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sergij07 [2.7K]
3 years ago
10

Two hundred undergraduate students were randomly selected from a university that has 47,000 students in total. Systolic blood pr

essure was tested on the 200 students. The sample mean is 118.0 mmHg and the sample standard deviation is 11.0 mmHg. Please construct a 95% confidence interval for the population mean of students' systolic blood pressure. Which one of the following results is the closest to your answer? (Hint: use 1.96 as the critical z-value)
a. [110.5, 125.5]
b. [112.5, 123.5]
c. [114.5, 121.5]
d. [116.5, 119.5]
Mathematics
1 answer:
Hoochie [10]3 years ago
6 0

Answer:

The  correct option is d

Step-by-step explanation:

From the question we are told that

    The  population size is  N  =  47000

    The  sample size is  n  =  200

     The  sample mean is   \= x  =  118.0 \ mmHg

     The  standard deviation is  \sigma  =  11.0 \ mmHg

     

Given that the confidence level is  95% then the level of significance can be calculated as

                     \alpha  =  100 -  95

                    \alpha  =  5 \%

                      \alpha  =  0.05

Next we obtain the critical value of  \frac{\alpha }{2} from  z-table , the value is  Z_{\frac{\alpha }{2}  } =  Z_{\frac{0.05 }{2}  } = 1.96

The reason we are obtaining critical value of   \frac{\alpha }{2} instead of    \alpha is because  

 \alpha represents the area under the normal curve where the confidence level interval (   1-\alpha) did not cover which include both the left and right tail while  

 \frac{\alpha }{2}  is just the area of one tail which what we required to calculate the margin of error .

NOTE: We can also obtain the value using critical value calculator (math dot armstrong dot edu)

  Generally the margin of error is mathematically represented as

            E  =  Z_{\frac{\alpha }{2} } *\frac{\sigma }{ \sqrt{n} }

substituting values

           E  =  1.96 *\frac{11.0  }{ \sqrt{200} }

           E  = 1.5245

The  95% confidence level interval is mathematically represented as

       \= x  -  E < \mu < \= x  +  E

substituting values

       118.0   -  1.5245 < \mu < 118.0   +  1.5245

        116.5< \mu < 119.5

         [116.5 , 119.5]

 

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