Question

Let be a random variable that represents red blood cell count (ABC) in Millions of cells per cui mitmeter of whole blood. Thenx has a distribution that is approximately normal. For the population of healthy female adults, suppose the mean of the distribution is about 4. Suppose that a female patient has taken sie laboratory blood tests over the past several months and that the account data sent to the patient's doctor are as follows 4.9 4.2 4,5 41 4.4 4.3 USE SALT Use a calculator with sample mean and standard deviation keys to find and sound your answers to two decimal places) X - 09 Do the given to indicate that the population mean Raccount for this patient is lower than 4.BOP US 0.05. (a) What is the level of significance State the null and alternate hypotheses | Họ là < 4.0; Wi: H = 4.4 Mo! - 4.8; M: 4.8 CHE 48 MM 48 (b) Whatsamling we will you use? Explain the rationale for your choice of aming on The student's were that has a stribution and is now The Student's since we assume that has a normal distribution and is unknown The standard normal, since we assume that has a normal distribution and is known The standard normal, since we assume that has a normal distribution and is unknown What is the aut of the sample testi (Round your answer to three decapaces) Estimate the value 0.250 0.1000 250 0.010.100 0.010.00 Pa 0010

Answer 1

The provided sample mean is X = 4.40 and the sample standard deviation is s = 0.283, and the sample size is n = 6. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: u = 4.80 Ha: u < 4.80 This corresponds to a left-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.05, and the critical value for a left-tailed testis te = -2.015. The rejection region for this left-tailed testis R=t:t< -2.015 (3) Test Statistics The t-statistic is computed as follows: X - μo ta 4.40 - 4.80 0.283/16 = -3.462 s/n (4) Decision about the null hypothesis Since it is observed that t = -3.462 <te= -2.015, it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value is p = 0.009, and since p= 0.009 < 0.05, it is concluded that the null hypothesis is rejected. (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean u is less than 4.80, at the 0.05 significance level. Confidence interval The 95% confidence interval is 4.103<u < 4.697.

Level of significance = 0.05

b. 2nd option is correct.

Test statistics = -3.462

P value < 0.010

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