Stat 2023 | Numerical analysis homework help

STAT 2023
1. 300 randomly selected light bulbs were tested in a laboratory, 170 lasted more than 500 hours.
Find a point estimate of the true proportion of all light bulbs that last more than 500 hours.
A) 45% B) 57% C) 75% D) 36%
2. In a marketing study about public opinion in favor of vanilla cereals. How many subjects must be
surveyed if you want, with a confidence level of 99%, that the estimation of the population
proportion in favor of vanilla cereals has a margin of error of 5%?
A) 821 B) 358 C) 664 D) 605
3. Use the given data to find the sample size required to estimate the population proportion p:
a) Margin of error 1%, 99% confidence level;
A) 16,580 B) 12,320 C) 13,250 D) 15,421
b) Margin of error: 1%; 95% confidence level; from a prior study: p = 27%
A) 6,520 B) 5,025 C) 7,572 D) 7,008
4. The tobacco industry closely monitors all surveys that involve smoking. One survey showed that
among 800 randomly selected subjects who completed four years of college, 430 of them smoke.
a) Find a point estimate of the population proportion of subjects who completed four years of
college who smoke.
A) 45% B) 36% C) 66% D) 54%
b) How many subjects must be surveyed if you want, with a confidence level of 99%, that the
estimation of the population proportion has a margin of error of 2%?
A) 2,520 B) 4,118 C) 5,310 D) 4,800
c) How many subjects must be surveyed if you want, with a confidence level of 99%, that the
estimation of the population proportion has a margin of error of 6%?
A) 403 B) 385 C) 458 D) 510
5. In a poll of 1000 likely voters, 560 say that the United States spends too little on fighting hunger at
home. Find a 95% confidence interval for the true proportion of voters who feel this way:
A) 0.53 < p < 0.59 B) 0.33 < p < 0.45 C) 0.39 < p < 0.64 D) 0.38 < p < 0.75
6. In a Gallup pool, 1025 randomly selected adults were surveyed and 29% of them said that they
used the Internet for shopping at least a few times a year.
Find the point estimate of the percentage of adults who use the Internet for shopping;
A) 26% B) 29% C) 33% D) 40%
7. A pilot study estimated that 40% of Americans aged 25 to 29 are unmarried. How large a sample is
necessary to estimate the true population proportion of unmarried Americans in this age group
within 2.5 percentage points with 90% confidence level?
A) 2,100 B) 1,200 C) 1,525 D) 1,039
8. A sociologist develops a test to measure attitude about public transportation. The sociologist
randomly selects 36 subjects to take the test; they got a mean score of 76.2 points. Assuming that
the population standard deviation is σ = 12.2.
a) Find a point estimate of the population mean:
A) 68.9 B) 26.8 C) 76.2 D) 12.2
b) Find the margin of error in the estimation of the true population mean, using 95% confidence
level:
A) 3.99 B) 2.89 C) 5.12 D)22.5
c) The 95% confidence interval to estimate the population mean is:
A) 53.21 < μ< 60.23 B) 48.23 < μ< 68.19 C) 53.82 <μ< 72.33 D) 72.21 <μ< 80.19
9. Roky Media Research wants to estimate the mean amount of time (in hours) that college students
spend watching television each weekend. A pilot study of 91 college students showed that the mean
time is 8 hours and the standard deviation 3.5 hours.
a) Find the best point estimate of the population mean time.
A) 12 B) 8 C) 9 D) 16
b) Find the margin of error in the estimation of the true population mean, using a confidence level
of 95%.
A) 0.08 B) 1.28 C) 0.73 D) 2.20
10. Indicate in which of the following cases the normal approximation to the Binomial distribution is
suitable:
a) n = 12 and p = 0.4
b) n = 10 and p = 0.6
c) n = 50 and p = 0.7
A) Only (a) B) only (b) C) (b) & (c) D) only (c)
11. There is an 80% chance that a prospective employer will check the educational background of a job
applicant. For 100 randomly selected job applicants:
a) Find the probability that exactly 90 have their educational backgrounds checked.
A) 0.5422 B) 0.1055 C) 0.3122 D) 0.2010
b) Find the probability that less than 87 have their educational backgrounds checked.
A) 0.9484 B) 0.8914 C) 0.0870 D) 0.4551







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