1) How much do wild mountain lions weigh? Adult wild mountain lions (18 months or older) captured and released for the first time in the San Andres Mountains gave the following weights (pounds):
68 | 108 | 127 | 127 | 60 | 64 |
Assume that the population of x values has an approximately normal distribution.
(a) Use a calculator with mean and sample standard deviation keys to find the sample mean weight x and sample standard deviation s. (Round your answers to one decimal place.)
x = | lb |
s = | lb |
(b) Find a 75% confidence interval for the population average
weight μ of all adult mountain lions in the specified
region. (Round your answers to one decimal place.)
lower limit | lb |
upper limit | lb |
2) Anystate Auto Insurance Company took a random sample of 376
insurance claims paid out during a 1-year period. The average claim
paid was $1505. Assume σ = $252.
Find a 0.90 confidence interval for the mean claim payment. (Round
your answers to two decimal places.)
lower limit | $ |
upper limit | $ |
Find a 0.99 confidence interval for the mean claim payment. (Round
your answers to two decimal places.)
lower limit | $ |
upper limit | $ |
3) Three experiments investigating the relation between need for cognitive closure and persuasion were performed. Part of the study involved administering a "need for closure scale" to a group of students enrolled in an introductory psychology course. The "need for closure scale" has scores ranging from 101 to 201. For the 66 students in the highest quartile of the distribution, the mean score was x = 175.10. Assume a population standard deviation of σ = 7.49. These students were all classified as high on their need for closure. Assume that the 66 students represent a random sample of all students who are classified as high on their need for closure. Find a 95% confidence interval for the population mean score μ on the "need for closure scale" for all students with a high need for closure. (Round your answers to two decimal places.)
lower limit | |
upper limit |
4) A research group conducted an extensive survey of 2848 wage and salaried workers on issues ranging from relationships with their bosses to household chores. The data were gathered through hour-long telephone interviews with a nationally representative sample. In response to the question, "What does success mean to you?" 1510 responded, "Personal satisfaction from doing a good job." Let p be the population proportion of all wage and salaried workers who would respond the same way to the stated question. Find a 90% confidence interval for p. (Round your answers to three decimal places.)
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upper limit |
5)Three-circle, red-on-white is one distinctive pattern painted on ceramic vessels of the Anasazi period found at an archaeological site. At one excavation, a sample of 169 potsherds indicated that 68 were of the three-circle, red-on-white pattern.
(a) Find a point estimate p̂ for the proportion of all
ceramic potsherds at this site that are of the three-circle,
red-on-white pattern. (Round your answer to four decimal
places.)
(b) Compute a 95% confidence interval for the population proportion
p of all ceramic potsherds with this distinctive pattern
found at the site. (Round your answers to three decimal
places.)
lower limit | |
upper limit |
Question 1
Values ( X ) | Σ ( X_{i}- X̅ )^{2} | |
68 | 592.1095 | |
108 | 245.4455 | |
127 | 1201.7801 | |
127 | 1201.7801 | |
60 | 1045.4423 | |
64 | 802.7759 | |
Total | 554 | 5089.3334 |
Mean X̅ = Σ Xi / n
X̅ = 554 / 6 = 92.3
Sample Standard deviation S_{X} = √ ( (Xi - X̅ )2 / n - 1
)
S_{X} = √ ( 5089.3334 / 6 -1 ) = 31.9
Part b)
Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.25 /2, 6- 1 ) = 1.301
92.3333 ± t(0.25/2, 6 -1) * 31.904/√(6)
Lower Limit = 92.3333 - t(0.25/2, 6 -1) 31.904/√(6)
Lower Limit = 75.4
Upper Limit = 92.3333 + t(0.25/2, 6 -1) 31.904/√(6)
Upper Limit = 109.3
75% Confidence interval is ( 75.4 , 109.3 )
Question 2
Confidence Interval :-
X̅ ± Z( α /2) σ / √ ( n )
Z(α/2) = Z (0.1 /2) = 1.64
1505 ± Z (0.1/2 ) * 252/√(376)
Lower Limit = 1505 - Z(0.1/2) 252/√(376)
Lower Limit = 1483.6236
Upper Limit = 1505 + Z(0.1/2) 252/√(376)
Upper Limit = 1526.3764
90% Confidence interval is ( 1483.62 , 1526.38
)
Part b)
Confidence Interval :-
X̅ ± Z( α /2) σ / √ ( n )
Z(α/2) = Z (0.01 /2) = 2.58
1505 ± Z (0.01/2 ) * 252/√(376)
Lower Limit = 1505 - Z(0.01/2) 252/√(376)
Lower Limit = 1471.5248
Upper Limit = 1505 + Z(0.01/2) 252/√(376)
Upper Limit = 1538.4752
99% Confidence interval is ( 1471.52 , 1538.48
)
Question 3
Confidence Interval :-
X̅ ± Z( α /2) σ / √ ( n )
Z(α/2) = Z (0.05 /2) = 1.96
175.1 ± Z (0.05/2 ) * 7.49/√(66)
Lower Limit = 175.1 - Z(0.05/2) 7.49/√(66)
Lower Limit = 173.293
Upper Limit = 175.1 + Z(0.05/2) 7.49/√(66)
Upper Limit = 176.907
95% Confidence interval is ( 173.29 , 176.91 )
Question 4
p̂ = 0.53
p̂ = 1 - p̂ = 0.47
n = 2848
p̂ ± Z(α/2) √( (p * q) / n)
0.53 ± Z(0.1/2) √( (0.53 * 0.47) / 2848)
Z(α/2) = Z(0.1/2) = 1.645
Lower Limit = 0.53 - Z(0.1) √( (0.5302 * 0.4698) / 2848) =
0.5148
upper Limit = 0.53 + Z(0.1) √( (0.5302 * 0.4698) / 2848) =
0.5456
90% Confidence interval is ( 0.515 , 0.546 )
( 0.515 < P < 0.546 )
Question 5
Part a)
p̂ = X / n = 68/169 = 0.4024
Part b)
p̂ = 0.402
p̂ = 1 - p̂ = 0.598
n = 169
p̂ ± Z(α/2) √( (p * q) / n)
0.402 ± Z(0.05/2) √( (0.402 * 0.598) / 169)
Z(α/2) = Z(0.05/2) = 1.96
Lower Limit = 0.402 - Z(0.05) √( (0.4024 * 0.5976) / 169) =
0.3284
upper Limit = 0.402 + Z(0.05) √( (0.4024 * 0.5976) / 169) =
0.4763
95% Confidence interval is ( 0.328 , 0.476 )
( 0.328 < P < 0.476 )
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