QCAA Specialist Mathematics Alternative Sequence Statistical inference

The population distribution ($X$) can take any shape, such as skewed or uniform, and is not required to be normally distributed.

The sampling distribution of the sample mean ($\bar{X}$) is only approximately normal if the sample size is large enough (Central Limit Theorem) or if the population itself is normal.

The population mean ($\mu$) is a fixed parameter. It represents the entire population and remains constant regardless of the specific samples drawn.

A confidence interval estimates a specific population parameter, not the shape or normality of the underlying population distribution.

This is a common misconception; a confidence interval does not guarantee it contains the true mean, but rather provides a range generated by a process that is successful a specified percentage of the time over repeated sampling.

A point estimate is a single specific value (such as the sample mean $\bar{x}$), whereas a confidence interval provides a range of values.

Incorrect. This value does not correspond to the correct probability and likely results from a calculation error when finding the standard error or z-score.

Incorrect. This is the probability that the sample mean is less than 4 ($P(Z < 0.96) \approx 0.83$). The question asks for the probability that the mean is more than 4.

Incorrect. This is the calculated z-score ($z \approx 0.96$), not the probability. You must use the standard normal distribution to find the area to the right of this z-score.

This value is obtained by subtracting the margin from 500 instead of adding it. This would be the correct mean if there was a 75% probability that the sample mean is less than 500 grams.

This is a calculation error, likely resulting from incorrectly using the 25% probability directly in the calculation instead of finding the corresponding z-score.

This value incorrectly uses the population standard deviation ($7.37$) instead of the standard error of the sample mean ($\frac{7.37}{\sqrt{20}}$) in the z-score formula.

While decreasing the confidence level narrows the interval, decreasing the sample size increases the standard error, which widens the interval. Therefore, this combination cannot guarantee a shorter width.

Decreasing the sample size increases the standard error, and increasing the confidence level increases the critical value. Both of these changes would actually result in a wider confidence interval, not a shorter one.

While increasing the sample size narrows the interval, increasing the confidence level requires a larger critical value, which widens the interval. This combination cannot guarantee a shorter width.

An 85% confidence interval is too narrow. The required confidence level must be at least 92.5% (based on a t-score of 1.836) for the interval to extend far enough to include the population mean of 157.0 cm.

A 90% confidence interval has a margin of error of approximately 2.49 cm. This gives a lower bound of 157.21 cm, which just misses the true population mean of 157.0 cm.

While a 99% confidence interval is wide enough to include the population mean, it is not the smallest confidence level among the choices that successfully does so.

Incorrect. This interval has a margin of error of ＄0.05, which would require a $z$-value of approximately 2.5 rather than the given 1.5.

Incorrect. This interval has a margin of error of ＄0.04, which would require a $z$-value of approximately 2.0 rather than the given 1.5.

Incorrect. This interval has a margin of error of ＄0.02, which would require a $z$-value of approximately 1.0 rather than the given 1.5.

This option incorrectly assigns the population mean to the sample mean ($\bar{x}$) and the sample mean to the population mean ($\mu$).

This option uses sample mean notation for both values, failing to recognize that 35.6 is a population mean ($\mu$) since it represents all workers.

This option incorrectly uses sample mean notation ($\bar{X}$) for the population mean of 35.6.

This incorrectly centers the confidence interval around the claimed population mean ($8.2$) instead of the observed sample mean ($\bar{x} = 8.1$).

This option incorrectly calculates the standard error by dividing the standard deviation by $n$ ($10$) instead of $\sqrt{n}$ ($\sqrt{10}$), resulting in an interval that is too narrow.

This incorrectly centers the interval on the claimed population mean ($8.2$) and incorrectly calculates the standard error as $\frac{s}{n}$ instead of $\frac{s}{\sqrt{n}}$.