|
Please show your work before checking the answer and explanations.
An Important Tip. Please draw a density curve and label all given in information on the curve before use the standard normal distribution table to find answers for any normal distribution related problems.
Find the confidence interval specified.
Problem 1. For a t-curve with df=20, find t0.01
A) 2.330
B) 2.528
C) 1.325
D) 2.539
View Answer
Ans : B
Explanation: Look up in a t table, use R, or TI-84. Area under the curve=0.99 and the df=20. On TI-84 invT(.99,20), which equals 2.528.
Problem 2. For a t-curve with df = 27, find the two t-values that divide the area under the curve into a middle 0.95 area and two outside areas of 0.025
A) 0, 2.052
B) -2.052, 2.052
C) -1.703, 1.703
D) 0, 1.703
View Answer
Ans : B
Explanation: We want to find the t scores that divide the area under the curve into an area of 0.95 and two outside areas of 0.025. So the area under the curve we want to calcualte is 0.025 with 27 degrees of freedom. Use a t-table, R, or TI-84 to calculate the area under 0.025 with 27 df. On TI-84 invT(0.025,27)=-2.052, as the curve is symetric the upper t value is the positive of -2.052. This can be seen by calculating the upper t value with an area of .975 and 27 df. invT(0.975, 27) is 2.052.
Find the confidence interval specifed. Assume that the population is normally distrubted. .
Problem 3. A laboratory tested twelve chicken eggs and found that the mean amount of cholesterol was 226 milligrams with s = 17.9 milligrams. Construct a 95% confidence interval for the true mean cholesterol content of all such eggs.
A) 216.7 to 235.3 milligrams
B) 214.7 to 237.3 milligrams
C) 214.6 to 237.4 milligrams
D) 214.5 to 237.5 milligrams
View Answer
Ans : C
Explanation: x̄=226, n=12, df=n-1=11, s=17.9, CI of 95% so (1-.95)/2=0.025 and t=t0.025=2.201. CI=(x̄-t0.025(s/√n), x̄+t0.025(s/√n)). Therefore CI=(226-2.201(17.9/√12), 226+2.201(17.9/√12)=(214.6,237.4).
Problem 4. )
A savings and loan association needs information concerning the checking account balances of its local customers. A random sample of 14 accounts was checked and yielded a mean balance of $664.14 and a standard deviation of $297.29. Find a 90% confidence interval for the true mean checking account balance for local customers.
A) $493.71 to $834.57
B) $523.43 to $804.85
C) $492.52 to $835.76
D) $455.65 to $872.63
View Answer
Ans : B
Explanation: x̄=664.14, n=14, df=n-1=13, s=297.29, CI of 90% so (1-.90)/2=0.05 and t=t0.05=1.771. CI=(x̄-t0.05(s/√n), x̄+t0.05(s/√n)). Therefore CI=(664.14-1.771(297.29/√14), 664.14+1.771(297.29/√14)=(523.43,804.85).
Problem 5.
The principal randomly selected six students to take an aptitude test. Their scores were:
81.6 72.0 81.1 86.4 70.2 83.1 Determine a 90% confidence interval for the mean score for all students.
A) 73.75 to 84.39
B) 84.49 to 73.65
C) 73.65 to 84.49
D) 84.39 to 73.75
View Answer
Ans : A
Explanation: x̄=(81.6+72.0+81.1+86.4+70.2+83.1)/6=79.07
[(81.6-79.07) 2+(72.0-79.07)2+(81.1-79.07)2+(86.4-79.07)2+(70.2-79.07)2+(83.1-79.07)2]/(6-1)=41.8609, s=√41.8609=6.47.
t0.05=2.015.
CI=(79.07-2.015(6.47/√6), 79.07+2.015(6.47/√6))=(73.75,84.39).
Use the one-proportion z-interval procedure to find the required confidence interval .
Problem 6. A researcher wishes to estimate the proportion of adults in the city of Darby who are vegetarian. In a random sample of 1423 adults from this city, the proportion that are vegetarian is 0.085. Find a 90% confidence interval for the proportion of all adults in the city of Darby that are vegetarians.
A) 0.0776 to 0.0924
B) 0.0755 to 0.0945
C) 0.0632 to 0.1068
D) 0.0728 to 0.0972
View Answer
Ans : D
Explanation: Given: p̂=0.085 and n=1423. 90% confidence interval so t0.05=1.645.
CI=(p̂-t0.05√((p̂(1-p̂))/n),p̂+t0.05√((p̂(1-p̂))/n))
CI=(0.085-1.645√(0.085(0.915)/1423),(0.085+1.645√(0.085(0.915)/1423)) therefore answer is 0.0728 to 0.0972.
Problem 7. A survey of 300 union members in New York State reveals that 112 favor the Republican candidate for governor. Construct the 98% confidence interval for the proportion of all New York State union members who favor the Republican candidate.
A) 0.316 to 0.430
B) 0.308 to 0.438
C) 0.304 to 0.442
D) 0.301 to 0.445
View Answer
Ans : B
Explanation:p̂=112/300 and n=300. 98% confidence interval so t0.01=2.326
CI=(p̂-t0.01√((p̂(1-p̂))/n),p̂+t0.01√((p̂(1-p̂))/n))
CI=(112/300-2.326√(112/300(47/75)/300),(112/300+2.326√(112/300(47/75)/300)) therefore answer is 0.308 to 0.438.
Problem 8. Of 106 patients selected at random from a clinic, 31 were found to have high blood pressure. Construct a 95% confidence interval for the percentage of all patients at this clinic that have high blood pressure.
A) 17.8% to 40.6%
B) 19.0% to 39.5%
C) 20.6% to 37.9%
D) 22.0% to 36.5%
View Answer
Ans : C
Explanation: p̂=31/106 and n=106. 95% confidence interval so t0.025=1.96
CI=(p̂-t0.025√((p̂(1-p̂))/n),p̂+t0.025√((p̂(1-p̂))/n))
CI=(31/106-1.96√(31/106(75/206)/106),(31/106+1.96√(31/106(75/106)/106)) therefore the interval is .206 to 0.379. The question asks for this as a percentage, multiple each by 100 and you get 20.6% to 37.9%.
|