The hemoglobin count (HC) in grams per 100 milliliters of whole blood is approxi
ID: 3226500 • Letter: T
Question
The hemoglobin count (HC) in grams per 100 milliliters of whole blood is approximately normally distributed with a population mean of 14 for healthy adult women. Suppose a particular female patient has had 15 laboratory blood tests during the past year. The sample readings showed an average HC of 12.09 with a standard deviation of 1.31. Does it appear that the population average HC for this patient is not 14? (a) State the null and alternative hypotheses: (Type "mu" for the symbol , e.g. mu > 1 for the mean is greater than 1, mu < 1 for the mean is less than 1, mu not = 1 for the mean is not equal to 1) H0 : HA : (b) Find the test statistic, t =
Explanation / Answer
Solution:-
Suppose a particular female patient has had 15 laboratory blood tests during the past year.
That is N = 15.
The sample readings showed an average HC of 12.09,
That is mean = 12.09
with a standard deviation of 1.31.
That is s = 1.31
Does it appear that the population average HC for this patient is not 14?
The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: mu = 14
Alternative hypothesis: mu 14
Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the sample mean is too big or if it is too small.
Formulate an analysis plan. For this analysis, let us assume the significance level is 0.05. The test method is a one-sample t-test.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = s / sqrt(n) = 1.31 / sqrt(15) = 0.33824
DF = n - 1 = 15 - 1 = 14
t = (x - mu) / SE = (12.09 - 14)/0.33824 = -5.6468
where s is the standard deviation of the sample, x is the sample mean, mu is the hypothesized population mean, and n is the sample size.
Since we have a two-tailed test, the P-value is the probability that the t statistic having 14 degrees of freedom is less than -5.6468 or greater than 5.6468.
We use the t Distribution Calculator to find P(t < -5.6468)
The P-Value is 0.00006
The result is significant at p < 0.05
Interpret results. Since the P-value is lesser than the significance level (0.05), we can reject the null hypothesis.
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