Question

1) We wish to estimate what percent of adult residents in a certain county are parents. Out of 600 adult residents sampled, 270 had kids. Based on this, construct a 90% confidence interval for the proportion p of adult residents who are parents in this county. Provide the point estimate and margin of error. Give your answers as decimals, to three places.

P= __  ± __

2) You measure 36 textbooks' weights, and find they have a mean weight of 57 ounces. Assume the population standard deviation is 5.6 ounces. Based on this, construct a 99% confidence interval for the true population mean textbook weight.
Give your answers as decimals, to two places

__  < μ < __

3) You measure 49 dogs' weights, and find they have a mean weight of 53 ounces. Assume the population standard deviation is 9.2 ounces. Based on this, determine the point estimate and margin of error for a 90% confidence interval for the true population mean dog weight.
Give your answers as decimals, to two places.

μ= __ ± __ ounces

Answer 1

Solution :

1) Given that,

n = 600

x = 270

Point estimate = sample proportion = = x / n = 270 / 600= 0.45

1 - = 1 - 0.45 = 0.55

At 90% confidence level

= 1 - 90%

=1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z_0.05 = 1.645

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.645 (((0.45 * 0.55) / 600 )

= 0.033

A 90% confidence interval for population proportion p is ,

± E  

= 0.45  ± 0.033

= ( 0.417, 0.483 )

2) Given that,

Point estimate = sample mean = = 57 ounces

Population standard deviation =    = 5.6 ounces

Sample size = n = 36

At 99% confidence level

= 1 - 99%  

= 1 - 0.99 =0.01

/2 = 0.005

Z/2 = Z_0.005 = 2.576

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

= 2.576 * ( 5.6 /  36 )

= 2.40

At 99% confidence interval estimate of the population mean is,

- E < < + E

57 - 2.40 <   < 57 + 2.40

( 54.60 <   < 59.40 )

3) Given that,

Point estimate = sample mean = = 53 ounces

Population standard deviation =    = 9.2 ounces

Sample size = n = 49

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z_0.05 = 1.645

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

= 1.645 * ( 9.2 /  49 )

= 2.16

At 90% confidence interval estimate of the population mean is,

  ± E

53 ± 2.16   

( 50.84, 55.16 )