Question 1

In a random sample of 500 people aged 20-24,22% were smokers.In a random sample of 450 people aged 25-29,14% were smokers.Construct a 95% confidence interval for the difference in smoking rates for the two groups.

Accepted Answer

D

Question 2

A university wishes to compare the proportions of female students in the science and social science programs.A random sample of 120 science students contained 43 females.A random sample of 200 social science students contained 132 females.Construct a 99% confidence interval for the difference in the proportions of female students in the science and social science programs at this university.

Accepted Answer

The confidence interval for the difference in proportions is calculated using the formula for the difference between two independent proportions. The sample proportions are 43/120 for science and 132/200 for social science. Using these, the 99% confidence interval is approximately (-0.44, -0.16).

Question 3

A survey of randomly chosen adults found that 30 of the 64 women and 49 of the 74 men follow regular exercise programs.Construct a 95% confidence interval for the difference in the proportions of women and men who have regular exercise programs.

Accepted Answer

The confidence interval for the difference between two proportions is calculated using the formula for the standard error of the difference between two proportions and the Z-score for the desired confidence level (in this case, 95%, which corresponds to a Z-score of approximately 1.96). The formula for the confidence interval is: $CI = (p_1 - p_2) \pm Z \cdot \sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}$, where $p_1$ and $p_2$ are the sample proportions, and $n_1$ and $n_2$ are the sample sizes for the two groups. Plugging in the given numbers, the confidence interval calculation leads to the interval (-0.356, -0.030), indicating that the difference in proportions of women and men who have regular exercise programs falls within this range with 95% confidence.

Question 4

Suppose the proportion of women who watch CBC News is $\mathrm { P } _ { \mathrm { W } }$ and the proportion of men who watch CBC News is $\mathrm { Pm }$ .The survey found a 95% confidence interval for $p _ { W } - p _ { m }$ is $( - 0.06,0.25 )$ Give an interpretation of this confidence interval.

Accepted Answer

The confidence interval for $p _ { W } - p _ { m }$ from $(-0.06, 0.25)$ suggests that we are 95% confident the true difference in proportions (women's proportion minus men's proportion) falls within this range, indicating women's viewership could be from 6% lower to 25% higher than men's.

Question 5

An Ontario study examined the effect of day care on behavior in toddlers.Randomly selected parents who had a toddler in full-time day care were asked if their child had behavioral problems.The researchers found that among 987 parents surveyed,212 said their child had behavioral problems.Among 349 randomly selected parents with a toddler at home,17 reported that their child had behavioral problems.

Accepted Answer

The answer of An Ontario study examined the effect of...

Question 6

A marketing survey involves product recognition in Ontario and British Columbia.Of 558 Ontario residents surveyed,193 knew the product while 196 out of 614 British Columbia residents knew the product.Construct a 99% confidence interval for the difference in the proportions of Ontario and British Columbia residents who knew the product.

Accepted Answer

The answer of A marketing survey involves product recognition in...

Question 7

A study was conducted to determine if patients recovering from knee surgery should receive physical therapy two or three times per week.Suppose $\text { P3 }$ represents the proportion of patients who showed improvement after one month of therapy three times a week and $\mathrm { P } 2$ represents the proportion of patients who showed improvement after one month of therapy twice a week.A 90% confidence interval for $\text { p3 - p2 }$ is $( 0.15,0.39 )$ Give an interpretation of this confidence interval.

Accepted Answer

The confidence interval (0.15, 0.39) for $p3 - p2$ suggests that we are 90% confident that the true difference in proportions (improvement with therapy three times a week minus improvement with therapy twice a week) lies between 0.15 and 0.39, indicating an improvement with more frequent therapy sessions.

Question 8

A university found it retained 25 students out of 352 in 2012 and 26 students out of 334 in 2013.Construct a 90% confidence interval for the difference in the proportions of students retained in 2012 and 2013.

Accepted Answer

The answer of A university found it retained 25 students...

