Please finish the non-graded practice problem. WILL thumbs up if correct. **Transcription for Educational Website****Study on Raynaud's Syndrome and Blood Circulation Issue:**Persons having Raynaud's syndrome are apt to suffer a sudden impairment of blood circulation in fingers and toes. In an experiment to study the extent of this impairment, each subject immersed a forefinger in water and the resulting heat output (cal/cm²/min) was measured. For \( m = 9 \) subjects with the syndrome, the average heat output was \( \bar{x} = 0.62 \), and for \( n = 9 \) nonsufferers, the average output was 2.08. Let \( \mu_1 \) and \( \mu_2 \) denote the true average heat outputs for the sufferers and nonsufferers, respectively. Assume that the two distributions of heat output are normal with \( \sigma_1 = 0.3 \) and \( \sigma_2 = 0.5 \).**(a) Hypothesis Testing:**Consider testing \( H_0 : \mu_1 - \mu_2 = -1.0 \) versus \( H_a : \mu_1 - \mu_2 < -1.0 \) at level 0.01.**Describe in words what \( H_a \) says:**- \( H_a \) says that the average heat output for sufferers is more than 1 cal/cm²/min below that of non-sufferers.**Calculate the test statistic and \( P \)-value.** (Round your test statistic to two decimal places and your \( P \)-value to four decimal places.)- \( z = \) [Input box]- \( P \)-value = [Input box]**State the conclusion in the problem context:**- Reject \( H_0 \). The data suggests that the average heat output for sufferers is more than 1 cal/cm²/min below that of non-sufferers. (This option is checked)---**(b) Probability of a Type II Error:**What is the probability of a type II error when the actual difference between \( \mu_1 \) and \( \mu_2 \) is \( \mu_1 - \mu_2 = -1.3 \)? (Round your answer to four decimal places.)- [Input box]**(c) Sample Size Calculation:**Assuming that \( m = n \), what sample sizes

Given information:a) Null hypothesis: Altenate hypothesis: (C) Alternative hypothesis: HA says that the average heat output for sufferers is more than 1cal/cm2/min below that of non-sufferers.These hypotheses constitute one-tailed test and significance level for the test is 0.05. Using the sample data, we compute the standard error (SE), and the z statistic test statistic (z). The P-value for this test is the probability that a z statistic less than -2.37.P-value = P(z < -2.37)Use the excel formula, "=NORM.S.DIST(-2.37,TRUE)" to determine the P-value. P-value = 0.0089Decision: Since the P-value (0.0089) is less than the significance level (0.01), we have to reject the null hypothesis. Conclusion: (C) Reject H0. The data suggest that the average heat output for sufferers is more than 1 cal/cm2/min below that of non-sufferers.b) Use the excel formula, "=NORM.S.INV(0.01)" to determine the critical value for z-score:zCritical = - 2.326Determine the rejection region using the normal distribution formula:Rejection region is that means we will accept the null hypothesis if .Use the excel formula, "=NORM.S.DIST(-0.77,TRUE)" for determining the P-values.The probability of a type II error when the actual difference between and is is 77.94%.c) The sample size required to ensure that when is 50 subjects.a) (C) Alternative hypothesis: HA says that the average heat output for sufferers is more than 1cal/cm2/min below that of non-sufferers.These hypotheses constitute one-tailed test and significance level for the test is 0.05. z = - 2.37Using the sample data, we compute the standard error (SE), and the z statistic test statistic (z). P-value = 0.0089Decision: Since the P-value (0.0089) is less than the significance level (0.01), we have to reject the null hypothesis. Conclusion: (C) Reject H0. The data suggest that the average heat output for sufferers is more than 1 cal/cm2/min below that of non-sufferers.b) The probability of a type II error when the actual difference between μ1 and is is 77.94%.c) The sample size required to ensure that when is 50 subjects.

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Statistics

Persons having Raynaud's syndrome are apt to suffer a sudden impairment of blood circulation in fingers and toes. In an experiment to study the extent of this impairment, each subject immersed a forefinger in water and the resulting heat output (cal/cm2/min) was measured. For m = 9 subjects with the syndrome, the average heat output was x = 0.62, and for n = 9 nonsufferers, the average output was 2.08. Let μ₁ and ₂ denote the true average heat outputs for the sufferers and nonsufferers, respectively. Assume that the two distributions of heat output are normal with ₁ = 0.3 and %₂=0.5. (a) Consider testing Ho: H₁-H₂ = -1.0 versus H₂: H₁-H₂ < -1.0 at level 0.01. Describe in words what He says, and then carry out the test. OH says that the average heat output for sufferers is the same as that of non-sufferers. ⒸH₂ says that the average heat output for sufferers is more than 1 cal/cm²/min below that of non-sufferers. OH, says that the average heat output for sufferers is less than 1 cal/cm2/min below that of non-sufferers. Calculate the test statistic and P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.) z = P-value= State the conclusion in the problem context. O Reject Ho. The data suggests that the average heat output for sufferers is the same as that of non-sufferers. O Fail to reject Ho. The data suggests that the average heat output for sufferers is less than 1 cal/cm²/min below that of non-sufferers. Reject Ho. The data suggests that the average heat output for sufferers is more than 1 cal/cm²/min below that of non-sufferers. O Fail to reject Ho. The data suggests that the average heat output for sufferers is the same as that of non-sufferers. (b) What is the probability of a type II error when the actual difference between ₁ and ₂ is ₁-₂=-1.3? (Round your answer to four decimal places.) (c) Assuming that m = n, what sample sizes are required to ensure that = 0.1 when μ₁ - H₂= -1.3? (Round your answer up to the nearest whole number.) subjects