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A simple random sample of pulse rates of 60 women from a normally distributed population results in a standard deviation of 1
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Answer #1

c) P-value = 2 * P(\chi^2 > 102.801)

= 2 * 0.00036

= 0.00072 = 0.001

d) Since the P-value is less than the significance level(0.001 < 0.10), so we should reject the null hypothesis.

Reject H0. There is sufficient evidence to warrant rejection of the claim that pulse rates of women have a standard deviation equal to 10 beats per minute.

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