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a_sh-v [17]
3 years ago
12

A simple random sample of 100 8th graders at a large suburban middle school indicated that 81% of them are involved with some ty

pe of after school activity. Find the 98% confidence interval that estimates the proportion of them that are involved in an after school activity.
Mathematics
1 answer:
Sophie [7]3 years ago
4 0

Answer:

The  interval is  0.7187  < p < 2.421

Step-by-step explanation:

From the question we are told that

      The  sample size is  n  = 100

       The  population  proportion is p  =  0.81

       The  confidence level is  C =  98%

The level of significance is mathematically evaluated as

     \alpha  =  100 -98

    \alpha  =  2%%

    \alpha  =  0.02

Here this level of significance represented the left and the right tail

The degree of  freedom is evaluated as

     df =  n-1

substituting value  

    df =  100 - 1

     df = 99

Since we require the critical value of one tail in order to evaluate the  98% confidence interval that estimates the proportion of them that are involved in an after school activity. we will divide the level of significance by 2

The  critical value of  \frac{\alpha}{2} and the evaluated degree of freedom is  

      t_{df , \alpha } =  t_{99 , \frac{0.02}{2}  }  = 2.33

this is obtained from the critical value table  

The standard error is mathematically evaluated as

             SE =  \sqrt{\frac{p(1-p )}{n} }  

substituting value  

           SE =  \sqrt{\frac{0.81(1-0.81 )}{100} }  

           SE = 0.0392  

The 98%  confidence interval is evaluated as

      p  - t_{df ,  \frac{\alpha }{2} } *  SE  < p <  p  + t_{df ,  \frac{\alpha }{2} }

substituting value  

     0.81  - 2.33  *  0.0392  < p <  0.81  +2.33 *  0.0392

      0.7187  < p < 2.421

     

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