Linear Regression / Confidence Bounds for Prediction Worksheet

**Page 1**

1.

In a market survey it was found that 72 of 198 people in a city use brand A soap. Find $\stackrel{\u02c6}{p}$ and $\stackrel{\u02c6}{q}$, where $\stackrel{\u02c6}{p}$ 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 |

Proportion of people that use brand A soap,

So,

Correct answer : (3)

2.

Select the incorrect statement(s).

I. If$n$ = 75 and $x$ = 40, then $\stackrel{\u02c6}{p}$ = 0.53 and $\stackrel{\u02c6}{q}$ = 0.47

II. If$n$ = 200 and $x$ = 60, then $\stackrel{\u02c6}{p}$ = 0.3 and $\stackrel{\u02c6}{q}$ = 0.7

III. If$n$ = 95 and $x$ = 35, then $\stackrel{\u02c6}{p}$ = 0.27 and $\stackrel{\u02c6}{q}$ = 0.73

I. If

II. If

III. If

a. | III only | ||

b. | I only | ||

c. | II and III only | ||

d. | I, II and III |

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

So, only statement III is incorrect.

Correct answer : (1)

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% |

Sample proportion,

For a 95% confidence interval, α = 1 - 0.95 = 0.05 and

[Use the standard normal distribution table.]

The 95% confidence interval for the true proportion is:

0.15 - 1.96

0.15 - 0.040 <

0.110 <

11% <

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

Correct answer : (1)

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% |

Sample proportion,

For 95% confidence interval, α = 1 - 0.95 = 0.05 and

[Use the standard normal distribution table.]

The 95% confidence interval for the true proportion is :

0.244 - 1.96

0.244 - 0.067 <

0.177 <

17.7% <

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%.

Correct answer : (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 |

[Use the standard normal distribution table.]

The maximum error of estimation allowed,

Minimum sample size,

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

Correct answer : (3)

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 |

[Use the standard normal distribution table.]

The maximum error of estimation allowed,

Sample size,

[Minimum sample size.]

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

Correct answer : (1)

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% |

Sample proportion,

For 95% confidence interval, α = 1 - 0.95 = 0.05 and

[Use the standard normal distribuiton table.]

The 95% confidence interval for the true proportion is:

[Formula.]

0.49 - 1.96

0.49 - 0.038 <

0.452 <

45.2% <

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

Correct answer : (1)

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% |

Sample proportion,

For 90% confidence interval, α = 1 - 0.9 = 0.1 and

[Use the standard normal distribution table.]

The 90% confidence interval for the true proportion is:

0.72 - 1.65

0.72 - 0.0165 <

0.7035 <

70.35% <

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%.

Correct answer : (4)

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% |

Sample proportion,

For 99% confidence interval, α = 1 - 0.99 = 0.01 and

[Use the standard normal distribution table.]

The 99% confidence interval for the true proportion is:

[Formula.]

0.65 - 2.58

0.65 - 0.112 <

0.538 <

53.8% <

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

Correct answer : (3)

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% |

Sample proportion,

For 90% confidence interval, α = 1 - 0.9 = 0.1 and

[Use the standard normal distribution table.]

The 90% confidence interval for the true proportion is:

0.58 - 1.65

0.58 - 0.086 <

0.494 <

49.4% <

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

Correct answer : (1)