Question

3.

a. According to a study, 75% of adults ages 18-29 years had broadband Internet access at home in 2011. A researcher wanted to estimate the proportion of undergraduate college students (18-23 years) with access, so she randomly sampled 181 undergraduates and found that 158 had access. Estimate the true proportion with 90% confidence. Round your answers to three decimal places. ___ < p < ____

b. it has been reported that 20.4% of incoming freshmen indicate that they will major in business or a related field. a random sample of 424 incoming college freshmen were asked their preference, and 81 replied that they were considering business as a major Estimate the true proportion of freshman business majors with 97% confidence. a survey of 50 students in grades 4 through 12 found 64% have classroom Wi-Fi access. Find the 90% confidence interval of the population proportion.

____ < p < ____

c. using table g, find the values for x^2 left and x^2 right of the following. when a = 0.02 and n =11

X^2 left =_____

x^2 right= _____

d. Find the 98% confidence interval for the variance and standard deviation of the ages of seniors at Oak Park college if a sample of 23 students has a standard deviation of 2.7 years. Assume the variable is normally distributed. Use the chi-square distribution table to find any chi-square values to three decimal places. ROUND TO ONE DECIMAL PLACE FOR FINAL ANSWER

____ < o^2 < ______

____ < o^2 < ______

e. A random sample of stock prices per share (in dollars) is shown. Find the 98% confidence interval for the variance and standard deviation for the prices. Assume the variable is normally distributed. ROUND TO THREE DECIMAL PLACES

46.12 10.87 40.25 60.50

28.00 28.25 6.94 45.12

13.62 53.81

_______ < o^2 < ______

______ < o < ______

f. estimate the variance in mean mathematics SAT scores by state, using the randomly selected scores listed below. Estimate the 95% confidence. 502 211 209 499 565 469 543 572 550 515

___ < o^2 < _____

Answer 1

a. According to a study, 75% of adults ages 18-29 years had broadband Internet access at home in 2011. A researcher wanted to estimate the proportion of undergraduate college students (18-23 years) with access, so she randomly sampled 181 undergraduates and found that 158 had access. Estimate the true proportion with 90% confidence. Round your answers to three decimal places. 0.832 < p <0.914

b. it has been reported that 20.4% of incoming freshmen indicate that they will major in business or a related field. a random sample of 424 incoming college freshmen were asked their preference, and 81 replied that they were considering business as a major Estimate the true proportion of freshman business majors with 97% confidence.

0.150<p<0.232

a survey of 50 students in grades 4 through 12 found 64% have classroom Wi-Fi access. Find the 90% confidence interval of the population proportion.

0.528 < p < 0.752

c. using table g, find the values for x^2 left and x^2 right of the following. when a = 0.02 and n =11

=3.609

= 22.618

d. Find the 98% confidence interval for the variance and standard deviation of the ages of seniors at Oak Park college if a sample of 23 students has a standard deviation of 2.7 years. Assume the variable is normally distributed. Use the chi-square distribution table to find any chi-square values to three decimal places. ROUND TO ONE DECIMAL PLACE FOR FINAL ANSWER

For variance: 4.0< <16.8

For Standard deviation 2.0< <4.1

e. A random sample of stock prices per share (in dollars) is shown. Find the 98% confidence interval for the variance and standard deviation for the prices. Assume the variable is normally distributed. ROUND TO THREE DECIMAL PLACES

46.12 10.87 40.25 60.50

28.00 28.25 6.94 45.12

13.62 53.81

For Variance 145.467 < < 1509.504

For SD 12.061 <   < 38.852

f. estimate the variance in mean mathematics SAT scores by state, using the randomly selected scores listed below. Estimate the 95% confidence. 502 211 209 499 565 469 543 572 550 515

----------------------------------------------------------Work outs----------------------------------------------------------------------------------a). We hve

90% Confidence interval :  

  

  

b).

We hve

97% Confidence interval :  

  

  

  

90% Confidence interval

  

  

  

d). The 98% confidence interval for variance is:

The confidence interval for SD

e). The data given:

	x	x^2
	46.12	2127.0544
	10.87	118.1569
	40.25	1620.0625
	60.5	3660.25
	28	784
	28.25	798.0625
	6.94	48.1636
	45.12	2035.8144
	13.62	185.5044
	53.81	2895.5161
Total	333.48	14272.5848

The sample variance is

  

The CI for sd:

f). The data given

	x	x^2
	502	252004
	211	44521
	209	43681
	499	249001
	565	319225
	469	219961
	543	294849
	572	327184
	550	302500
	515	265225
Total	4635	2318151