Question

1-We wish to estimate what percent of adult residents in a certain county are parents. Out of 200 adult residents sampled, 76 had kids. Based on this, construct a 95% 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- If n=16n=16, ¯x=48x¯=48, and s=20s=20, construct a confidence interval at a 90% confidence level. Assume the data came from a normally distributed population.

Give your answers to one decimal place.

< μ <

3- You measure 30 turtles' weights, and find they have a mean weight of 31 ounces. Assume the population standard deviation is 9.1 ounces. Based on this, determine the point estimate and margin of error for a 95% confidence interval for the true population mean turtle weight.

Give your answers as decimals, to two places

μ=± ounces

4-

You intend to estimate a population mean μμ from the following sample.

84.1	80	70.1	83.4
82.5	100	88.8	68.2
69.7	80.3	66.6	74.2
94.3	86	73.1	98.6
74.6	88.2	83.7	78.1
66.7	71.8	78.1	79.1
83.5	84.2	74.8	76.6
96.9	83

Find the 99.9% confidence interval. Enter your answer accurate to two decimal place (because the sample data are reported accurate to one decimal place).

<μ<

Answer 1

1-We wish to estimate what percent of adult residents in a certain county are parents. Out of 200 adult residents sampled, 76 had kids. Based on this, construct a 95% 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.

We have n > 30 that is it large

x: No. of residents having kids

x= 76 n = 200

Sample proportion = 0.38 ( x / n)

We will approximate it to normal distribution.

95% confidence interval for population proportion

Where the alpha = 1- 95% = 0.05

Using the normal distribution percentage tables

= 1.96

The interval is

2- If n=16n=16, ¯x=48x¯=48, and s=20s=20, construct a confidence interval at a 90% confidence level. Assume the data came from a normally distributed population.

We do not have the population SD so we will use the t-dist

90% confidence interval for population mean

Where alpha = 1 - 90% = 0.10

Critical value at 0.10

= 1.753 ...................t-dist tables p = 0.05, df = 15

90% interval

3- You measure n =30 turtles' weights, and find they have a mean weight of = 31ounces. Assume the population standard deviation is = 9.1 ounces. Based on this, determine the point estimate and margin of error for a 95% confidence interval for the true population mean turtle weight.

We have the population SD so we will use normal distribution to calculate the critical values

95% confidence interval for population mean

Where alpha = 1 - 95% = 0.05

Critical value at 0.05

= 1.96 ...................normal percentage tables with p = 0.025

95% interval

4-

You intend to estimate a population mean μμ from the following sample.

84.1	80	70.1	83.4
82.5	100	88.8	68.2
69.7	80.3	66.6	74.2
94.3	86	73.1	98.6
74.6	88.2	83.7	78.1
66.7	71.8	78.1	79.1
83.5	84.2	74.8	76.6
96.9	83

Here again we will use the t-dist since we do not have population SD

n = 30  
Find the 99.9% confidence interval. Enter your answer accurate to two decimal place (because the sample data are reported accurate to one decimal place).

99.9% confidence interval for population mean

Where alpha = 1 - 99.9% = 0.001

Critical value at 0.001

= 3.659 ...................t-dist tables p = 0.0005, df = 29

99.9% interval