 36 mangoes. Now set the confidence level and the sample size the farm manager describes.
You have found both. The same 425 and the same $s$ give two very different intervals, because $z$ and $n$ both changed.
Work through the solution one mark at a time
Try each step yourself before you reveal it.
Find the midpoint and the half-width of the interval.
$$\begin{aligned}\bar{x}&=\frac{414.5+435.5}{2}\\&=\frac{850}{2}=425\\[4pt]E&=\frac{435.5-414.5}{2}\\&=\frac{21}{2}=10.5\end{aligned}$$
An approximate confidence interval for the population mean has the form $\left(\bar{x}-E,\ \bar{x}+E\right)$, where $E=z\dfrac{s}{\sqrt{n}}$ is the margin of error. The interval is symmetric about $\bar{x}$. So its midpoint is the sample mean and its half-width is the margin of error.
Marker: correctly determines the sample mean and the margin of error
Use the interval of values for single mangoes. First you need the $z$-value for the middle 95%.
The stem gives you two different intervals, and you need to tell them apart. $(414.5,\ 435.5)$ is about the population mean, so its width is set by the spread of sample means, $\dfrac{s}{\sqrt{n}}$. “Approximately 95% of all mangoes … between 362 grams and 488 grams” is about single mangoes, so its width is set by $s$ itself.
For a normal distribution, the middle 95% of values lie within $z$ standard deviations of the mean.
State the 95% $z$-value directly.
$$z=1.96$$
$z=1.96$, because 95% of the mangoes lie within these values. This is the standard 95% value, so no calculator step is needed.
So the interval of values is
$$\left(\bar{x}-1.96s,\ \bar{x}+1.96s\right)=(362,\ 488)$$
The half-width of the interval of values is $1.96s$, and the only unknown in it is $s$. Solve for $s$.
$$\begin{aligned}1.96s&=488-425=63\\s&=\frac{63}{1.96}\\&=32.142857\ldots\\&\approx32.14\end{aligned}$$
On your calculator, $63\div1.96=32.1429$. Store the full value in memory and use the stored value from here on. Write it down as 32.14.
“Approximately 95% of values lie within 2 standard deviations of the mean” is also accepted. Then $2s=63$ and $s=31.5$. Carried through, that gives $n\approx34.6$ and a final lower limit of about 416.8, so the verdict is the same.
If you use $\dfrac{s}{\sqrt{n}}$ for this interval of values, you do not get this mark, but follow-through applies to the marks after it.
Marker: correctly determines the sample standard deviation from the interval of values
Put $E$, $z$ and $s$ into the margin of error formula. It becomes an equation in $n$.
$$1.96\times\frac{s}{\sqrt{n}}=10.5$$
The sample size is never stated, but you need it, because the new sample is “64 more” than the original. The margin of error formula is the only place $n$ appears, and you now know $E$, $z$ and $s$. You are using the confidence interval formula backwards.
The numerator $1.96s$ is exactly 63, from Step 2. Substitute 63 directly and solve.
$$\begin{aligned}\frac{63}{\sqrt{n}}&=10.5\\\sqrt{n}&=\frac{63}{10.5}=6\\n&=6^2=36\end{aligned}$$
Using 63 avoids any rounding error from $s$.
Marker: correctly determines the size of the original sample
The 64 mangoes are added to the original 36, so the new sample contains all of them.
$$\begin{aligned}n_{\text{new}}&=36+64=100\\\sqrt{100}&=10\end{aligned}$$
Now state the 99% $z$-value directly.
$$z=2.58$$
$z=2.58$, because a 99% interval holds the middle 99% of the distribution. Like 1.96 for 95%, this is a standard value and can be stated directly.
The sample mean and standard deviation are unchanged. Only $z$ (now 2.58) and $n$ (now 100) change.
$$\begin{aligned}E_{99}&=2.58\times\frac{s}{\sqrt{100}}\\&=2.58\times\frac{32.142857\ldots}{10}\\&=8.292857\ldots\\&\approx8.29\end{aligned}$$
Enter $2.58\times s\div10$ with the stored $s$. Your calculator gives 8.29286.
For $z$, 2.58 is accepted, and 2.576 is accepted too.
Marker: correctly determines the margin of error for the 99% confidence interval
The interval is $\bar{x}\pm E_{99}$, centred at the unchanged sample mean, 425.
$$\begin{aligned}425-8.29&=416.71\\425+8.29&=433.29\end{aligned}$$
$$\begin{aligned}&\text{99\% CI}\\&\quad=(416.71,\ 433.29)\text{ grams}\end{aligned}$$
Give the limits to 2 decimal places, one more than the given data. Any reasonable rounding gives the same verdict, because 415 is 1.71 g below the lower limit.
Using $z=2.576$ gives $(416.72,\ 433.28)$. This is accepted too.
Marker: correctly determines the approximate 99% confidence interval
The claim is that the new 99% interval does not contain 415 g. Check where 415 lies relative to the limits.
$$\begin{aligned}&415<416.71\\&416.71-415=1.71\text{ g}\end{aligned}$$
415 g is below the lower limit, so it is not inside $(416.71,\ 433.29)$.
Then say explicitly whether that supports the claim. Write it as a sentence: “415 g is below the lower limit of the interval, so the interval does not contain 415 g. The claim is reasonable.”
The decision is the mark. Say explicitly whether the interval supports the claim.
Marker: evaluates the reasonableness of the claim using a suitable comment
Each phrase of the question gave you something
The engine is inversion with a representation switch. The sample size is hidden, so you reach it by running the confidence interval formula backwards, using an $s$ taken from a different kind of interval: one about individual masses, not means. Because the sample grows by adding 64, that inversion is compulsory. The verdict then depends on two opposing effects. The higher confidence level widens the interval and the larger sample narrows it.
If I had to guess where you might lose marks…
- Reading “64 more mangoes” as a new sample of 64. Then $E=2.58\times\dfrac{32.14}{\sqrt{64}}\approx10.37$ gives $(414.63,\ 435.37)$. That contains 415, so you reach the wrong verdict.
- Not updating $n$. The same goes for arguing “99% intervals are wider than 95% intervals, and $(414.5,\ 435.5)$ already contains 415”. With $n=36$, $E\approx13.82$ gives $(411.18,\ 438.82)$, which contains 415: the wrong verdict. The larger sample outweighs the higher confidence level.
- Using the half-width 63 as $s$. If you forget to divide by 1.96, then $\sqrt{n}=11.76$, $n\approx138$, the new $n$ is 202 and the interval is about $(413.6,\ 436.4)$. That contains 415: the wrong verdict.
- The wrong $z$-value for 99%. Keeping 1.96 gives $E=6.3$ and $(418.7,\ 431.3)$. Using the 98% central value 2.33 (all of the 0.01 placed in one tail) gives $(417.51,\ 432.49)$. The verdict survives, but you lose marks 4 and 5.
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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