Question

A study takes a simple random sample from a population of full-term infants. The standard deviation of birth weights in this population is 2 pounds. Calculate 95% confidence intervals for the mean (u) for samples in which: Question 21. n=81 and x = 6.1 pounds A. (5.7 to 6.5) B. (6.3 to 7.7) C. (3.6 to 6.8) D. (4.5 to 7.1) Question 22. n= 36 and x = 7.0 pounds A. (5.7 to 6.5) B. (6.3 to 7.7) C. (3.6 to 6.8) D. (4.5 to 7.1)

Answer 1

Given:

$\sigma$=2

Ans21 :

n=81  $\bar x$=6.1

Z$\alpha$/2 at 95% CI is = 1.96

Margin of error = ME = Z$\alpha$/2* ($\sigma$ /$\sqrt$n)

= 1.96 * (2 / $\sqrt$ 81)

= 0.4

At 95% confidence interval estimate of the population mean is,

$\bar x$ - ME < $\mu$ < $\bar x$ + <ME

6.1 - 0.4 < $\mu$ < 6.1 + 0.4

5.7 < $\mu$ < 6.5

(5.7 , 6.5)

Option A is correct

Ans22:

n=36  $\bar x$=7

Margin of error = ME = Z$\alpha$/2* ($\sigma$ /$\sqrt$n)

= 1.96 * (2 / $\sqrt$ 36)

= 0.7

At 95% confidence interval estimate of the population mean is,

$\bar x$ - ME < $\mu$ < $\bar x$ +ME

7.0 - 0.7 < $\mu$ < 7.0 + 0.7

6.3 < $\mu$ < 7.7

(6.3 , 7.7)

#Option B is correct

Ans23

n=9 $\bar x$ =5.8

Margin of error = ME = Z$\alpha$/2* ($\sigma$ /$\sqrt$n)

= 1.96 * (2 / $\sqrt$ 9)

= 1.3

At 95% confidence interval estimate of the population mean is,

$\bar x$ - ME < $\mu$ < $\bar x$ +M E

5.8 - 1.3 < $\mu$ < 5.8 + 1.3

4.5 < $\mu$ < 7.1

(4.5 , 7.1)

Option D is correct

Ans24

n=9 $\bar x$ =5.8

Z$\alpha$/2 at 90% CI is = 1.645

Margin of error = ME = Z$\alpha$/2* ($\sigma$ /$\sqrt$n)

= 1.645 * (2 / $\sqrt$ 9)

= 1.1

At 95% confidence interval estimate of the population mean is,

$\bar x$ - ME < $\mu$ < $\bar x$ +M E

5.8 - 1.1 < $\mu$ < 5.8 + 1.1

4.7< $\mu$ < 6.9

(4.70 , 6.90)

Option C is correct