A simple random sample of 56 men results in a standard deviation of 10.09 beats per minute. Use the sample results to test the claim that men have pulse rates that have a standard deviation equal to the population standard deviation of 10.63 beats per minute at the 5% significance level. Assume that the sample is from a normally distributed population. What are the correct hypotheses? (Select the correct symbols and values.) Hy: Select an answer beats per minute Original claim - Seect an ansver ♥ H,: Select an answer beats per minute Enter the critical values, along with the significance level and degrees of freedom xa,ar, below the graph. (Graph is for illustration only. No need to shade.) X- Distribution 012 01 0.08 0.06 0.04 0.02 (Round to three decimal places.) Test Statistic - | (Round to three decimal piaces.) Decision: [Select an answer Conclusion: [Select an answer have a standard deviation equal to the population standard deviation of 10.63 beats per minute. v the claim that men have pulse rates that Probability 19

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A simple random sample of 56 men results in a standard deviation of 10.09 beats per minute. Use the
sample results to test the claim that men have pulse rates that have a standard deviation equal to the
population standard deviation of 10.63 beats per minute at the 5% significance level. Assume that the
sample is from a normally distributed population.
What are the correct hypotheses? (Select the correct symbols and values.)
Hg: Select an answer
beats per minute Original claim = Select an answer
H;: Select an answer
beats per minute
Enter the critical values, along with the significance level and degrees of freedom xg.dr, below the
graph. (Graph is for illustration only. No need to shade.)
(a,df)
X- Distribution
0.12
01
0.08
0.06
0.04
0.02
(Round to three decimal places.)
Test Statistic =
(Round to three decimal places.)
Decision: Select an answer
Conclusion: [Select an answer
have a standard deviation equal to the population standard deviation of 10.63 beats per minute.
the claim that men have pulse rates that
EET
EOT
26
19
Probability
Transcribed Image Text:A simple random sample of 56 men results in a standard deviation of 10.09 beats per minute. Use the sample results to test the claim that men have pulse rates that have a standard deviation equal to the population standard deviation of 10.63 beats per minute at the 5% significance level. Assume that the sample is from a normally distributed population. What are the correct hypotheses? (Select the correct symbols and values.) Hg: Select an answer beats per minute Original claim = Select an answer H;: Select an answer beats per minute Enter the critical values, along with the significance level and degrees of freedom xg.dr, below the graph. (Graph is for illustration only. No need to shade.) (a,df) X- Distribution 0.12 01 0.08 0.06 0.04 0.02 (Round to three decimal places.) Test Statistic = (Round to three decimal places.) Decision: Select an answer Conclusion: [Select an answer have a standard deviation equal to the population standard deviation of 10.63 beats per minute. the claim that men have pulse rates that EET EOT 26 19 Probability
Expert Solution
Step 1

The claim of the test is that men have pulse rates that have a standard deviation equal to the population standard deviationσ of 10.63 beats per minute. The hypothesis is,

Null hypothesis:

H0:σ=10.63 beats per minutes

Alternative hypothesis:

H1:σ10.63 beats per minutes

The original claim is men have pulse rates that have a standard deviation equal to the population standard deviation of 10.63 beats per minute. That is, H0.

Thus, the original claim is H0.

Step 2

The sample size is 56. The significance level is 5%. The degrees of freedom is,

df=n-1=56-1=55

The degrees of freedom is 55.

Critical value for χL2 :

The right-tail area is,

1+α2=1+0.952=0.975

The critical value of at 55 degrees of freedom can be obtained using the excel formula “=CHISQ.INV.RT(0.975,55)”. The critical value is 36.398.

The value of χL2 is 36.398.

Critical value for χU2 :

The right-tail area is,

1-α2=1-0.952=0.025

The critical value of at 55 degrees of freedom can be obtained using the excel formula “=CHISQ.INV.RT(0.025,55)”. The critical value is 77.380.

The value of χU2 is 77.380.

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