The mean and standard deviation of a random sample of n measurements are equal to and , respectively. a. Find a % confidence interval for if n. b. Find a % confidence interval for if n. c. Find the widths of the confidence intervals found in parts a and b. What is the effect on the width of a confidence interval of quadrupling the sample size while holding the confidence coefficient fixed?

Complete Question

The complete question is shown on the first uploaded image

a

b

c

Generally the width at  n =  49 is mathematically represented as

Generally the width at  n =  196 is mathematically represented as

d

The correct option is E

Step-by-step explanation:

From the question we are told  that

   The sample mean is  

    The standard deviation is  

Generally given that the confidence level is 95% then the level of significance is  

=>    

Generally from the normal distribution table the critical value  of   is  

Considering question a

From the question  n  =  49

Generally the margin of error is mathematically represented as  

=>  

=>  

Generally 95% confidence interval is mathematically represented as  

=>    

Considering question b

From the question  n  =  196

Generally the margin of error is mathematically represented as  

=>  

=>  

Generally 95% confidence interval is mathematically represented as  

=>  

Considering question c

Generally the width at  n =  49 is mathematically represented as

Generally the width at  n =  196 is mathematically represented as

Now when the sample size is quadrupled i.e from n = 49 to  n =  196  

The width of the  confidence interval  decrease by 2 from 1.904 to  0.9374  

c

step-by-step explanation:

step-by-step explanation: so we look at the formula un = a + (n-1)d

un is the amount of seats

n is the nth term aka row number

a is the original sequence/row number

d is the difference between each term

so in this a=22 d=6 and n can remain as n because we are simply formjng a rule

lets put these in the equation

un=22+(n-1)(6)

this is your answer to part a

however

lets double check that this is correct, we'll try putting it into the second term

un=a+(n-1)d

u2=22+(2-1)(6)

u2=22+6

u2=28

as the question said the second row has 28 seats so now we know the formula we made is correct.

as for part b

un=a+(n-1)

its asking us to find a row with 100 seats.

remember un was the amount of seata so all we do is substitute this in.

un=100 a=22 d=6

un=a+(n-1)d

100=22+(n-1)(6)

so now we simplify to find n which is the row number.

100-22=22+(n-1)(6)-22

78=(n-1)(6)

78/6=(n-1)(6)/6

13=n-1

13+1=n-1+1

14=n

so now we know that in the 14th row there is 100 seats.

hope this and that you understood! its my first time using the app or anyone out on here : p its also me to revise so a bunch! ask more questions if you dont understand ☺