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Practice — Statistics bridge for Further Mathematics • 0.12 Confidence intervals foundations

1. Tail areas for a confidence level
Difficulty: ★☆☆
2204

A two-sided \(98\%\) normal confidence interval leaves equal probability in each tail of the standard normal distribution. What is the probability in each tail?

Hint
Subtract the confidence level from one, then share the remaining probability between the two tails.
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2. Margin of error
Difficulty: ★☆☆
2205

A normal population has known standard deviation \(20\). An independent random sample has size \(100\). Using \(z^*=1.96\), what is the margin of error of a \(95\%\) confidence interval for the population mean?

Hint
Multiply the standard error of the sample mean by the stated critical value.
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3. Interpreting confidence through repeated sampling
Difficulty: ★☆☆
2206

Using a valid method, a student calculates a \(95\%\) confidence interval \((18.2,21.8)\) for a population mean. Which statement gives the repeated-sampling interpretation of the method?

Hint
Ask what changes when the sampling procedure is repeated and what remains fixed.
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4. A known-variance confidence interval
Difficulty: ★★☆
2207

An independent random sample of \(49\) observations from a normal population has mean \(34\). The population standard deviation is known to be \(7\). Using \(z^*=1.96\), find a \(95\%\) confidence interval for the population mean.

Hint
The second step after finding the standard error is to place equal margins around the sample mean.
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5. Changing the confidence level
Difficulty: ★★☆
2208

Component lifetimes are normal with known standard deviation \(12\) hours. An independent random sample of \(16\) has mean \(180\) hours. Using \(z^*=1.645\), calculate a \(90\%\) confidence interval for the population mean, giving endpoints to 3 decimal places.

Hint
Use the given 90% critical value with the standard error, not with the standard deviation of individual lifetimes.
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6. Sample size and interval width
Difficulty: ★★☆
2209

Normal confidence intervals use the same confidence level and the same known population standard deviation. If the independent random sample size increases from \(25\) to \(100\), what is the ratio of the new interval width to the old interval width?

Hint
In the width formula, only the square root of the sample size changes.
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7. A large-sample estimated-variance interval
Difficulty: ★★☆
2210

An independent random sample has \(n=144\), \(\bar x=62\) and sample standard deviation \(s=24\), where \(s^2\) uses denominator \(n-1\). Population variance is unknown, and a large-sample normal approximation is suitable. Using \(z^*=2.576\), find an approximate \(99\%\) confidence interval for the population mean.

Hint
The standard error is estimated from \(s\), and the interval must use the supplied 99% critical value.
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8. Recovering sample size from an interval
Difficulty: ★★☆
2211

A normal population has known standard deviation \(9\). An independent random sample gives the \(95\%\) interval \((47.06,52.94)\), calculated using \(z^*=1.96\). Treat these displayed endpoints as the exact values obtained with that critical value. What was the sample size?

Hint
Take half the interval width to obtain the margin, then rearrange the margin formula.
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9. Minimum sample size for a full-width target
Difficulty: ★★★
2212

A normal population has known standard deviation \(11\). Independent random observations will be taken. Using \(z^*=1.645\), what is the smallest sample size giving a two-sided \(90\%\) confidence interval with full width at most \(5\)?

Hint
The full width is twice the margin of error. Rearrange the width inequality and round upwards.
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10. An interval from sample summaries
Difficulty: ★★★
2213

An independent random sample has \(n=100\), \(\sum x_i=4200\) and \(\sum x_i^2=179964\). The population variance is unknown, and a large-sample normal approximation is suitable. Use the unbiased variance estimate and \(z^*=1.96\) to find an approximate \(95\%\) confidence interval for the population mean. Give endpoints to 3 decimal places.

Hint
Obtain the mean and unbiased variance from the summaries, then use \(s/\sqrt n\) in the normal interval.
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