on (Testing H, when Gis unknown Letxbe a random variable that represents hemoglo
ID: 3230675 • Letter: O
Question
on (Testing H, when Gis unknown Letxbe a random variable that represents hemoglobin count (HC) in grams per 100 mililiters of whole blood. Then x has a distribution that is approximately normal, with population mean of about 14 for healthy adult women. Suppose that a female patient has taken 10 laboratory blood tests during the past year. The Hc data sent to the patient's doctor are as follows. 15 18 16 19 14 11 14 18 16 11 The sample data gives a sample average of 15.20 with a sample standard deviation of 2.78. Does the information indicate that the population average HC for this patient is higher than 14? Use the 0.01 level of significance. write the null hypothesis Write the alternative hypothesis Show your work for the sample test statistic or no credit. P-interval Do you have sufficient evidence to support the claim that the population average Hc for this patient is higher than 14? What is the basis for this decision? (Use the P-interval value as part of your answer.)Explanation / Answer
Solution:-
Given, HC data -
15, 18, 16, 19, 14, 11, 14, 18, 16, 11
Mean = 15.2
Standard deviation = 2.78
State the hypotheses.
Null hypothesis: >= 14
Alternative hypothesis: < 14
Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.01. The test method is a one-sample t-test.
SE = s / sqrt(n) = 2.78 / sqrt(10) = 0.879
DF = n - 1 = 10 - 1 = 9
t = (x - ) / SE = (15.2 - 14)/0.879 = 1.36
12.34303 x 18.05697
Using the formula, Z value = (X - µ) /
Where, X = Standardized Random Variable, µ = Sample Mean, = Sample Standard Deviation.
Z = (12.34303 - 15.2 ) / 2.78
Z = -2.85697 / 2.78
Z = -1.0276870503597
Z = (18.05697 - 15.2 ) / 2.78
Z = 2.85697 / 2.78
Z = 1.0276870503597
So, -1.027687 < z < 1.027687
0.151975 < p < 0.152045
Interpret results. Since the P-value is greater than the significance level (0.01), we cannot reject the null hypothesis. That is the population average HC for this patient is higher than 14.
3).
State the hypotheses.
Null hypothesis: = 16.4, storm is same as the average peak wave height of 16.4
Alternative hypothesis: 16.4, storm is different from the average peak wave height of 16.4
Using the formula,
Z value = (X - µ) /
Where, X = Standardized Random Variable, µ = Sample Mean, = Sample Standard Deviation.
z = (17 - 16.4) / 3.5 = 0.1714
The corresponding P-Value is 0.431955
No, we don't have sufficient evidence to support the claim.
Interpret results. Since the P-value (0.431955) is greater than the significance level (0.01), we cannot reject the null hypothesis. That is storm is same as the average peak wave height of 16.4
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