Question

In: Math

Suppose a 90% confidence interval for the mean salary of college graduates in a town in...

Suppose a 90% confidence interval for the mean salary of college graduates in a town in Mississippi is given by [$45,783, $57,017]. The population standard deviation used for the analysis is known to be $13,700. [You may find it useful to reference the z table.]

a. What is the point estimate of the mean salary for all college graduates in this town?

b. Determine the sample size used for the analysis. (Round "z" value to 3 decimal places and final answer to the nearest whole number.)

Solutions

Expert Solution

Solution :

Given that,

A ) Lower confidence interval = $45,783

Upper confidence interval = $57,017

  = (Lower confidence interval + Upper confidence interval ) / 2

= (45,783 + 57,017) / 2

= 102800 / 2

= 51400

= 51400

Margin of error = ME = Upper confidence interval -

= 57,017 - 51400

= 5617

standard deviation = = $13,700

B ) margin of error = E = $ 5617

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z/2 = Z0.05 = 1.645

Sample size = n = ((Z/2 * ) / E)2

= ((1.645 * 13700) / 5617)2

= 16.088

Sample size = 16


Related Solutions

Suppose a 95% confidence interval for the mean salary of college graduates in a town in...
Suppose a 95% confidence interval for the mean salary of college graduates in a town in Mississippi is given by [$34,321, $41,279]. The population standard deviation used for the analysis is known to be $14,200. a. What is the point estimate of the mean salary for all college graduates in this town? b. Determine the sample size used for the analysis. (Round "z" value to 3 decimal places and final answer to the nearest whole number.)
   Suppose a 99% confidence interval for the mean salary of college graduates in a town...
   Suppose a 99% confidence interval for the mean salary of college graduates in a town in Mississippi is given by [$39,986, $48,414]. The population standard deviation used for the analysis is known to be $14,700. a. What is the point estimate of the mean salary for all college graduates in this town?   Point estimate    b. Determine the sample size used for the analysis.   Sample size   
Suppose we want to estimate the mean salary μ of all college graduates. We take a...
Suppose we want to estimate the mean salary μ of all college graduates. We take a sample of 25 graduates and the sample average is $39,000 with a sample standard deviation of $10,000. We construct a 95% confidence interval for the true average salary. What is the upper bound of the confidence interval i.e. what is the upper confidence limit? $39,000 $43,128 $41,000 $42.422
Suppose you constructed a 90% confidence interval for the true average GPA of California community college...
Suppose you constructed a 90% confidence interval for the true average GPA of California community college students. If all of the sample data were to be held constant, how would this confidence interval compare to an 80% confidence interval based on the same data? A random sample of 100 students was taken to determine interest in switching to a quarter system instead of semesters. Out of the 100 students, 73 stated that they were interested in switching to a quarter...
A 90 % confidence interval (a t interval) for the mean lives (in minutes) of Kodak...
A 90 % confidence interval (a t interval) for the mean lives (in minutes) of Kodak AA batteries is ( 410, 470 ). Assume that this result is based on a sample of size 15 . 2) What is the value of the sample standard deviation? 66.2785 65.9677 86.6694 86.3843 3) Construct the 99% confidence interval. (389.2961,490.7039) (395.2976,484.7024) (414.8193,465.1807) (417.1658,462.8342) Tries 0/2 4) If the confidence interval (414.0575 ,465.9425) is obtained from the same sample data, what is the degree...
A 90 % confidence interval (a t interval) for the mean lives (in minutes) of Kodak...
A 90 % confidence interval (a t interval) for the mean lives (in minutes) of Kodak AA batteries is ( 400, 440 ). Assume that this result is based on a sample of size 25 . 1) What is the value of the sample mean? Incorrect 440 Correct: 420 Incorrect 415 Incorrect 425 Incorrect 410 You are correct. Your receipt no. is 159-3963 Previous Tries 2) What is the value of the sample standard deviation? 58.4494 75.8820 75.9679 58.5432 Tries...
A 90 % confidence interval (a t interval) for the mean lives (in minutes) of Kodak...
A 90 % confidence interval (a t interval) for the mean lives (in minutes) of Kodak AA batteries is ( 480, 520 ). Assume that this result is based on a sample of size 25 a. value of sample standard deviation b. 99% confidence interval c. If the confidence interval (482.6156 ,517.3844) is obtained from the same sample data, what is the degree of confidence?
The annual salary of fresh college graduates is thought to be normally distributed with a mean...
The annual salary of fresh college graduates is thought to be normally distributed with a mean of $45,000 and standard deviation of $8000. Do the following. (a) What is the z −score of the salary of $55,000? (10 points) (b) If you randomly select such a graduate, what is the probability that he/she will be earning a salary of $55,000 or less? (Use z −score and Excel function to calculate this) (10 points) (c) If you randomly select such a...
The mean starting salary for college graduates in spring of 2018 was $43,200. Assume that the...
The mean starting salary for college graduates in spring of 2018 was $43,200. Assume that the distribution of starting salaries follows the normal distribution with a standard deviation of $3500. What percent of the graduates have starting salaries: A.) Less than $38,000? B.) More than $45,000? C.) Between $38,000 and $45,000?
A sample is selected to find a 90% confidence interval for the average starting salary. Here...
A sample is selected to find a 90% confidence interval for the average starting salary. Here are the sample statistics: n = 31, x ̄ = $43, 780, s = $1, 600. a). Find the t− score used in the calculation of the confidence interval. b). Build a 90% confidence interval for the mean starting salary. c). Based on the result of part b), could we make a conclusion that the mean staring salary is below $45, 000? Explain your...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT