Question

In: Statistics and Probability

1. A sample of 100 results in 27 successes. a. Calculate the point estimate for the...

1. A sample of 100 results in 27 successes.
a. Calculate the point estimate for the population proportion of successes. (Do not round intermediate calculations. Round your answer to 3 decimal places.)  

b. Construct 95% and 90% confidence intervals for the population proportion. (Round intermediate calculations to at least 4 decimal places. Round "z" value and final answers to 3 decimal places.)

95% -

90%-

c. Can we conclude at 95% confidence that the population proportion differs from 0.330?  

  • No, since the confidence interval does not contain the value 0.330.

  • No, since the confidence interval contains the value 0.330.

  • Yes, since the confidence interval does not contain the value 0.330.

  • Yes, since the confidence interval contains the value 0.330.


d. Can we conclude at 90% confidence that the population proportion differs from 0.330?

  • No, since the confidence interval contains the value 0.330.

  • No, since the confidence interval does not contain the value 0.330.

  • Yes, since the confidence interval contains the value 0.330.

  • Yes, since the confidence interval does not contain the value 0.330.

Solutions

Expert Solution

Solution :

Given that,

n = 100

x = 27

a)

Point estimate = sample proportion = = x / n = 27 / 100 = 0.270

1 - = 1 - 0.270 = 0.730

b)

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z/2 = Z 0.05 = 1.645

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

= 1.645 * (((0.270 * 0.730) / 100)

= 0.0730

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.270 - 0.0730 < p < 0.270 + 0.0730

0.1970 < p < 0.3430

(0.1970 , 0.3430)

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96

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

= 1.96 * (((0.270 * 0.730) / 100)

= 0.0870

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.270 - 0.0870 < p < 0.270 + 0.0870

0.1830< p < 0.3570

(0.1830 , 0.3570)

c)

Yes, since the confidence interval contains the value 0.330

d)

Yes, since the confidence interval contains the value 0.330


Related Solutions

A random sample of 100 observations results in 70 successes. [You may find it useful to...
A random sample of 100 observations results in 70 successes. [You may find it useful to reference the z table.] a. Construct the a 95% confidence interval for the population proportion of successes. (Round intermediate calculations to at least 4 decimal places. Round "z" value and final answers to 3 decimal places.)   Confidence interval to b. Construct the a 95% confidence interval for the population proportion of failures. (Round intermediate calculations to at least 4 decimal places. Round "z" value...
A sample of 120 results in 30 successes. [You may find it useful to reference the...
A sample of 120 results in 30 successes. [You may find it useful to reference the z table.]    a. Calculate the point estimate for the population proportion of successes. (Do not round intermediate calculations. Round your answer to 3 decimal places.)   b. Construct 95% and 90% confidence intervals for the population proportion. (Round intermediate calculations to at least 4 decimal places. Round "z" value and final answers to 3 decimal places.)   confidence intervals 95% Confidence intervals 90% c. Can we...
A random sample of 140 observations results in 119 successes. a. Construct the a 95% confidence...
A random sample of 140 observations results in 119 successes. a. Construct the a 95% confidence interval for the population proportion of successes. (Round intermediate calculations to at least 4 decimal places. Round "z" value and final answers to 3 decimal places.)   b. Construct the a 95% confidence interval for the population proportion of failures. (Round intermediate calculations to at least 4 decimal places. Round "z" value and final answers to 3 decimal places.)
A sample of 115 results in 46 successes. [You may find it useful to reference the...
A sample of 115 results in 46 successes. [You may find it useful to reference the z table.]    a. Calculate the point estimate for the population proportion of successes. (Do not round intermediate calculations. Round your answer to 3 decimal places.)   Point estimate b. Construct 90% and 99% confidence intervals for the population proportion. (Round intermediate calculations to at least 4 decimal places. Round "z" value and final answers to 3 decimal places.)   Confidence Level Confidence Interval 90% to 99%...
A random sample of 27 observations is used to estimate the population mean. The sample mean...
A random sample of 27 observations is used to estimate the population mean. The sample mean and the sample standard deviation are calculated as 113.9 and 20.40, respectively. Assume that the population is normally distributed. [You may find it useful to reference the t table.] a. Construct the 90% confidence interval for the population mean. (Round intermediate calculations to at least 4 decimal places. Round "t" value to 3 decimal places and final answers to 2 decimal places.) b. Construct...
A random sample of 27 observations is used to estimate the population variance. The sample mean...
A random sample of 27 observations is used to estimate the population variance. The sample mean and sample standard deviation are calculated as 44 and 4.5, respectively. Assume that the population is normally distributed. (You may find it useful to reference the appropriate table: chi-square table or F table) a. Construct the 95% interval estimate for the population variance. (Round intermediate calculations to at least 4 decimal places and final answers to 2 decimal places.) b. Construct the 99% interval...
A random sample of 130 observations results in 78 successes. [You may find it useful to...
A random sample of 130 observations results in 78 successes. [You may find it useful to reference the z table.] a. Construct the an 95% confidence interval for the population proportion of successes. (Round intermediate calculations to at least 4 decimal places. Round "z" value and final answers to 3 decimal places.) b. Construct the an 95% confidence interval for the population proportion of failures. (Round intermediate calculations to at least 4 decimal places. Round "z" value and final answers...
A random sample of 120 observations results in 90 successes. [You may find it useful to...
A random sample of 120 observations results in 90 successes. [You may find it useful to reference the z table.] a. Construct the 90% confidence interval for the population proportion of successes. (Round intermediate calculations to at least 4 decimal places. Round "z" value and final answers to 3 decimal places.)   Confidence interval _____ to _____ b. Construct the 90% confidence interval for the population proportion of failures. (Round intermediate calculations to at least 4 decimal places. Round "z" value...
A random sample of 20 observations results in 11 successes. [You may find it useful to...
A random sample of 20 observations results in 11 successes. [You may find it useful to reference the z table.] a. Construct the an 90% confidence interval for the population proportion of successes. (Round intermediate calculations to at least 4 decimal places. Round "z" value and final answers to 3 decimal places.)   Confidence interval to Construct the an 90% confidence interval for the population proportion of failures. (Round intermediate calculations to at least 4 decimal places. Round "z" value and...
1. You want to obtain a sample to estimate a population proportion. At this point in...
1. You want to obtain a sample to estimate a population proportion. At this point in time, you have no reasonable preliminary estimation for the population proportion. You would like to be 95% confident that you estimate is within 1% of the true population proportion. How large of a sample size is required? n = 2.Based on historical data in Oxnard college, we believe that 38% of freshmen do not visit their counselors regularly. For this year, you would like...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT