# 4 Test Standard Deviation10Hypothesized ValueActual Estimate12.5436146DFChiSquareTest Statistic 229.7176 Chisq1.0000

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Testing Claims About Variation. In Exercises 5–16, test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. Assume that a simple random sample is selected from a normally distributed population.

Pulse Rates of Women Repeat the preceding exercise using the pulse rates of women listed in Data Set 1 “Body Data” in Appendix B. For the sample of pulse rates of women, n = 147 and s = 12.5. See the accompanying JMP display that results from using the original list of pulse rates instead of the summary statistics. (Hint: The bottom three rows of the display provide P -values for a two-tailed test, a left-tailed test, and a right-tailed test, respectively.) What do the results indicate about the effectiveness of using the range rule of thumb with the “normal range” from 60 to 100 beats per minute for estimating σ  in this case? help_outlineImage Transcriptionclose4 Test Standard Deviation 10 Hypothesized Value Actual Estimate 12.5436 146 DF ChiSquare Test Statistic 229.7176 <.0001 Test Min PValue Prob < Chisq Prob > Chisq 1.0000 <0001* fullscreen
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Step 1

The null and the alternative hypotheses are,

H0: σ=10 (The standard deviation of beats per minute is equal to ten.)

H1: σ ≠10 (The standard deviation of beats per minute is not equal to ten.)

From the given output, the test statistic is identified and its value is 229.7176.

The P-value is also identified from the given output and its value is almost equal to 0.

Step 2

Decision rule:

Reject the null hypothesis H0 when the P-value is lesser than the level of significance α. Otherwise, do not reject the null hypothesis.

Conclusion:

The P-value is 0 and the level of significance is considered as 0.01.

The P-value is lesser than the level of si...

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