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AP Statistics : Estimation

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Example Question #5 :Confidence Intervals

Subject	Horn Length (in)	Subject	Horn Length (in)
1	19.1	11	11.6
2	14.7	12	18.5
3	10.2	13	28.7
4	16.1	14	15.3
5	13.9	15	13.5
6	12.0	16	7.7
7	20.7	17	17.2
8	8.6	18	19.0
9	24.2	19	20.9
10	17.3	20	21.3

The data above represents measurements of the horn lengths of African water buffalo that were raised on calcium supplements. Construct a 95% confidence interval for the population mean for horn length after supplments.

Possible Answers:

(14.57 in, 18.28 in)

(18.84 in, 16.21 in)

(14.47 in, 18.56 in)

(21.81 in, 11.23 in)

(13.99 in, 19.07 in)

Correct answer:

(13.99 in, 19.07 in)

Explanation:

First you must calculate the sample mean and sample standard deviation of the sample.

$\bar{x}=16.53$

$\sigma=5.29$

Because we do not know the population standard deviation we will use the t-distribution to calculate the confidence intervals. We must use standard error in this formula because we are working with the standard deviation of the sampling distribution.

Formula:

$\bar{x} \pm t\cdot \frac{s}{\sqrt{n}}$

To find the appropriate t-value for 95% confidence interval:

$\alpha =0.05, \frac{0.05}{2}=0.025$

df=n-1= 20-1 = 19

Look up $t_{\frac{\alpha }{2}, 19}$ in t-table and the corresponding t-value = 2.093.

Thus the 95% confidence interval is:

$16.53 \pm 2.093\left(\frac{5.29}{\sqrt{19}}\right)= (13.99 mm Hg, 19.07 mm Hg)$

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Example Question #1 :How To Find Confidence Intervals For A Mean

The population standard deviation is 7. Our sample size is 36.

What is the 95% margin of error for:

1) the population mean

2) the sample mean

Possible Answers:

1) 15.554

2) 3.656

1) 13.720

2) 2.287

1) 14.567

2) 4.445

1) 11

2) 3

1) 12.266

2) 3.711

Correct answer:

1) 13.720

2) 2.287

Explanation:

For 95% confidence, Z = 1.96.

1) The population M.O.E. =

$1.96 \times 7 = 13.720$

2) The sample standard deviation =

$7\div\sqrt{36}=1.167$

The sample M.O.E. =

$1.96\times1.167=2.287$

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Example Question #7 :Confidence Intervals

The $95\%$ confidence interval created for the difference in means between two training programs for middle distance college runners is $\small \left ( -4.5, 6.5\right )$ . The variable being measured is the improvement in seconds of mile times over the course of a season. One program has more speed work and intervals, while the other focuses more on distance training.

What does the confidence interval tell us about the difference in the two programs?

Possible Answers:

The mean improvement of $\small 1$ second $\small \left ( (6.5-4.5)\times 0.5 = 1\right )$ is too small to matter, so reject the null. There is no evidence that one program is better at reducing miles times.

$\small 6.5$ is greater than $\small \left | -4.5\right |$ , so reject the null. This is evidence that one program is better at reducing mile times than the other.

Zero is in the interval, so do not reject the null. There is no evidence that one program is better than the other.

Zero is not in the interval, so reject the null. This is evidence that one program is significantly better at reducing mile times.

The confidence interval is large, so one program is clearly better at reducing mile times than the other.

Correct answer:

Zero is in the interval, so do not reject the null. There is no evidence that one program is better than the other.

Explanation:

For there to be a statistically significant difference in the training programs, the 95% confidence interval cannot include zero. $\small \left ( -4.5, 6.5\right )$ includes zero, so we can't say that one program is significantly better than the other.

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Example Question #8 :Confidence Intervals

Possible Answers:

Correct answer:

Explanation:

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Example Question #9 :Confidence Intervals

Possible Answers:

Correct answer:

Explanation:

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Example Question #10 :Confidence Intervals

In a simple random sample of 1000 people, 261 are left-handed. Find a $95\%$ confidence interval for the true proportion of left-handed people in the entire population.

Possible Answers:

$0.261\pm 1.64\sqrt{\frac{(0.261)(0.739)}{1000}}$

$0.261\pm \sqrt{\frac{(0.261)(0.739)}{1000}}$

The sample is not sufficiently large. A confidence interval cannot be used.

$0.261\pm1.96 \sqrt{\frac{(0.261)(0.739)}{1000}}$

$0.261\pm 1.64\sqrt{\frac{(0.25)(0.75)}{1000}}$

Correct answer:

$0.261\pm1.96 \sqrt{\frac{(0.261)(0.739)}{1000}}$

Explanation:

Step 1: Determine the approporiate formula. In this case, the statistic is a proportion because the question says 261 people out of 1000 are left-handed. We want to use a confidence interval for proportions. The equation is

$\hat{p}\pm z^{*} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

Step 2: $\hat{p}$ , or the sample proportion is equal to $\frac{261}{1000} = 0.261$

Step 3: The questions asks for a 95% confidence interval. You can assume a normal distribution because of the large sample size. Therefore, in a normal distribution, 95% of the data is contained within $z=\pm 1.96$ .

$z^{*}=1.96$

Step 4: Substitue all the values into the equation to get the answer.

$0.261\pm1.96 \sqrt{\frac{(0.261)(0.739)}{1000}}$

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Example Question #51 :Ap Statistics

A health insurance executive suspects that 50% of insurance applications are incomplete (i.e. critical data is missing from the application). He wants to gather a sample to estimate the true proportion. He wants to be within 5% of the true mean with 95% confidence.

What sample size does he need?

Possible Answers:

375

385

395

405

365

Correct answer:

385

Explanation:

$n=pq\times(Z/E)^{2}$

$n=.5\times.5\times(1.96/.05)^{2} = 384.16 = 385$

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Example Question #1 :Confidence Intervals And Regression

You estimate a regression model with $\beta = 3.4$ and SE = .024 , where $\beta$ is the beta coefficient andis the standard error. Construct 95% confidence intervals for $\beta$ .

Possible Answers:

$\left [ 3.35,3.45 \right ]$

$\left [ 3.4,3.45 \right ]$

$\left [ 3.1,3.45 \right ]$

Correct answer:

$\left [ 3.35,3.45 \right ]$

Explanation:

To construct 95% confidence intervals for $\beta$ , we simply take the coefficient and add/subtract $1.96\times the\ standard\ error\ of\ \beta$ . This is because $\beta$ is assumed to follow a symmetrical distribution (the normal), and 95% of the values in the sampling distribution are contained within 1.96 standard errors of $\beta$ .

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Example Question #2 :Confidence Intervals And Regression

Which of the following statements are correct about confidence intervals?

Possible Answers:

The width of a confidence interval decreases as the sample size increases and increases as the confidence level increases.

The width of a confidence interval does not change as the sample size increases and increases as the confidence level increases.

The width of a confidence interval increases as the sample size increases and increases as the confidence level increases.

The width of a confidence interval decreases as the sample size increases and increases as the confidence level decreases.

The width of a confidence interval increases as the sample size increases and increases as the confidence level decreases.

Correct answer:

The width of a confidence interval decreases as the sample size increases and increases as the confidence level increases.

Explanation:

Larger samples give narrower intervals. We are able to estimate a population proportion more precisely with a larger sample size.

As the confidence level increases the width of the confidence interval also increases. A larger confidence level increases the chance that the correct value will be found in the confidence interval. This means that the interval is larger.

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Example Question #3 :Confidence Intervals And Regression

You are asked to create a $95\%$ confidence interval with a margin of error no larger thanwhile sampling from a normally distributed population with a standard deviation of. What is the minimum required sample size?

Possible Answers:

Correct answer:

Explanation:

Keep in mind that the margin of error for a confidence interval based on a normal population is equal to $z^{*}\left(\frac{\sigma}{\sqrt{n}}\right)$ , where $z^{*}$ is the-score corresponding to the desired confidence level.

From the problem, we can tell that $z^{*} = 1.96$ and $\sigma = 5$ . We can then solve foralgebraically:

$3 = 1.96\left(\frac{5}{\sqrt{n}}\right)$

$\frac{3}{1.96} = \frac{5}{\sqrt{n}}$

$\frac{3}{1.96}\cdot \frac{1}{5} = \frac{1}{\sqrt{n}}$

$n =\left ( \frac{3}{1.96 \cdot 5} \right )^{-2} = 10.671$

The minimum sample size is 10.671 rounded up, which is. If you are unsure on problems like these, you can check the margin of error for your answer rounded down and then rounded up (in this case, for n = 10 and n = 11 .)

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AP Statistics : Estimation

Study concepts, example questions & explanations for AP Statistics

All AP Statistics Resources

Example Questions

Example Question #5 :Confidence Intervals

Example Question #1 :How To Find Confidence Intervals For A Mean

Example Question #7 :Confidence Intervals

Example Question #8 :Confidence Intervals

Example Question #9 :Confidence Intervals

Example Question #10 :Confidence Intervals

Example Question #51 :Ap Statistics

Example Question #1 :Confidence Intervals And Regression

Example Question #2 :Confidence Intervals And Regression

Example Question #3 :Confidence Intervals And Regression

All AP Statistics Resources

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