Question 8
A random sample of 40 students completed a puzzle. The sample mean time was 41.62 seconds and the sample standard deviation was 2.30 seconds.
An approximate 90% confidence interval for the population mean time (seconds) is
Not quite. You did not divide by $\sqrt{n}$, giving $41.62\pm1.645\times2.30$.
Not quite. You used the 95% value $z=1.96$ instead of $1.645$.
That’s it. Step through the hints below to see why.
Not quite. You used $z=1.282$ from invNorm$(0.9)$, which is an 80% interval.
Use the 1-sample $z$-interval with $s$ in place of $\sigma$.
Enter the 1-sample $z$-interval with $\sigma=2.30$, $\bar{x}=41.62$, $n=40$ and C-level $0.90$.
Output: $(41.0218,\ 42.2182)$
$n=40$ is large and $\sigma$ is unknown.
The 1-sample $z$-interval gives $(41.0218,\ 42.2182)$.
The approximate 90% confidence interval is $(41.02,\ 42.22)$ seconds, option (C).
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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