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Zarrin [17]
3 years ago
9

Legalization of marijuana, Part I. The 2010 General Social Survey asked 1,578 US res- idents: \Do you think the use of marijuana

should be made legal, or not?" 61% of the respondents said it should be made legal.44 (a) Is 61% a sample statistic or a population parameter? Explain. (b) Construct a 95% condence interval for the proportion of US residents who think marijuana should be made legal, and interpret it in the context of the data. (c) A critic points out that this 95% condence interval is only accurate if the statistic follows a normal distribution, or if the normal model is a good approximation. Is this true for these data? Explain.(d) A news piece on this survey's ndings states, \Majority of Americans think marijuana should be legalized." Based on your condence interval, is this news piece's statement justied?
Mathematics
1 answer:
mixer [17]3 years ago
4 0

Answer:

Step-by-step explanation:

a) Sample statistics are used to estimate population value. Since 48% is a sample proportion, therefore, it is a sample statistic.

b) For 95% confidence level, z* = 1.96.

\hat{p}\pm z^* \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}= 0.61\pm 0.61\sqrt{\frac{0.61(1-0.61)}{1578}}=0.61\pm 0.024 \ or (0.586, 0.634).

We are 95% confident that the true proportion of US residents who think marijuana should be made legal lies between 58.6% and 63.4%.

c)

\\np=1578(0.61)=962.58

\\n(1-p)=1578(1-0.61)=615.42

Since both np and n(1-p), are at least 10, the normal model is a good approximation for these data.

d) As the lower limit of confidence interval is less than 0.5, less than 50% population is also a plausible value of true proportion. This means the statement "Majority of Americans think marijuana should be legalized" is not justified.

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matrenka [14]

Answer:

0.03593

Step-by-step explanation:

We solve using z score formula

z = (x-μ)/σ, where

x is the raw score = 17,340

μ is the population mean = 15,000

σ is the population standard deviation = 1300

z score:

z = 17340 - 15000/1300

z =1.8

Probability value from Z-Table:

P(x<17340) = 0.96407

P(x >17340) = 1 - P(x<17340)

= 1 - 0.96407

= 0.03593

The probability that the life span of the monitor will be more than 17,340 hours is 0.03593

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irina [24]

Answer:

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