Questions (i) and (ii). Please! Let x be a random variable that represents red blood cell count (RBC) in millions of cells per cubic millimeter of whole blood. Then x has a distribution that is approximately normal. For the population of healthy female adults, suppose the mean of the x distribution is about 4.78. Supposethat a female patient has taken six laboratory blood tests over the past several months and that the RBC count data sent to the patient's doctor are as follows.4.9 4.2 4.5 4.1 4.4 4.3A USE SALT(1) Use a calculator with sample mean and standard deviation keys to find x and s. (Round your answers to two decimal places.)=5(ii) Do the given data indicate that the population mean RBC count for this patient is lower than 4.78? Use a = 0.05.(a) What is the level of significance?State the null and alternate hypotheses.Ο Η,: μ < 4.73; Η , : μ= 4.78O H: = 4.78; H,: u> 4.7sO H,: u> 4.78; H,: = 4.78O H,: u = 4.78; H: u < 4.78Ο Η, μ= 478; Η , , μ# 4,78(b) What sampling distribution will you use? Explain the rationale for your choice of sampling distribution.O The standard normal, since we assume that x has a normal distribution and o is unknown.O The Student's t, since we assume that x has a normal distribution and o is unknown.O The standard normal, since we assume that x has a normal distribution and o is known.O The Student's t, since we assume that x has a normal distribution and o is known.What is the value of the sample test statistic? (Round your answer to three decimal places.)

Name: Questions (i) and (ii). Please! Let x be a random variable that repres
Uploaded: 2024-03-20T06:38:44.000Z
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Description: Questions (i) and (ii). Please! Let x be a random variable that represents red blood cell count (RBC) in millions of cells per cubic millimeter of whole blood. Then x has a distribution that is approximately normal. For the population of healthy fema

Here, the n=6 observations are (4.9,4.2,4.5,4.1,4.4,4.3). Then the sample mean and sample variance is given by x=1n∑i=1nxi=4.9+4.2+4.5+4.1+4.4+4.36=2646=4.4s2=1n-1∑i=1n(xi-x)2=15(∑i=1nxi2-nx2)=15(116.56-116.16)=0.45=0.08Hence, sample standard deviation is s=0.08=0.283≈0.28