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AP Statistics : Confidence Intervals and Mean

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Example Question #1 : Confidence Intervals

Suppose you have a normally distributed variable with known variance. How many standard errors do you need to add and subtract from the sample mean so that you obtain 95% confidence intervals?

Possible Answers:

1.98

1.96

2.035

1.645

1.99

Correct answer:

1.96

Explanation:

To obtain 95% confidence intervals for a normal distribution with known variance, you take the mean and add/subtract $1.96\times standard\ error$ . This is because 95% of the values drawn from a normally distributed sampling distribution lie within 1.96 standard errors from the sample mean.

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Example Question #1 : Confidence Intervals

An automotive engineer wants to estimate the cost of repairing a car that experiences a 25 MPH head-on collision. He crashes 24 cars, and the average repair is $11,000. The standard deviation of the 24-car sample is $2,500.

Provide a 98% confidence interval for the true mean cost of repair.

Possible Answers:

$11,000\pm987.6$

$11,000\pm1,275.5$

$11,000\pm1,366.7$

$11,000\pm1,456.6$

$11,000\pm1,188.77$

Correct answer:

$11,000\pm1,275.5$

Explanation:

Standard deviation for the samle mean:

$2500/\sqrt{24}=510.2$

Since n < 30, we must use the t-table (not the z-table).

The 98% t-value for n=24 is 2.5.

$11,000\pm2.5\times510.2=11,000\pm1275.5$

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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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AP Statistics : Confidence Intervals and Mean

Study concepts, example questions & explanations for AP Statistics

All AP Statistics Resources

Example Questions

Example Question #1 : Confidence Intervals

Example Question #1 : Confidence Intervals

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

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