Question

Please show all work, step by step:
a. The critical value
b. the error bound
c. The minimum and maximum numbers of the interval. On interpretations include information about the specific problem.

1.   As part of an environmental studies class project, students measured the circumference of a random sample of 45 blue spruce trees near Brainard Lake, Colorado. The sample mean circumference was x = 29.8 inches. Assume that o is known to be 7.2 inches.
a. Find a 90% confidence interval for the population mean circumference of all blue spruce trees near this lake.
b. Interpret the meaning of the confidence interval in the context of this problem.

2. James is self employed and sells cookware at home parties. She wants to estimate the average amount a client spends at each party. A random sample of 35 receipts gave a mean of x = $34.70 with standard deviation s = $4.85.
a. Find a 99% confidence interval for the average amount spent by all clients.
b. Interpret the meaning of the confidence interval in the context of this problem.

3. How long does it take to commute from home to work? It depends on several factors, including routes, traffic and time of departure. The data below are results (in minutes) from a random sample of eight trips.
27. 38. 30. 42. 24. 37. 30. 39.
a. What are the sample mean x and the sample standard deviation s?

b. Use these data to create a 98% confidence interval for the population mean time of the commute.

A random sample of 19 rainbow trout caught at Brainard Lake x = 11.9 inches with sample standard deviation o= 2.8 inches.
Find a 95% confidence interval for the population mean length of all rainbow trout in this Lake.

b. Interpret the meaning of the confidence interval in the context of this problem.

5. A random sample of 78 students was interviewed, and 59 students said that they would vote for Stella Joh as student body president.

a. Let p represent the proportion of all students at this college who will vote for Stella. Find a point estimate p for p.
b. Find a 98% confidence interval for p.

6. A random sample of students was asked for the number of semester hours they are taking this semester. The standard deviation was found to be o = 4.7 semester hours.
a. How many students should be included in the sample to be 90% sure that the sample mean x is within 1 semester hour of the population mean u for all students at this college.?

What percentage of college students owns a cellular phone? Let p be the proportion of college students that own a cellular phone.
a. If no preliminary study is made to estimate p, how large a sample a sample is needed to be 95% sure that a point estimate p will be within a distance of 0.08 from p.

b. A preliminary study shows that approximately 38%of college students own cellular phones. How large a sample is needed to be 95% sure that a point estimate p will be within a distance of 0.08 from p.

Answer 1

Solution:

1. Given x = 29.8, σ = 7.2, n = 45, 90% Confidece interval = 1.645

a. 90% Confidence interval for the population mean = 29.8 +/- 1.645*7.2/sqrt(45)
= (28.03 , 31.57)
  
b. interpretation: we are 90% Confident the true mean is between 28.03 and 31.57
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2. Given that x = 34.70,s = 4.85, n = 35, 99% Confidence interval
df = 34, t = 2.728

a. 99% Confidence interval for the mean = X +/- t*s/sqrt(n)
= 34.70 +/- 2.728*4.85/sqrt(35)
= (32.46 , 36.94)

b. interpretation: we are 99% Confident the true mean is between 32.46 and 36.94
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3. Given that samples 27,38,30,42,24,37,30,39
n = 8, 98% Confidence interval
df = 7, t = 2.998

a. mean x = 33.375 = 33.38 (rounded)

sample standard deviation s = 6.4573

b. 98% Confidence interval for the population mean = X +/- t*s/sqrt(n)
= 33.38 +/- 2.998*6.4573/sqrt(8)
= (26.54 , 40.22)
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4. Given that n = 19, x = 11.9, σ = 2.8, 95% Confidence for Z = 1.96

a. 95% Confidence interval for the population mean = 11.9 +/- 1.96*2.8/sqrt(19)
= (10.6 , 13.2)
b. interpretation: we are 95% Confident the true mean is between 10.6 and 13.2

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5. Given that x = 59, n = 78
p = x/n = 59/78 = 0.76

a. The point estimate of p = 0.76

b. 98% Confidence interval for the population proportion = p +/- Z*sqrt(pq/n)
= 0.76 +/- 2.33*Sqrt(0.76*0.24/78)
= (0.6473 , 0.8727)