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AP Statistics : AP Statistics

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Example Questions

Example Question #1 : Confidence Intervals

300 hundred eggs were randomly chosen from a gravid female salmon and individually weighed. The mean weight was 0.978 g with a standard deviation of 0.042. Find the 95% confidence interval for the mean weight of the salmon eggs (because it is a large n, use the standard normal distribution).

Possible Answers:

(1.042 g, 0.827 g)

(-1.96 g, 1.96 g)

(0.974 g, 0.982 g)

(0.563 g, 0.294 g)

(0.973 g, 0.9827 g)

Correct answer:

(0.973 g, 0.9827 g)

Explanation:

Because we have such a large sample size, we are using the standard normal or z-distribution to calculate the confidence interval.

Formula:

$\bar{x} \pm z \cdot \frac{\sigma}{\sqrt{n}}$

$\bar{x}= 0.978 g$

$\sigma= 0.042$

We must find the appropriate z-value based on the given $\alpha$ for 95% confidence:

$1-\frac{\alpha}{2}= 1-\left(\frac{0.05}{2}\right)= 0.975$

Then, find the associated z-score using the z-table for

$z _{0.975} = 1.96$

Now we fill in the formula with our values from the problem to find the 95% CI.

$0.978 \pm 1.96 \left( \frac{0.042}{\sqrt{300}}\right)= (0.973 g, 0.9827 g)$

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Example Question #2 : Confidence Intervals

A sample of n=56 observations of 0₂ consumption by adult western fence lizards gave the following statistics:

Mean= 249 mm^3/g/hr

s= 24 mm^3/g/hr

Find the $90\%$ confidence limit for the mean 0₂ consumption by adult western fence lizards.

Possible Answers:

(187.34 mm^3/g/hr, 225.73 mm^3/g/hr)

(119.23 mm^3/g/hr, 219.23 mm^3/g/hr)

(242.57 mm^3/g/hr, 255.42 mm^3/g/hr)

(273 mm^3/g/hr, 225 mm^3/g/hr)

(243.63 mm^3/g/hr, 254.36 mm^3/g/hr)

Correct answer:

(243.63 mm^3/g/hr, 254.36 mm^3/g/hr)

Explanation:

Because we are only given the sample standard deviation we will use the t-distribution to calculate the confidence interval.

Appropriate Formula:

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

Now we must identify our variables:

$\bar{x}=249$

s=24

n=56

We must find the appropriate t-value based on the given $\alpha$

t-value at 90% confidence:

$t_{\frac{\alpha }{2}, n-1}= t _{\frac{0.10}{2}}= t_{0.05, 55}$

Look up t-value for 0.05, 55 , so t-value= ~ 1.6735

90% CI becomes:

$249 \pm 1.6735\left(\frac{24}{\sqrt{56}}\right)$

= (243.63 mm^3/g/hr, 254.36 mm^3/g/hr)

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Example Question #3 : 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:

(21.81 in, 11.23 in)

(13.99 in, 19.07 in)

(14.47 in, 18.56 in)

(14.57 in, 18.28 in)

(18.84 in, 16.21 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 : Confidence Intervals

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) 14.567

2) 4.445

1) 13.720

2) 2.287

1) 11

2) 3

1) 15.554

2) 3.656

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

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

Zero is in the interval, so do not reject the null. There is no evidence that one program is better 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 #4 : Confidence Intervals

Possible Answers:

Correct answer:

Explanation:

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Example Question #2 : Confidence Intervals

Possible Answers:

Correct answer:

Explanation:

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Example Question #51 : Ap Statistics

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:

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

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

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

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

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

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 #1 : How To Find Confidence Intervals For A Proportion

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:

385

405

395

365

375

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 #51 : Inference

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

Possible Answers:

$\left [ 3.1,3.45 \right ]$

$\left [ 3.35,3.45 \right ]$

$\left [ 3.4,3.45 \right ]$

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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AP Statistics : AP Statistics

Study concepts, example questions & explanations for AP Statistics

All AP Statistics Resources

Example Questions

Example Question #1 : Confidence Intervals

Example Question #2 : Confidence Intervals

Example Question #3 : Confidence Intervals

Example Question #1 : Confidence Intervals

Example Question #1 : Confidence Intervals

Example Question #4 : Confidence Intervals

Example Question #2 : Confidence Intervals

Example Question #51 : Ap Statistics

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

Example Question #51 : Inference

All AP Statistics Resources

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