Question

# please answer all the questions 1. A Confidence Interval (CI) for a population characteristic is: An...

1. A Confidence Interval (CI) for a population characteristic is:

1. An alternative to a Point Estimate
2. An interval of plausible values for a population characteristic
3. An interval of ranges for a large sample that is unbiased
4. A population of critical values

Directions: Find each specified point estimate or confidence interval.

2. A survey of teachers reports that a point estimate for the mean amount of money spent each week on lunch is \$18.00. If the margin of error for a 95% confidence interval for the mean amount of money spent each week on lunch by all teachers is \$1.70, construct a 95% confidence interval for the mean amount of money spent each week on lunch for all teachers.

1. 16.30<μ<19.70
2. 18.00<μ<19.70
3. 16.30<μ<16.30
4. 18.30<μ<19.70

Directions: Calculate the margin of error (E) of a confidence interval for the population mean at the given level of confidence.

3.      n = 134,     σ = 0.27,      c = 0.99

1. 0.06
2. 0.90
3. 0.99
4. 0.27

Directions: Construct a confidence interval for the population mean at the given level of confidence.

4.   n = 64,     σ = 8.01,      c = 0.90,      x = 90.40

1. 90.75<μ<92.05
2. 64.00<μ<90.40
3. 88.75<μ<92.05
4. 64.75<μ<92.05

Directions: Calculate the minimum sample size needed to construct a confidence interval with the desired characteristics.

5.    E = 2,        σ = 12.10,       c = 0.90

1. Need a minimum sample size of 90
2. Need a minimum sample size of 30
3. Need a minimum sample size of 10
4. Need a minimum sample size of 100

1. A Confidence Interval (CI) for a population characteristic is:

1. An interval of plausible values for a population characteristic

$2.~95\%~C.I.~for~mean~amount~of~money~spent~each~week~on~lunch~for\\ all~teachers=(18.00-1.70,~18.00+1.70)=(16.3,~19.7).\\ Option~A:~16.30<\mu<19.70.\\ 3.~Margin~of~error=z_{0.005}\times \frac{\sigma}{\sqrt{n}}=0.06\\ where,~z_{0.005}=2.5758\\ Option~A:~0.06.\\ 4.~90\%~C.I.~for~\mu~is~\\ \bar{x}-z_{0.05}\times \frac{\sigma}{\sqrt{n}}< \mu< \bar{x}+z_{0.05}\times \frac{\sigma}{\sqrt{n}}\\ or,~90.40-1.6449*8.01/8< \mu< 90.40+1.6449*8.01/8\\ or,~88.75< \mu< 92.05\\ where,~z_{0.05}=1.6449.\\ Option~C.~88.75< \mu< 92.05.\\ 5.~E=z_{0.05}\times \frac{\sigma}{\sqrt{n}}\Rightarrow n=(1.6449*12.10/2)^2\approx 100.\\ Option~D.$

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