A simple random sample of pulse rates of 25 women from a normally distributed population results in a standard deviation of 12.7 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 pullse rates of women 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.&nbsp;H0:σ≠10&nbsp;beats per minute,&nbsp;H1:σ=10&nbsp;beats per minuteB.&nbsp;H0:σ=10&nbsp;beats per minute,&nbsp;H1:σ≠10&nbsp;beats per minuteSNKC.&nbsp;H0:σ=10&nbsp;beats per minute,&nbsp;H1:σ&lt;10&nbsp;beats per minuteD.&nbsp;H0:σ≥10&nbsp;beats per minute,&nbsp;H1:σ&lt;10&nbsp;beats per minuteb. Compute the test statistic.χ2=□(Round to three decimal places as needed.)c. Find the P-value of the test statistic.The P-value of the test statistic is&nbsp;□(Round to three decimal places as needed.)d. State the conclusion.'Reject'/'Fail to reject'&nbsp;H0&nbsp;'is'/'is not' sufficiert 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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A simple random sample of pulse rates of 25 women from a normally distributed population results in a standard deviation of 12.7 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 pullse rates of women 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.&amp;nbsp;H0:σ≠10&amp;nbsp;beats per minute,&amp;nbsp;H1:σ=10&amp;nbsp;beats per minuteB.&amp;nbsp;H0:σ=10&amp;nbsp;beats per minute,&amp;nbsp;H1:σ≠10&amp;nbsp;beats per minuteSNKC.&amp;nbsp;H0:σ=10&amp;nbsp;beats per minute,&amp;nbsp;H1:σ&amp;lt;10&amp;nbsp;beats per minuteD.&amp;nbsp;H0:σ≥10&amp;nbsp;beats per minute,&amp;nbsp;H1:σ&amp;lt;10&amp;nbsp;beats per minuteb. Compute the test statistic.χ2=□(Round to three decimal places as needed.)c. Find the P-value of the test statistic.The P-value of the test statistic is&amp;nbsp;□(Round to three decimal places as needed.)d. State the conclusion.&#039;Reject&#039;/&#039;Fail to reject&#039;&amp;nbsp;H0&amp;nbsp;&#039;is&#039;/&#039;is not&#039; sufficiert evidence to warrant rejection of the claim that pulse rates of women have a standard deviation equal to 10 beats per minute.

Tavolillaqra · Accepted Answer

a). The null hypothesis states that population standard deviation is equal to 10.The alternative hypothesis states that population standard deviation is not equal to 10.Thus, the correct option is:b). H0:σ=10 beats per minute, H1:σ≠10 beats per minuteb). The formula of chi-square test statistic is:χ2=(n−1)s2σ2Given:n=25σ=10s=12.7The calculation for chi-square test statistic is as follows:χ2=(25−1)⋆12.72102=3870.96100=38.709Thus, the value of chi-square test statistic is 38.709.c). The degree of freedom is as follows:df=n-1=25-1=24Observe df=24 in the chi-square table to obtain the p-value. The answer is 0.029.Thus, p-value =0.029

massifyc9 · Accepted Answer

d). Level of significance=0.10If the p-value is less than the significance level, we Reject the null hypothesis.As 0.029&amp;lt;0.10, we REJECT the null hypothesis. Hence we have sufficient evidence to reject the null hypothesis.Thus, the correct options are:1. Reject;2. Is

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