Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.† Suppose a small group of 12 Allen's hummingbirds has been under study in Arizona. The average weight for these birds is x = 3.15 grams. Based on previous studies, we can assume that the weights of Allen's hummingbirds have a normal distribution, with σ = 0.20 gram.

(a)

Find an 80% confidence interval for the average weights of Allen's hummingbirds in the study region. What is the margin of error? (Round your answers to two decimal places.)

lower limitupper limitmargin of error

(b)

What conditions are necessary for your calculations? (Select all that apply.)

σ is knownσ is unknownn is largenormal distribution of weightsuniform distribution of weights

(c)

Interpret your results in the context of this problem.

There is a 20% chance that the interval is one of the intervals containing the true average weight of Allen's hummingbirds in this region.The probability that this interval contains the true average weight of Allen's hummingbirds is 0.20.    There is an 80% chance that the interval is one of the intervals containing the true average weight of Allen's hummingbirds in this region.The probability that this interval contains the true average weight of Allen's hummingbirds is 0.80.The probability to the true average weight of Allen's hummingbirds is equal to the sample mean.

(d)

Find the sample size necessary for an 80% confidence level with a maximal margin of error E = 0.09 for the mean weights of the hummingbirds. (Round up to the nearest whole number.)

hummingbirds

Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.† Suppose a small group of 19 Allen's hummingbirds has been under study in Arizona. The average weight for these birds is x = 3.15 grams. Based on previous studies, we can assume that the weights of Allen's hummingbirds have a normal distribution, with σ = 0.26 gram.

(a) Find an 80% confidence interval for the average weights of Allen's hummingbirds in the study region. What is the margin of error? (Round your answers to two decimal places.)

(b) What conditions are necessary for your calculations? (Select all that apply.)

σ is knownσ is unknownnormal distribution of weightsn is largeuniform distribution of weights

(c) Interpret your results in the context of this problem.

The probability to the true average weight of Allen's hummingbirds is equal to the sample mean.The probability that this interval contains the true average weight of Allen's hummingbirds is 0.80.    There is an 80% chance that the interval is one of the intervals containing the true average weight of Allen's hummingbirds in this region.There is a 20% chance that the interval is one of the intervals containing the true average weight of Allen's hummingbirds in this region.The probability that this interval contains the true average weight of Allen's hummingbirds is 0.20.

(d) Find the sample size necessary for an 80% confidence level with a maximal margin of error E = 0.15 for the mean weights of the hummingbirds. (Round up to the nearest whole number.)
hummingbirds

Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.† Suppose a small group of 19 Allen's hummingbirds has been under study in Arizona. The average weight for these birds is x = 3.15 grams. Based on previous studies, we can assume that the weights of Allen's hummingbirds have a normal distribution, with σ = 0.32 gram.

(a) Find an 80% confidence interval for the average weights of Allen's hummingbirds in the study region. What is the margin of error? (Round your answers to three decimal places.)

(b) What conditions are necessary for your calculations? (Select all that apply.)

σ is known uniform distribution of weights normal distribution of weights n is large σ is unknown

(c) Interpret your results in the context of this problem.

The probability to the true average weight of Allen's hummingbirds is equal to the sample mean.

The probability that this interval contains the true average weight of Allen's hummingbirds is 0.80.    

There is an 80% chance that the interval is one of the intervals containing the true average weight of Allen's hummingbirds in this region.

There is a 20% chance that the interval is one of the intervals containing the true average weight of Allen's hummingbirds in this region.

The probability that this interval contains the true average weight of Allen's hummingbirds is 0.20.

(d) Find the sample size necessary for an 80% confidence level with a maximal margin of error E = 0.15 for the mean weights of the hummingbirds. (Round up to the nearest whole number.)
                            hummingbirds

Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.† Suppose a small group of 17 Allen's hummingbirds has been under study in Arizona. The average weight for these birds is x = 3.15 grams. Based on previous studies, we can assume that the weights of Allen's hummingbirds have a normal distribution, with σ = 0.38 gram. (a) Find an 80% confidence interval for the average weights of Allen's hummingbirds in the study region. What is the margin of error? (Round your answers to two decimal places.) lower limit upper limit margin of error (b) What conditions are necessary for your calculations? (Select all that apply.) n is large normal distribution of weights σ is unknown σ is known uniform distribution of weights (c) Interpret your results in the context of this problem. The probability that this interval contains the true average weight of Allen's hummingbirds is 0.80. The probability that this interval contains the true average weight of Allen's hummingbirds is 0.20. The probability to the true average weight of Allen's hummingbirds is equal to the sample mean. There is an 80% chance that the interval is one of the intervals containing the true average weight of Allen's hummingbirds in this region. There is a 20% chance that the interval is one of the intervals containing the true average weight of Allen's hummingbirds in this region. (d) Find the sample size necessary for an 80% confidence level with a maximal margin of error E = 0.12 for the mean weights of the hummingbirds. (Round up to the nearest whole number.) hummingbirds

Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.† Suppose a small group of 14 Allen's hummingbirds has been under study in Arizona. The average weight for these birds is x = 3.15 grams. Based on previous studies, we can assume that the weights of Allen's hummingbirds have a normal distribution, with σ = 0.20 gram. (a) Find an 80% confidence interval for the average weights of Allen's hummingbirds in the study region. What is the margin of error? (Round your answers to two decimal places.) lower limit upper limit margin of error (b) What conditions are necessary for your calculations? (Select all that apply.) normal distribution of weights uniform distribution of weights n is large σ is known σ is unknown (c) Interpret your results in the context of this problem. The probability to the true average weight of Allen's hummingbirds is equal to the sample mean. The probability that this interval contains the true average weight of Allen's hummingbirds is 0.20. There is an 80% chance that the interval is one of the intervals containing the true average weight of Allen's hummingbirds in this region. There is a 20% chance that the interval is one of the intervals containing the true average weight of Allen's hummingbirds in this region. The probability that this interval contains the true average weight of Allen's hummingbirds is 0.80. (d) Find the sample size necessary for an 80% confidence level with a maximal margin of error E = 0.15 for the mean weights of the hummingbirds. (Round up to the nearest whole number.) ___hummingbirds

Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.† Suppose a small group of 14 Allen's hummingbirds has been under study in Arizona. The average weight for these birds is x = 3.15 grams. Based on previous studies, we can assume that the weights of Allen's hummingbirds have a normal distribution, with σ = 0.22 gram.

(a)

Find an 80% confidence interval for the average weights of Allen's hummingbirds in the study region. What is the margin of error? (Round your answers to two decimal places.)

lower limitupper limitmargin of error

(b)

What conditions are necessary for your calculations? (Select all that apply.)

σ is knownσ is unknownn is largeuniform distribution of weightsnormal distribution of weights

(c)

Interpret your results in the context of this problem.

There is an 80% chance that the interval is one of the intervals containing the true average weight of Allen's hummingbirds in this region.There is a 20% chance that the interval is one of the intervals containing the true average weight of Allen's hummingbirds in this region.    The probability to the true average weight of Allen's hummingbirds is equal to the sample mean.The probability that this interval contains the true average weight of Allen's hummingbirds is 0.20.The probability that this interval contains the true average weight of Allen's hummingbirds is 0.80.

(d)

Find the sample size necessary for an 80% confidence level with a maximal margin of error E = 0.10 for the mean weights of the hummingbirds. (Round up to the nearest whole number.)

hummingbirds

Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.† Suppose a small group of 20 Allen's hummingbirds has been under study in Arizona. The average weight for these birds is x = 3.15 grams. Based on previous studies, we can assume that the weights of Allen's hummingbirds have a normal distribution, with σ = 0.34 gram.

(a) Find an 80% confidence interval for the average weights of Allen's hummingbirds in the study region. What is the margin of error? (Round your answers to two decimal places.)

lower limit:

upper limit:

margin of error:

(b) What conditions are necessary for your calculations? (Select all that apply.)

uniform distribution of weights

σ is unknown

n is large

normal distribution of weights

σ is known

(c) Interpret your results in the context of this problem.

The probability to the true average weight of Allen's hummingbirds is equal to the sample mean.

The probability that this interval contains the true average weight of Allen's hummingbirds is 0.80.

The probability that this interval contains the true average weight of Allen's hummingbirds is 0.20.

There is a 20% chance that the interval is one of the intervals containing the true average weight of Allen's hummingbirds in this region.

There is an 80% chance that the interval is one of the intervals containing the true average weight of Allen's hummingbirds in this region.

(d) Find the sample size necessary for an 80% confidence level with a maximal margin of error E = 0.09 for the mean weights of the hummingbirds. (Round up to the nearest whole number.)

1) A population of values has a normal distribution with μ = 97.3 and σ = 21.5 .

You intend to draw a random sample of size n = 42 .

A) Find the probability that a single randomly selected value is greater than 107.3. P(X > 107.3) =

Round to 4 decimal places.

B) Find the probability that the sample mean is greater than 107.3. P( ¯¯¯ X > 107.3) =

Round to 4 decimal places.

Answers obtained using exact z-scores or z-scores rounded to 2 decimal places are accepted.

2) Company XYZ know that replacement times for the quartz time pieces it produces are normally distributed with a mean of 11.3 years and a standard deviation of 1 years.

A) Find the probability that a randomly selected quartz time piece will have a replacement time less than 8.7 years? P(X < 8.7 years) = Enter your answer accurate to 4 decimal places.

Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

B) If the company wants to provide a warranty so that only 1.5% of the quartz time pieces will be replaced before the warranty expires, what is the time length of the warranty?

warranty = years Enter your answer as a number accurate to 1 decimal place.

Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.† Suppose a small group of 18 Allen's hummingbirds has been under study in Arizona. The average weight for these birds is x = 3.15 grams. Based on previous studies, we can assume that the weights of Allen's hummingbirds have a normal distribution, with σ = 0.30 gram.

(a) Find an 80% confidence interval for the average weights of Allen's hummingbirds in the study region. What is the margin of error? (Round your answers to two decimal places.)

(b) What conditions are necessary for your calculations? (Select all that apply.)

uniform distribution of weightsσ is unknownσ is knownn is largenormal distribution of weights

(c) Interpret your results in the context of this problem.

The probability that this interval contains the true average weight of Allen's hummingbirds is 0.20.The probability to the true average weight of Allen's hummingbirds is equal to the sample mean.    There is a 20% chance that the interval is one of the intervals containing the true average weight of Allen's hummingbirds in this region.There is an 80% chance that the interval is one of the intervals containing the true average weight of Allen's hummingbirds in this region.The probability that this interval contains the true average weight of Allen's hummingbirds is 0.80.

(d) Find the sample size necessary for an 80% confidence level with a maximal margin of error E = 0.12 for the mean weights of the hummingbirds. (Round up to the nearest whole number.)
hummingbirds

Consider the field experiment with politicians that we discussed in the lecture. The experiment is based on a modified dictator game. In Treatment 1 (i.e. T1), nature plays with high probability (equal to 0.8) and randomly assigns the endowment either to the politician-dictator or to the recipient. The politician--dictator plays with complementary probability (and knows, when making a decision, that this decision will be implemented); in contrast, a recipient who receives zero (or the full endowment), will not know whether the dictator or nature is responsible. In Treatment 2 (i.e. T2), the probability that nature intervenes is very low (equal to 0.1). Final results are published and seen by all participants in the room.

Suppose a politician-dictator wants to keep 90% of the given endowment and give 10% of the endowment to the recipient. She knows there is a norm of equal-sharing and not giving 50% would cost her a social image loss. Is she still able to give 10% of the endowment without revealing her identity to anyone in the experiment?