1A) Let x represent the hemoglobin count (HC) in grams per 100 milliliters of wh
ID: 3055484 • Letter: 1
Question
1A)
Let x represent the hemoglobin count (HC) in grams per 100 milliliters of whole blood. The distribution for HC is approximately normal with ? = 14 for healthy adult women. Suppose that a female patient has taken 12 laboratory blood samples in the last year. The HC data sent to her doctor is listed below. We would like to know if the data indicates this patient has significantly high HC compared to the population.
22,19,14,19,15,16,21,22,21,14,20,20
Give the p-value and interpret the results.
a) p = .0562; Based on 5% significance level, I will fail to reject the null hypothesis and conclude this patient does not have a high HC level.
b) p = 0.0001; Based on 5% significance level, I will reject the null hypothesis and conclude this patient has a high HC level.
c) p = 0.0001; Based on 5% significance level, I will fail to reject the null hypothesis and conclude this patient does not have a high HC level.
d) p = .1053; Based on 5% significance level, I will fail to reject the null hypothesis and conclude this patient does not have a high HC level.
e) p = 0.0003; Based on 5% significance level, I will reject the null hypothesis and conclude this patient has a high HC level.
1B) In a test of significance, assuming the null hypothesis is true, the probability of observing the test statistic extreme or more extreme than the observed test statistic (in the way of the alternative hypothesis) is
a) the probability the null hypothesis is true.
b) the probability the null hypothesis is false.
c) the p-value.
d) the level of significance ?.
e) None of the above
Explanation / Answer
1A)
e) p = 0.0003; Based on 5% significance level, I will reject the null hypothesis and conclude this patient has a high HC level.
Explanantion:
To test H0: µ<=14 versus H1: µ>14(claim)
The sample mean=Xbar=18.454
Sample standard deviation=s=2.899
n=11
µ0=14
The test statistic=?n(Xbar-µ0)/s=?11*(18.454-14)/2.899 =5.096
The test statistic ~t distribution with (n-1)=10 degrees of freedom.
The p value for the test statistic=5.5428 with degrees of freedom=10 and right tailed test is=0.0003.
1B) the level of significance ?.
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