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GRE Subject Test: Math : Confidence Intervals & Hypothesis Testing

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Example Questions

Example Question #1 : 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:

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

(187.34 mm^3/g/hr, 225.73 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 #1 : 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:

(18.84 in, 16.21 in)

(21.81 in, 11.23 in)

(14.57 in, 18.28 in)

(13.99 in, 19.07 in)

(14.47 in, 18.56 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 #11 : Estimation

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.

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

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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GRE Subject Test: Math : Confidence Intervals & Hypothesis Testing

Study concepts, example questions & explanations for GRE Subject Test: Math

All GRE Subject Test: Math Resources

Example Questions

Example Question #1 : Confidence Intervals

Example Question #1 : Confidence Intervals

Example Question #11 : Estimation

All GRE Subject Test: Math Resources

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