# Linear Regression / Confidence Bounds for Prediction Worksheet

Linear Regression / Confidence Bounds for Prediction Worksheet
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1.
In a market survey it was found that 72 of 198 people in a city use brand A soap. Find $\stackrel{ˆ}{p}$ and $\stackrel{ˆ}{q}$, where $\stackrel{ˆ}{p}$ is the proportion of people that use brand A soap.
 a. 0.72 and 1.26 b. 2.75 and 1.57 c. 0.36 and 0.64 d. 0.64 and 0.36

#### Solution:

Number of people use brand A soap, x = 72 and n = 198

Proportion of people that use brand A soap, pˆ = xn = 72198 = 0.364 or 36.4 %

qˆ = n - xn = 198 - 72198 = 126198 = 0.636 or 63.6 %

So, pˆ = 0.36 and qˆ = 0.64

2.
Select the incorrect statement(s).
I. If $n$ = 75 and $x$ = 40, then $\stackrel{ˆ}{p}$ = 0.53 and $\stackrel{ˆ}{q}$ = 0.47
II. If $n$ = 200 and $x$ = 60, then $\stackrel{ˆ}{p}$ = 0.3 and $\stackrel{ˆ}{q}$ = 0.7
III. If $n$ = 95 and $x$ = 35, then $\stackrel{ˆ}{p}$ = 0.27 and $\stackrel{ˆ}{q}$ = 0.73

 a. III only b. I only c. II and III only d. I, II and III

#### Solution:

n = 95 and x = 35

pˆ = xn = 3595 = 0.37

qˆ = 1 - p = 1 - 0.37 = 0.63

All the other cases were also checked in the similar manner and they were found to be correct.

So, only statement III is incorrect.

3.
A sample of 300 tickets purchased during vacation at a travelling center included 45 children's tickets. Find the 95% confidence interval for the true proportion of children, who were traveling during the vacation.
 a. 11% < $p$ <19% b. 13% < $p$ <17% c. 12.9% < $p$ <17.1% d. 15% < $p$ < 85%

#### Solution:

The problem is to find 95% confidence interval for the true proportion (p) of children, who were traveling during the vacation.

Sample proportion, pˆ = 45300 = 0.15 and qˆ = 1 - 0.15 = 0.85

For a 95% confidence interval, α = 1 - 0.95 = 0.05 and z/2 = 1.96
[Use the standard normal distribution table.]

The 95% confidence interval for the true proportion is:
pˆ - z/2 pˆqˆn < p < pˆ + z/2 pˆqˆn

0.15 - 1.96(0.15)(0.85)300 < p < 0.15 +1.96(0.15)(0.85)300

0.15 - 0.040 < p < 0.15 + 0.040

0.110 < p < 0.190

11% < p < 19%

So, we are fairly 95% confident that the true percentage of children traveling during the vacation is between 11% and 19%.

4.
A recent study of 160 people of a locality in a city found that 39 people were obese. Find the 95% confidence interval of the population proportion of individuals living in that city who are obese.
 a. 17.7% < $p$ < 31.1% b. 17% < $p$ < 31% c. 21% < $p$ < 27.8% d. 24.4% < $p$ < 75.6%

#### Solution:

The problem is to find 95% confidence inteval of the population proportion of individuals living in a city who are obese.

Sample proportion, pˆ = xn = 0.244 and qˆ = 1 - p = 0.756

For 95% confidence interval, α = 1 - 0.95 = 0.05 and z/2 is 1.96
[Use the standard normal distribution table.]

The 95% confidence interval for the true proportion is :
pˆ - z/2 pˆqˆn < p < pˆ + z/2 pˆqˆn

0.244 - 1.96(0.244)(0.756)160 < p < 0.244 + 1.96(0.244)(0.756)160

0.244 - 0.067 < p < 0.244 + 0.067

0.177 < p < 0.311

17.7% < p < 31.1%

So, we are fairly 95% confident that the proportion of individuals living in a city who are obese is between 17.7% and 31.1%.

5.
A researcher wishes to estimate, with 99% confidence the proportion of executives who have a bank account in 'ABC' bank. A previous study shows that 38% of those interviewed had a bank account in that bank. The researcher wishes to be accurate within 2% of the true proportion. Find the minimum sample size necessary.
 a. 1490 b. 31 c. 3921 d. 924

#### Solution:

For 99% confidence interval, zα/2 is 2.58.
[Use the standard normal distribution table.]

The maximum error of estimation allowed, E = 0.02

pˆ = 38% = 0.38, qˆ = 1 - pˆ = 1 - 0.38 = 0.62

Minimum sample size, n = pˆ qˆ (Zα/2E)2

n = (0.38) (0.62) (2.580.02)2 = 3920.62 3921

So, the minimum sample sample size required for being 99% confident that the proportion of executives who have a bank account in 'ABC' bank is 3921

6.
A recent study of 50 people in a city found that 70% of the people have less than average incomes for the city. How many people should be surveyed to estimate the true proportion of people having an income of less than average with a 99% confidence interval 4% wide?
 a. 874 b. 375 c. 65 d. 416

#### Solution:

For 99% confidence interval, zα/2 is 2.58.
[Use the standard normal distribution table.]

