Question

You measure 42 dogs' weights, and find they have a mean weight of 60 ounces. Assume the population standard deviation is 3.7 ounces. Based on this, determine the point estimate and margin of error for a 95% confidence interval for the true population mean dog weight.

Give your answers as decimals, to two places
_±_ ounces

In a survey, 32 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $41 and standard deviation of $9. Construct a confidence interval at a 95% confidence level.

Give your answers to one decimal place.

_ ±_

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 60

Population standard deviation = = 3.7

Sample size = n = 42

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

Margin of error = E = Z_/2* ( /n)

= 1.96 * (3.7 / 42)

= 1.12

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

- E < < + E

60 - 1.12 < < 60 + 1.12

58.88 < < 61.12

(58.88 , 61.12 ) ounces

Point estimate = sample mean = = $41

Population standard deviation = = $9

Sample size = n = 32

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

Margin of error = E = Z_/2* ( /n)

= 1.96 * (9 / 32)

= 3.1

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

- E < < + E

41 - 3.1 < < 41 + 3.1

37.9 < < 44.1

(37.9 , 44.1 )