Question 9

A new manager,hired at a large warehouse,was told to reduce the employee sick leave.The manager introduced a new incentive program for employees with perfect attendance.The manager later compares the sick leave rate prior to the program starting (65 out of 250 employees were sick)to the sick leave rate one month into the program (60 out of 253 employees were sick).Construct a 98% confidence interval for the difference in the proportions of sick employees prior to the program and one month into the program.

Accepted Answer

We can use the two-proportion z-test formula to calculate the confidence interval.

p1 = 65/250 = 0.26 (proportion of sick employees prior to the program)
p2 = 60/253 = 0.24 (proportion of sick employees one month into the program)
n1 = 250
n2 = 253

pooled proportion:
p = (x1 + x2) / (n1 + n2) = (65 + 60) / (250 + 253) = 0.253

standard error:
SE = sqrt(p*(1-p)*(1/n1 + 1/n2)) = sqrt(0.253 * 0.747 * (1/250 + 1/253)) = 0.034

z-score (for 98% confidence interval):
z = 2.33

margin of error:
ME = z * SE = 2.33 * 0.034 = 0.079

confidence interval:
(p1 - p2) ± ME = (0.26 - 0.24) ± 0.079 = 0.02 ± 0.079

Therefore, the 98% confidence interval for the difference in proportions is (-0.0668, 0.1125). The best choice is D.

Question 10

Suppose the proportion of first year students at a particular university who purchased used textbooks in the past year is $\mathrm { P } 1$ and the proportion of second year students at the university who purchased used textbooks in the past year is p2.A study found a 95% confidence interval for p1 - p2 is $( 0.232,0.421 )$ Give an interpretation of this confidence interval.

Accepted Answer

The confidence interval suggests that the proportion of used textbook purchases for first-year students could be up to 42.1% higher than the proportion for second-year students, or up to 23.2% lower. Since the confidence interval includes negative values, this means that it's also possible that second-year students are purchasing more used textbooks than first-year students. Therefore, option B is the best choice as it represents the interpretation of the confidence interval in the correct direction.

Question 11

In Canada,a person with a body mass index (BMI)of 30 more is considered to be "obese." A random sample of 100 Canadian men revealed 26 to be obese.A random sample of 100 Canadian women revealed 21 to be obese.Construct a 90% confidence interval for the difference in the proportions of obese Canadian men and women.

Accepted Answer

The answer of In Canada,a person with a body mass...

Question 12

In a random sample of 300 Canadian women,68% watch CBC News.In a random sample of 200 Canadian men,56% watch CBC News.Construct a 98% confidence interval for the difference in the proportions of women and men who watch CBC News.

Accepted Answer

We can use the two sample z-test for the difference in proportions:
z = (0.68 - 0.56) / sqrt[(0.68 * 0.32 / 300) + (0.56 * 0.44 / 200)] = 3.06
Using a standard normal distribution table or calculator, the p-value is approximately 0.002, which is less than 0.02 (1 - 0.98). Therefore, we reject the null hypothesis and conclude that there is a significant difference in the proportions of women and men who watch CBC News.
The 98% confidence interval can be calculated as:
(0.68 - 0.56) ± 2.33 * sqrt[(0.68 * 0.32 / 300) + (0.56 * 0.44 / 200)]
= 0.12 ± 0.053
= (0.067, 0.173)
The best choice is (C) (0.017, 0.223), which is the closest range to (0.067, 0.173) among the options given.

Question 13

A survey of randomly selected university students found that 83 of the 110 first year students and 87 of the 120 second year students surveyed had purchased used textbooks in the past year.Construct a 98% confidence interval for the difference in the proportions of first and second year university students who purchased used textbooks.

Accepted Answer

To construct a 98% confidence interval for the difference in proportions, we first calculate the sample proportions: $p_1 = \frac{83}{110}$ for first year students and $p_2 = \frac{87}{120}$ for second year students. Then, we use the formula for the confidence interval for two proportions, which includes the critical value for the desired confidence level (which for 98% confidence is approximately 2.33, using the Z-distribution), the sample proportions, and the sample sizes. The calculation involves finding the standard error of the difference in proportions and then applying the critical value to find the margin of error. The correct interval, given the calculations, is (-0.105, 0.164), which suggests that the true difference in proportions could range from -10.5% to 16.4%, indicating that there is no significant difference between the proportions of first and second year students who purchased used textbooks.

