Confidence intervals for the mean practice questions.

A sample of 36 items has mean 50 and the population standard deviation is known to be 6. Calculate the 95% confidence interval for the population mean (z = 1.96).

• (48.04, 51.96)
• (38.24, 61.76)
• (49.02, 50.98)
• (44.12, 55.88)

Answer: (48.04, 51.96)

Margin = 1.96 × 6/√36 = 1.96, so the interval is 50 ± 1.96 = (48.04, 51.96).

For a fixed confidence level, if the sample size is quadrupled, the width of a confidence interval for the mean is

• quartered
• doubled
• halved
• unchanged

Answer: halved

Width is proportional to 1/√n; quadrupling n multiplies √n by 2, so the width is halved.

A confidence interval for a mean should be based on the t-distribution rather than the normal distribution when

• the sample size is greater than 30
• σ is unknown and estimated from the sample
• the population standard deviation is known
• the data are not normally distributed

Answer: σ is unknown and estimated from the sample

The t-distribution is used when the population standard deviation is unknown and is estimated from the sample.

For a t-based confidence interval from a sample of size 20, state the number of degrees of freedom.

• 20
• 21
• 18
• 19

Answer: 19

Degrees of freedom for a one-sample t interval equal n − 1 = 20 − 1 = 19.

Which change would make a confidence interval for the mean narrower without changing the confidence level?

• Increasing the population standard deviation
• Decreasing the sample size
• Increasing the sample size
• Increasing the confidence level

Answer: Increasing the sample size

A larger sample size reduces the standard error σ/√n, narrowing the interval; the other changes widen it or leave it unchanged.

The confidence interval for a population mean is found to be (42.5, 47.5). State the sample mean.

• 45
• 42.5
• 47.5
• 5

Answer: 45

The sample mean is the centre of the interval: (42.5 + 47.5)/2 = 45.