The maximum error of estimation allowed, E = 0.04

pˆ = 70% = 0.70, qˆ = 1 - pˆ = 1 - 0.70 = 0.30

Sample size, n = pˆ qˆ (Zα/2E)2
[Minimum sample size.]

n = (0.70) (0.30) (2.580.04)2 = 873.65 874

So, to be 99% confident 874 people are to be surveyed.

7.
To predict the outcome of a referendum an opinion poll is conducted. A random sample of 678 people on the electoral roll are asked, " Will you vote for $A$? " Their answers are recorded as either "Yes" or "No". If 333 people say "Yes", then find a 95% confidence interval for the true population proportion of "Yes" votes.
 a. 45.2% < $p$ < 52.8% b. 45.9% < $p$ < 52.1% c. 4.52% < $p$ < 5.28% d. 47.1% < $p$ < 50.9%

#### Solution:

The problem is to find 95% confidnce interval for the true population proportion(p) of "Yes" votes.

Sample proportion, pˆ = xn = 333678 = 0.49 and qˆ = 1 - p = 1 - 0.49 = 0.51

For 95% confidence interval, α = 1 - 0.95 = 0.05 and z/2 is 1.96
[Use the standard normal distribuiton table.]

The 95% confidence interval for the true proportion is:
pˆ - z/2 pˆqˆn < p < pˆ + z/2 pˆqˆn
[Formula.]

0.49 - 1.96(0.49)(0.51)678 < p < 0.49 + 1.96(0.49)(0.51)678

0.49 - 0.038 < p < 0.49 + 0.038

0.452 < p < 0.528

45.2% < p < 52.8%

So, we are fairly 95% confident that the true proportion of "Yes" votes lies between 45.2% and 52.8%.

8.
A survey of 2,000 people in a city found 72% of them take bread and butter as their breakfast. Find the 90% confidence interval of the true proportion of people taking bread and butter as their breakfast.
 a. 71% < $p$ < 73% b. 71.2% < $p$ < 72.8% c. 28% < $p$ < 72% d. 70.35% < $p$ < 73.65%

#### Solution:

The problem is to find 90% confidence interval for the true proportion (p) of people taking bread and butter as their breakfast.

Sample proportion, pˆ = 72% = 0.72 and qˆ = 1 - pˆ = 1 - 0.72 = 0.28.

For 90% confidence interval, α = 1 - 0.9 = 0.1 and z/2 = 1.65
[Use the standard normal distribution table.]

The 90% confidence interval for the true proportion is:
pˆ - z/2 pˆqˆn < p < pˆ + z/2 pˆqˆn

0.72 - 1.65(0.72)(0.28)2000 < p < 0.72 + 1.65(0.72)(0.28)2000

0.72 - 0.0165 < p < 0.72 - 0.0165

0.7035 < p < 0.7365

70.35% < p < 73.65%

So, we are fairly 90% confident, that the true percentage of people taking bread and butter as their breakfast is between 70.35% and 73.65%.

9.
A survey of 120 families showed that at least 78 had a computer at home. Find the 99% confidence interval of the true proportion of families who own a home computer.
 a. 64.5% < $p$ < 65.5% b. 5.38% < $p$ < 7.62% c. 53.8% < $p$ < 76.2% d. 60.6% < $p$ < 69.4%

#### Solution:

The problem is to find the true proportion of families who own a home computer

Sample proportion, pˆ = 78120 = 0.65 and qˆ = 1 - pˆ = 1 - 0.65 = 0.35

For 99% confidence interval, α = 1 - 0.99 = 0.01 and z/2 = 2.58
[Use the standard normal distribution table.]

The 99% confidence interval for the true proportion is:
pˆ - z/2 pˆqˆn < p < pˆ + z/2 pˆqˆn
[Formula.]

0.65 - 2.58(0.65)(0.35)120 < p < 0.65 - 2.58(0.65)(0.35)120

0.65 - 0.112 < p < 0.65 + 0.112

0.538 < p < 0.762

53.8% < p < 76.2%

So, we are fairly 99% confident, that the families who own a home computer is between 53.8% and 76.2%.

10.
A survey of 90 recent fatal traffic accidents showed that 52 were alcohol-related. Find the 90% confidence interval of the true proportion of alcohol-related accidents.
 a. 49.4% < $p$ < 66.6% b. 50% < $p$ < 66% c. 0.49% < $p$ < 0.66% d. 52.8% < $p$ < 63.2%

#### Solution:

The problem is to find 90% confidence interval for the true proportion of alcohol-related accidents.

Sample proportion,pˆ = 5290 = 0.58 and qˆ = 1 - pˆ = 1 - 0.58 = 0.42

For 90% confidence interval, α = 1 - 0.9 = 0.1 and z/2 is 1.65
[Use the standard normal distribution table.]

The 90% confidence interval for the true proportion is:
pˆ - z/2 pˆqˆn < p < pˆ + z/2 pˆqˆn

0.58 - 1.65(0.58)(0.42)90 < p < 0.58 + 1.65(0.58)(0.42)90

0.58 - 0.086 < p < 0.58 + 0.086

0.494 < p < 0.666

49.4% < p < 66.6%

So, we are fairly 90% confident that the alcohol-related accidents is between 49.4% and 66.6%.