Question 14

A researcher wished to test the claim that the rate of defectives among the computers of two different manufacturers is the same.She selected two independent random samples and found that 1.5% of 400 computers from manufacturer A were defective and 3.5% of 200 computers from manufacturer B were defective.

Accepted Answer

The answer of A researcher wished to test the claim...

Question 15

The weight loss franchise home office wanted to compare the results of two centres.A centre in the city provided a weight loss for 152 out of 200 participants.A centre in the suburb provided a weight loss for 109 out of 140 participants.Construct a 98% confidence interval for the difference in the proportions of participants in the city weight loss centre and suburban weight loss centre.

Accepted Answer

The correct confidence interval can be calculated using the formula for the difference in proportions. Given the data, the proportions for the city and suburb centers are 152/200 and 109/140, respectively. Using a 98% confidence level requires a Z-score that corresponds to the confidence level (approximately 2.33 for 98%). The calculation involves the sample proportions, sample sizes, and the Z-score, leading to the interval (-0.1263, 0.0891). This interval suggests that, with 98% confidence, the difference in success rates between the two centers falls within this range.

Question 16

In words,how would you find the standard deviation of the difference between two independent random variables?

Accepted Answer

The formula for the standard deviation of the difference between two independent random variables is the square root of the sum of their variances. This can be written as σdiff = √(σX^2 + σY^2), where σX and σY are the standard deviations of the two random variables X and Y, respectively.

Question 17

A poll of randomly selected Canadians between the ages of 20 and 29 reports that 35 of 410 men and 59 of 398 women suffered from insomnia at least once a week during the past year.

Accepted Answer

The answer of A poll of randomly selected Canadians between...

Question 18

Suppose the proportion of women who follow a regular exercise program is $\mathrm { P } _ { \mathrm { W } }$ and the proportion of men who follow a regular exercise program is $\mathrm { Pm }$ .A study found a 98% confidence interval for $p _ { w } - p _ { m }$ is ( $- 0.028,0.114 )$. Give an interpretation of this confidence interval.

Accepted Answer

The confidence interval for $p_w - p_m$ being (-0.028, 0.114) means we are 98% confident that the true difference in proportions (women minus men) falls within this range, indicating the proportion of women following a regular exercise program could be up to 2.8% less or up to 11.4% more than that of men.

Question 19

A researcher wishes to determine whether the proportion of Canadian women who smoke differs from the proportion of Canadian men who smoke.He wants to test the hypothesis $\mathrm { H } _ { 0 } : \mathrm { p } _ { 1 } = \mathrm { p } _ { 2 }$ where $\mathrm { p } _ { 1 }$ represents the proportion of Canadian women who smoke and $\mathrm { P } _ { 2 }$ represents the proportion of Canadian men who smoke.He randomly selects 100 married couples.Among the 100 women in the sample are 21 smokers.Among the 100 men are 29 smokers.Are the assumptions for a two-sample z-test for two population proportions met? If not,which assumption is violated and why?

Accepted Answer

The answer of A researcher wishes to determine whether the...

Question 20

A marketing survey involves product recognition in Ontario and British Columbia.Suppose the proportion of Ontario residents who recognized a product is $\mathrm { P } 1$ and the proportion of British Columbia residents who recognized the product is $\mathrm { P } 2$ .The survey found a 98% confidence interval for $p _ { 1 } - p _ { 2 }$ is $( - 0.023 , - 0.019 )$ Give an interpretation of this confidence interval.

Accepted Answer

The confidence interval tells us that we can be 98% confident that the true proportion of British Columbia residents who recognized the product is between 1.9% and 2.3% higher than the proportion of Ontario residents who recognized the product. This interpretation is consistent with the given information about the proportions in each province. Options A, B, and E are incorrect because they make claims that are not supported by the information given or the confidence interval provided. Option D is also incorrect because it makes a broad and unsupported claim about all possible samples, rather than the specific confidence interval obtained from this survey.

Quiz 19: Comparing Two Proportions