Math  /  Data & Statistics

Questionvarince
A simple random sample of 31 men from a normally distributed population results in a standard deviation of 10.7 beats per minute. The normal range of pulse rates of adults is typically given as 60 to 100 beats per minute. If the range rule of thumb is applied to that normal range, the result is a standard deviation of 10 beats per minute. Use the sample results with a 0.10 significance level to test the claim that pulse rates of men have a standard deviation equal to 10 beats per minute. Complete parts (a) through (d) below. a. Identify the null and alternative hypotheses. Choose the correct answer below. A. H0:σ10H_{0}: \sigma \neq 10 beats per minute B. H0:σ=10H_{0}: \sigma=10 beats per minute H1:σ=10H_{1}: \sigma=10 beats per minute H1:σ<10H_{1}: \sigma<10 beats per minute C. H0:σ=10H_{0}: \sigma=10 beats per minute D. H0:σ10H_{0}: \sigma \geq 10 beats per minute H1:σ10H_{1}: \sigma \neq 10 beats per minute H1:σ<10H_{1}: \sigma<10 beats per minute b. Compute the test statistic. x2=x^{2}=\square (Round to three decimal places as needed.)

Studdy Solution

STEP 1

What is this asking? We're checking if the standard deviation of men's pulse rates is actually 10 beats per minute, given a sample and a significance level of 0.10. Watch out! Don't mix up standard deviation with variance!
Also, remember that we're dealing with a *chi-squared* test since it's about standard deviation.

STEP 2

1. Set up the hypotheses
2. Calculate the test statistic

STEP 3

Our **null hypothesis** is that the standard deviation σ\sigma is equal to **10** beats per minute.
We write this as H0:σ=10H_0: \sigma = 10.
This is what we're trying to *disprove*!

STEP 4

The **alternative hypothesis** is that the standard deviation σ\sigma is *not* equal to **10** beats per minute.
We write this as H1:σ10H_1: \sigma \neq 10.
This is what we suspect might be true if our sample data provides strong enough evidence against the null hypothesis.
So, the correct answer for part (a) is *C*.

STEP 5

The formula for our test statistic is χ2=(n1)s2σ2 \chi^2 = \frac{(n-1) \cdot s^2}{\sigma^2} where nn is the **sample size**, ss is the **sample standard deviation**, and σ\sigma is the **hypothesized population standard deviation**.

STEP 6

We know that n=31n = \mathbf{31}, s=10.7s = \mathbf{10.7}, and σ=10\sigma = \mathbf{10}.
Let's plug these values into our formula: χ2=(311)(10.7)2(10)2 \chi^2 = \frac{(\mathbf{31}-1) \cdot (\mathbf{10.7})^2}{(\mathbf{10})^2}

STEP 7

First, let's subtract one from the sample size: 311=3031 - 1 = 30. χ2=30(10.7)2(10)2 \chi^2 = \frac{30 \cdot (\mathbf{10.7})^2}{(\mathbf{10})^2} Next, square the sample standard deviation: (10.7)2=114.49(10.7)^2 = 114.49. χ2=30114.49100 \chi^2 = \frac{30 \cdot 114.49}{100} Now, square the population standard deviation: (10)2=100(10)^2 = 100. Then, multiply the numerator: 30114.49=3434.730 \cdot 114.49 = 3434.7. χ2=3434.7100 \chi^2 = \frac{3434.7}{100} Finally, divide to get our **chi-squared test statistic**: χ2=34.347 \chi^2 = 34.347

STEP 8

a) The correct choice is C. H0:σ=10H_0: \sigma = 10 beats per minute and H1:σ10H_1: \sigma \neq 10 beats per minute. b) The test statistic is χ2=34.347\chi^2 = 34.347.

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