Question


1) A survey of 400 non-fatal accidents showed that 173 involved faulty equipment. Find a point estimate for p, the population
0 0
Add a comment Improve this question Transcribed image text
Answer #1

Solution :

Given that,

n = 400

x = 173

\hat p = x / n = 173 / 400 = 0.433

Point estimate = 0.433

Add a comment
Know the answer?
Add Answer to:
1) A survey of 400 non-fatal accidents showed that 173 involved faulty equipment. Find a point...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
  • A survey of 700 non-fatal accidents showed that 163 involved the use of a cell phone....

    A survey of 700 non-fatal accidents showed that 163 involved the use of a cell phone. Find a point estimate for p, the population proportion of non-fatal accidents that involved the use of a cell phone.

  • a survey of 400 non fatal accidents showed that 109 involved uninsured drivers. Construct a 99%...

    a survey of 400 non fatal accidents showed that 109 involved uninsured drivers. Construct a 99% confidence interval for the proportion of fatal accidents that involved uninsured drivers.

  • Beview 1JA survey of 700 non-fatal accidents showed that 183 involved faulty equipment. Find a point...

    Beview 1JA survey of 700 non-fatal accidents showed that 183 involved faulty equipment. Find a point estimate for p, the population proportion of accidents that involved faulty equipment. 2) An article a Florida newspaper reported on the topics that teenagers most want to discuss with their pare nts. The findings, the results of a poll, showed that 46% would like more discussion about the family's financial situation, 37% would like to talk about school, and 30% would like to talk...

  • 2) A survey of 700 non-fatal accidents showed that 167 involved uninsured drivers. Construct a 99%...

    2) A survey of 700 non-fatal accidents showed that 167 involved uninsured drivers. Construct a 99% confidence interval for the proportion of fatal accidents that involved uninsured drivers. Show set up. Answer is all decimal places displayed on calculator 3) A private opinion poll is conducted for a politician to determine what proportion of the population favors adding more national parks. How large a sample is needed in order to be 95% confident that the sample proportion will not differ...

  • How would I slove these using calculator functions 1. A survey of 700 non-fatal accidents showed...

    How would I slove these using calculator functions 1. A survey of 700 non-fatal accidents showed that 2. A researcher at a major clinic wishes to estimate 163 involved uninsured drivers. Construct a 99% | the proportion of the adult population of the United confidence interval for the proportion of non-fat States that has sleep deprivation. How large a accidents that involved uninsured drivers. (Round 1 sample is needed in order to be 95% confident that to three decimal places...

  • VULJHUIV 17 A survey of 700 non-fatal accidents in the United States showed that about 23%...

    VULJHUIV 17 A survey of 700 non-fatal accidents in the United States showed that about 23% involved uninsured drivers. The margin of error is 3 percentage points with a 95% confidence. Does the confidence interval support the claim that at least 25% of non-fatal accidents in the United States involve uninsured drivers. Why? No; the interval does not support this claim because the interval contains values that are above 25% O No; the interval does not support this claim because...

  • A researcher wishes to estimate the average blood alcohol concentration​ (BAC) for drivers involved in fatal...

    A researcher wishes to estimate the average blood alcohol concentration​ (BAC) for drivers involved in fatal accidents who are found to have positive BAC values. He randomly selects records from 51 such drivers in 2009 and determines the sample mean BAC to be 0.16 ​g/dL with a standard deviation of 0.070 ​g/dL. Complete parts​ (a) through​ (d) below. c) Determine and interpret a​ 90% confidence interval for the mean BAC in fatal crashes in which the driver had a positive...

  • According to a certain government agency for a large country, the proportion of fatal traffic accidents...

    According to a certain government agency for a large country, the proportion of fatal traffic accidents in the country in which the driver had a positive blood alcohol concentration (BAC) is 0.39 Suppose a random sample of 104 traffic fatalities in a certain region results in 52 that involved a positive BAC Does the sample evidence suggest that the region has a higher proportion of traffic fatalities involving a positive BAC than the country at the c = 0.05 level...

  • please answer part a and b A particular report from 2004 classified 718 fatal bicycle accidents...

    please answer part a and b A particular report from 2004 classified 718 fatal bicycle accidents according to the month in which the accident occurred, resulting in the accompanying table. Number of Month Accidents Month January February 36 31 March 44 April May 60 78 June 75 July August September 96 85 63 October 66 November December 43 41 1/12, T2 1/12, . . . , T12 = 1/12, where T1 is the proportion of fatal (a) Use the given...

  • According to a certain government agency for a large country, the proportion of fatal traffic accidents...

    According to a certain government agency for a large country, the proportion of fatal traffic accidents in the country in which the driver had a positive blood alcohol concentration (BAC) is 0.33. Suppose a random sample of 114 traffic fatalities in a certain region results in 48 that involved a positive BAC. Does the sample evidence suggest that the region has a higher proportion of traffic fatalities involving a positive BAC than the country at the a = 0.01 level...

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT