Question

11.You measure the weight of 37 turtles, and find they have a mean weight of 52 ounces. Assume the population standard deviation is 10.9 ounces. Based on this, what is the maximal margin of error associated with a 92% confidence interval for the true population mean turtle weight.
Give your answer as a decimal, to two places

12.In a survey, 18 people were asked how much they spent on their child's last birthday gift. The results were roughly shaped as a normal curve with a mean of $45 and standard deviation of $8. Find the margin of error at a 80% confidence level.

13. In a survey, 20 people were asked how much they spent on their child's last birthday gift. The results were roughly shaped as a normal curve with a mean of $34 and standard deviation of $12. Construct a confidence interval at a 80% confidence level.

Give your answers to one decimal place.

.................. ± ..............

Answer 1

11) Solution

Given,

= 52 ....... Sample mean

   = 10.9 ........Sample standard deviation

n = 37 ....... Sample size

Note that, Population standard deviation() is known..So we use z distribution.    c = 92% = 0.92

= 1- c = 1- 0.92 = 0.08

  /2 = 0.08 2 = 0.04 and 1- /2 = 0.960

Search the probability 0.960 in the Z table and see corresponding z value

= 1.751 for the 92% Confidence interval

The margin of error is given by

E =    *

= 1.751 * (10.9/37)

E = 3.14 is the maximal margin of error.

12)solution

Given,

n = 18 (sample size)

= 45 (sample mean)

s = 8 (sample SD)

c = 80% = 0.80 (confidence level)

   = 1- c = 1- 0.80 = 0.20

  /2 = 0.10

Since the population SD is unknown , we use t distribution.

d.f. = n - 1= 18 -1 = 17

     =    =    = 1.333 (using t table)

The margin of the error is

E = *

= 1.333 * (8/18)

= 2.51 is the margin of error.

13)Solution:

Given,

n = 20 (sample size)

= 34 (sample mean)

s = 12    (sample SD)

c = 80% = 0.80 (confidence level)

   = 1- c = 1- 0.80 = 0.20

  /2 = 0.10

Since the population SD is unknown , we use t distribution.

d.f. = n - 1 = 20 -1 = 19

     =    =    = 1.328   (using t table)

The margin of the error is

E = *

= 1.328 * (12/20)

= 3.6

Now confidence interval is

     margin of error

i.e 34 3.56