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BBUNDERSTAT12 8.2.017.S.
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PRACTICE ANOTHER
Let $x$ be a random variable that represents red blood cell count ( $R B C$ ) 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 4.66 . Suppose that 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.
$\begin{array}{lllllllll}4.9 & 4.2 & 4.5 & 4.1 & 4.4 & 4.3\end{array}$
¿ USE SALT
(1) Use a calculator with sample mean and standard deviation keys to find $\bar{x}$ and s. (Round your answers to two decimal places.)
\[
\begin{array}{l}
\bar{x}= \\
s=
\end{array}
\]
(ii) Do the given data indicate that the population mean $\mathrm{RBC}$ count for this patient is lower thon 4.66 ? Use $\alpha=0.05$.
(a) What is the level of significance?

Step 1 ：(i) To find the sample mean $\bar{x}$, we add up all the values and divide by the number of values. For the standard deviation s, we use the formula for the sample standard deviation, which is the square root of the variance. The variance is the average of the squared differences from the mean. The sample mean $\bar{x}$ is calculated as follows: $\bar{x} = (4.9 + 4.2 + 4.5 + 4.1 + 4.4 + 4.3) / 6 = 4.4$ The standard deviation s is calculated as follows: $s = \sqrt{[(4.9-4.4)^2 + (4.2-4.4)^2 + (4.5-4.4)^2 + (4.1-4.4)^2 + (4.4-4.4)^2 + (4.3-4.4)^2] / (6-1)}$ $s = \sqrt{0.25 + 0.04 + 0.01 + 0.09 + 0 + 0.01 / 5}$ $s = \sqrt{0.08} = 0.28$ (rounded to two decimal places) (ii) To determine whether the population mean RBC count for this patient is lower than 4.66, we can perform a one-sample t-test. The null hypothesis is that the population mean is 4.66, and the alternative hypothesis is that the population mean is less than 4.66. The level of significance is $\alpha = 0.05$. This means that we are willing to accept a 5% chance of rejecting the null hypothesis when it is true. To perform the t-test, we calculate the t-value as follows: $t = (\bar{x} - \mu) / (s / \sqrt{n})$ $t = (4.4 - 4.66) / (0.28 / \sqrt{6})$ $t = -2.79$ (rounded to two decimal places) We then compare the t-value to the critical t-value for a one-tailed test with 5 degrees of freedom (n-1) and a significance level of 0.05. The critical t-value is approximately -1.476. Since our calculated t-value is less than the critical t-value, we reject the null hypothesis and conclude that the data indicate that the population mean RBC count for this patient is lower than 4.66.