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ASHA 777 [7]
3 years ago
11

A simple random sample of size n is drawn from a population that is known to be normally distributed. The sample variance, s2, i

s determined to be 19.8.
(a) Construct a 95% confidence interval for s2 if the sample size, n, is 10.
(b) Construct a 95% confidence interval for s2 if the sample size, n, is 25. How does increasing the sample size affect the width of the interval?
(c) Construct a 99% confidence interval for s2 if the sample size, n, is 10. Compare the results with those obtained in part (a). How does increasing the level of confidence affect the width of the confidence interval?
Mathematics
1 answer:
g100num [7]3 years ago
3 0

Answer:

whenever we increase level of confidence for the same sample size we find that confidence interval becomes wider.

Step-by-step explanation:

Given that a random sample of size n  is drawn from a population that is known to be normally distributed. The sample variance, s2, is determined to be 19.8

Std error of sample would be =\sqrt{\frac{s^2}{n} } =0.890 for n =25(19.8-2.065*0.890,19.8+2.065*0.890)\\=(17.962, 21.638)

Margin of error = critical value * std error

a) Here n <30.  (95%)So we use t critical value for df 9 is used for confidence interval

t critical = 2.262

95% confidence interval =(19.8-2.262*1.483,19.8+2.262*1.483)\\=(16.445, 23.155)[

b) Here n <30.  (95%) So t critical for df 24 is used.

t critical = 2.065

Confidence interval = (17.962, 21.638)

c) For 99% t critical for df 9 is used.

t critical=3.250

Std error = 1.483

confidence interval = (14.980, 24.620)

a) whenever we increase level of confidence for the same sample size we find that confidence interval becomes wider.

Similarly for the same confidence level, if sample size is increases, confidence interval becomes narrower.

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The answer is:      "  3(m − 2)  "  .

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  →  The "factorized version" of the binomial expression, " 3m − 6 " ,  is:

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                →   "  3(m − 2) " .

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Explanation:

To solve:

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<u>Note</u>:  We are given the expression:  " 3m − 6 " ;

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Note that this expression is a "binomial expression" — which means there are two (2) terms — in this case:   "3m" and "6" .

Since only one (1) of these 2 (two) terms has a variable, and the remaining terms is a "constant" (non-zero integer), write these 2 (two) terms as:

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 2)  the other term:  "6" .

→  With the numbers:  "3 and 6" , we can factor out a "3" ;  

          →  {since:  " 6 ÷ 3 = 2 "} ;

 →   So;  given:  

               " 3m − 6 " ;  

 →   We can "factor out" a "3" ; as follows:

           →   Take the first term:  " 3m " :

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           →   Take the second term:  "  6 " :

                    "  6 = 3 * (what value?) ;

                 → divide each of this equation by "3" ;

    to isolate the "missing value" on one side of the equation; & to solve for the "missing value" ; as follows:

        →           "  6 / 3 = [ 3 * (what value?) ] / 3  ;

                    →  to get:  " 2  =  "(the missing value)"  ;

    →   So;    "  6 = 3 * (what value?) ;

           →     " 6 = 3 * 2 " .  

           →   Take the second term:  "  6 " :

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So:    " 3m − 6 "  =  (3 * m) − (3 * 2) " .  

Factor out a "3" ;  as follows:

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    →  " 3m − 6 "  =  (3 * m) − (3 * 2) " ;

                           =   " 3(m − 2)  "

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Let us check our answer:

__________________________________

Note the "distributive property of multiplication"  :

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        a(b + c) = ab +  ac ;  

        a(b − c) = ab −  ac ;

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So:  Take our answer:  " 3(m − 2) " .

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" 3(m − 2)  = ?  (3*m)  −  (3*2)  " ??  ;

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 We have:

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     →  " 3(m − 2)  = ?  (3*m)  −  (3*2)  " ??  ;

First let us rewrite this equation;  substitute "4" for "m" ; as follows:

     →   " 3(4 − 2)  = ?  (3*4)  −  (3*2)  " ??  ;

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And calculate;  to see if each side of the equation is equal ; i.e. to see if the equation holds true; as follows:

     →   " 3(2)  = ?  (12)  −  (6)  " ??  ;

     →   "  6  = ?  (6)  " ??  ;

     →   "  6 = 6 " !  Yes!  The equation holds true!

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Hope this answer was helpful!

Best wishes!

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