Question

The Fox TV network is considering replacing one of its prime-time crime investigation shows with a new family-oriented comedy show. Before a final decision is made, network executives commission a sample of 490 viewers. After viewing the comedy, 420 indicated they would watch the new show and suggested it replace the crime investigation show.   

a. Estimate the value of the population proportion. (Round the final answer to 3 decimal places.)

Estimate of proportion           

b. Compute the standard error of the proportion. (Round the final answer to 3 decimal places.)

Standard error of the proportion             

c. Develop a 95% confidence interval for the population proportion. (Round the final answers to 3 decimal places.)

The confidence interval is between  and  .

d. Interpret your findings. (Round the final answers to the nearest whole number.)

We are reasonably sure the population proportion is between  and  %.

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A sample of 15 observations is selected from a normal population where the population standard deviation is 12. The sample mean is 47.  

a. Determine the standard error of the mean. (Round the final answer to 3 decimal places.)

The standard error of the mean is  .

b. Determine the 95% confidence interval for the population mean. (Round the z-value to 2 decimal places. Round the final answers to 3 decimal places.)

The 95% confidence interval for the population mean is between  and  .

Answer 1

Solution :

Given that n = 490 , x = 420

a.
=> The population proportion p = x/n = 420/490 = 0.857

=> q = 1 - p = 0.143

b.   
=> The standard error of the proportion = sqrt(p*q/n)

= sqrt(0.857*0.143/490)

= 0.016

c.
=> for 95% confidence interval , Z = 1.96

=> The 95% confidence interval for the population proportion is

=> p +/- Z*sqrt(p*q/n)

=> 0.857 +/- 1.96*0.016

=> (0.8256 , 0.8883)

=> (0.826 , 0.888) (rounded)

=> (82.6% , 88.8%)

=> (83% , 89%) (nearest whole number)

=> The confidence interval is between 0.826 and 0.888   

d. => We are reasonably sure the population proportion is between 83% and 89%

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Solution :

Given that n = 15 , population standard deviation σ = 12 , sample mean x-bar = 47

a.
=> The standard error of the mean is = σ/sqrt(n)

= 12/sqrt(15)

= 3.098 (rounded)

b. => for 95% confidence interval , Z = 1.96

=> The 95% confidence interval for the population mean is

=> x-bar +/- Z*σ/sqrt(n)

=> 47 +/- 1.96*3.098

=> (40.9279 , 53.0721)

=> (40.928 , 53.072) (rounded)

=> The 95% confidence interval for the population mean is between 40.928 and 53.072