Question

In: Statistics and Probability

A random sample of 150 visitors traveling in Hawaii found that 14% of them hiked the...

A random sample of 150 visitors traveling in Hawaii found that 14% of them hiked the Legendary Na Pali Coast. Create a 94% confidence interval for the population proportion of visitors hiking the Na Pali Coast. Enter the lower and upper bounds for the interval in the following boxes, respectively. You may answer using decimals rounded to four places or a percentage rounded to two. Make sure to use a percent sign if you answer using a percentage.  

Solutions

Expert Solution

Solution :

Given that,

n = 150

Point estimate = sample proportion = = 0.14

1 - = 1 - 0.14 = 0.86

At 94% confidence level

= 1 - 94%

=1 - 0.94 =0.06

/2 = 0.03

Z/2 = Z0.03  = 1.881

Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 1.881 (((0.14 * 0.86) / 150)

= 0.0533

A 94% confidence interval for population proportion p is ,

± E   

= 0.14  ± 0.0533

= ( 0.0867, 0.1933 )

= ( 8.67%, 19.33% )


Related Solutions

A random sample of 150 WCC students found that 68 of them were Latino/a. a) At...
A random sample of 150 WCC students found that 68 of them were Latino/a. a) At the 5% level of significance can we prove that over 40% of all WCC students are Latino/a? State and test appropriate hypotheses. State conclusions. b) Find the p-value of the test in (a). c) If an error was made in (a), what type was it?
A random sample of 150 companies found that 75 of them use social-media tools and technologies....
A random sample of 150 companies found that 75 of them use social-media tools and technologies. Determine the LOWER confidence limit (LCL) for an 80% confidence interval for the population proportion.
A random sample of 150 men found that 88 of the men exercise regularly, while a...
A random sample of 150 men found that 88 of the men exercise regularly, while a random sample of 200 women found that 130 of the women exercise regularly. a. Based on the results, construct and interpret a 95% confidence interval for the difference in the proportions of women and men who exercise regularly. b. A friend says that she believes that a higher proportion of women than men exercise regularly. Does your confidence interval support this conclusion? Explain.
A random sample of 500 students were selected and it was found that 345 of them...
A random sample of 500 students were selected and it was found that 345 of them were satisfied with their field. Construct a 95% confidence interval for the true proportion of students who are satisfied with their field. (i)Identify n, p̂, q̂, and ??/2. ? = p̂ = q̂ = ??/2 = (ii)Calculate the margin of error, E. (iii)Construct the 95% confidence interval
In a random sample of 450 light bulbs, 150 of them burned out in less than...
In a random sample of 450 light bulbs, 150 of them burned out in less than 1000 hours. For those light bulbs: Estimate the value of the population proportion. (Round your answers to 3 decimal places.) Population proportion:……? Develop a 95% confidence interval for the population proportion. (Round your answers to 3 decimal places.) Confidence interval:……? and …..?
A random sample of 124 women over the age of 15 found that 3.68% of them...
A random sample of 124 women over the age of 15 found that 3.68% of them have been divorced. A random sample of 290 men over the age of 15 found that 5.86% have been divorced. Assuming normality and using a 95% significance level, test the claim that the proportion of divorced women is different than the proportion of divorced men. For this scenario, provide a hypothesis test with all six steps and provide both the critical value and the...
A random sample of 45 restaurant servers found the average tips per shift were $150. Assume...
A random sample of 45 restaurant servers found the average tips per shift were $150. Assume the sample standard deviation is $25. Using a 99% confidence level, calculate the following: Confidence Interval Group of answer choices A. (139.96; 160.04) B. (141.76; 161.12) C (140.09; 162.28) D. (143.56; 167.90) The average percentage of cities in the US that have snow on Christmas Day is 35%. A recent sample of 150 cities found that 21 had snow on Christmas Day last year....
USING THE TRADITIONAL 5-STEP PROCESS (NO P VALUE) --> A random sample of 450 visitors to...
USING THE TRADITIONAL 5-STEP PROCESS (NO P VALUE) --> A random sample of 450 visitors to a local shopping mall found that 90 were there to just walk the halls for exercise. Is this sufficient evidence to conclude that more than 15% of visitors to the mall were there simply for the exercise? Let α = 0.05. --> My grandfather, Roscoe, needed a new band saw for his carpentry workshop, and asked me to call two different stores to check...
To estimate the mean number of visitors(per/week) to the W.I.U Main Library, a random sample of...
To estimate the mean number of visitors(per/week) to the W.I.U Main Library, a random sample of 12 weeks selected. The sample mean is found to be 2509 visitors with a sample standard deviation of 630 visitors. (a) Estimate the mean number of visitors per week to the WIU library with a 95% confidence interval. (b) What is the critical value and degrees of freedom in above confidence interval? (c) What does this interval mean?
A random sample of 40 bags of flour is weighed. It is found that the sample...
A random sample of 40 bags of flour is weighed. It is found that the sample mean is 453.08 grams. All weights of the bags of flour have a population standard deviation of 5.42 grams. What is the lower limit of an 80% confidence interval for the population mean weight? Select one: A. No solution B. 451.983 grams C. 452.360 grams D. 452.405 grams After interviewing a sample of 100 residents of New Taipei City, you find that the mean...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT