##### Question

In: Computer Science

# Construct a 95% confidence interval for an experiment that found the sample mean temperature for a...

1. Construct a 95% confidence interval for an experiment that found the sample mean temperature for a certain city in August was 101.82, with a population standard deviation of 1.2. There were 6 samples in this experiment.
1. A group of 10, foot - surgery patients had a mean weight of 240 pounds. The sample standard deviation was 25 pounds. Find a confidence interval for a sample for the true mean weight of all foot – surgery patients. Find a 95% Confidence Interval.
1. 510 people applied to the Bachelor’s in Elementary Education program at Florida State College. Of those applicants, 57 were men. Find the 90% Confidence Interval of the true proportion of men who applied to the program.

1. A study revealed that 65% of men surveyed supported the war in Afghanistan and 33% of women supported the war. If 100 men and 75 women were surveyed, find the 90% confidence interval for the data’s true difference in proportions.

## Solutions

##### Expert Solution

1)

Confidence interval(CI) for normal distribution is (-z*/, +z*/)

here =101.82

n=6, =1.2

and Z value will be 1.96 which corresponds to (100-95)/2 %= 2.5%

z*/= 0.96

Therefore CI will be (101.82-0.96,101.82+0.96)= (100.86,102.78)

2)

Here we will use a t-distribution because n<30 and population mean is given as well .

Confidence interval(CI) for t-distribution is (-t*s/, +t*s/)

here =240

n=10, s=25

df=n-1 =10-1=9

By looking up in t-distribution table we find the value for df=9 and 2.5% value we get t=2.262

t*s/= 17.88

Therefore CI will be (240-17.88,240+17.88)= (222.12,257.88)

3)

CI for population proportion is( , )

n=510

Here alpha=1-0.9 =0.1

Z value for alpha/2 (.1/2=.05) is 0.13 ( By looking in the z table)

=57/510=0.112

1-=1-.112=0.888

=0.00182

Therefore CI will be (0.112-0.00182,0.112+0.00182)=(0.11018,0.11382)

4)
CI for difference in proportions is

( , )

n1=100,n2=75

=0.65

=0.33

= 1-0.65=0.35

=1-.33=0.67

Since it is the same 90% confidence interval as in previous sum z value will be 0.13

=0.0098

=0.65-0.33=0.32

Therefore CI will be (0.32-0.0098,0.32+0.0098)=(31.9902,32.0098)

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