3. When 14 different second-year medical students at Bellevue Hospital measured

Question

3. When 14 different second-year medical students at Bellevue Hospital measured the systolic blood pressure of the same person, they obtained the results listed below (in mmHg). Assuming the distribution follows a normal curve, use a 0.05 significance level to test the claim that the mean blood pressure level is less than 140 mmHg. What is your conclusion? If hypertension is defined to be a blood pressure greater than or equal to 140 mmHg, can it be safely concluded that the person does not have hypertension? 138 130 135 140 120 125 120 130 130 144 143 140 130 150

Explanation / Answer

Solution:

Firstly, you should find the mean of the values to get the sample mean, X-bar.
X-bar = 133.93

Sample size n = 14
Population standard deviation = 10

To conduct the hypothesis test:

State your hypotheses:
H0: = 140
H1: < 140 ...... H1 can also be denoted as HA (with A as a subscript), depending on your textbook

If blood pressure levels X are normally distributed, then the sampling distribution of the mean X-bar will also be normally distributed.

The significance level is 0.05, so = 0.05.
Since the population standard deviation is known, we can use the z-distribution.
Since this is a one-tailed test, with in the left tail, the critical value is -z0.05 = -1.645.

Decision rule:
If based on test-statistic: Reject H0 if z < -z0.05 = -1.645
If based on p-value: Reject H0 if p < = 0.05

Calculate the test-statistic:
z = ( X-bar - ) / ( / n ) ......... ( / n ) is the standard error, or standard deviation of the sampling distribution of the mean
is evaluated under the null hypothesis H0: = 140
z = ( 133.93 - 140 ) / ( 10 / 14 ) = -2.27
Based on the calculated test-statistic, z = -2.27 < -1.645 so reject H0.

To calculate the p-value, find the probability corresponding to your test-statistic, using z tables or software such as Excel.
P(z < -2.27) = 0.0116
If p = 0.0116 < 0.05, so reject H0.

There is sufficient evidence to suggest that the mean is less than 140, at a 5% level of significance.

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