ua casselman stats 260 exam 2

An analyst at Carnival Cruise Lines is interested in the time it takes for a room service order to go from time of order to time of delivery. In order to do this, she takes a sample of 49 random guests and finds that their average time to delivery was 11.2 minutes and that the sample standard deviation (s) was 3.5 minutes. Calculate a 95% Confidence Interval around the mean.

(10.195, 12.206)

What is the appropriate t-statistic for a 90% Confidence Interval from a sample of size 14?

1.771

Suppose GRE (Graduate Admissions) scores are known to be normally distributed with a mean (μ) of 1500 and a population standard deviation (σ) of 200. At what score will the top 10.03% (P=.1003) of all GRE test takers make that score or more?

1756

Approximately ___________ of the data falls between ± 1 Standard Deviation of the mean.

68%

Approximately ___________ of the data falls between ± 2 Standard Deviation of the mean.

95%

Approximately ____________of the data falls between ± 3 Standard Deviation of the mean.

99.7%

If P(-Z <= X <= Z) = .7580, ±Z= HINT: This is the middle 75.8% of the area of the Normal Distribution, between what two Z values will it fall? (Round to the nearest Z-value).

+/- 1.17

P(0 <= Z <= .13)

.0517

Suppose GRE (Graduate Admissions) scores are known to be normally distributed with a mean (μ) of 1500 and a population standard deviation (σ) of 200. With a score of 1800 or more?

.0668

Suppose GRE (Graduate Admissions) scores are known to be normally distributed with a mean (μ) of 1500 and a population standard deviation (σ) of 200. What is the probability that someone that takes the GRE will achieve a score of 1300 or less?

.1587

For a Binomial Distribution where n=5 and p=.2, calculate P(X=2)?

.2048

P(Z >= 0.13)

.4483

P(Z <= 0)

.5000

Suppose GRE (Graduate Admissions) scores are known to be normally distributed with a mean (μ) of 1500 and a population standard deviation (σ) of 200. With a score between 1300 and 1600?

.5328

P(Z <= 0.13)

.5517

P(-1.2 <= Z <= .58)

.6039

If P(X <= Z) = .7995, Z=

.84

Which of the following is NOT true about the t-distribution? (2 points) a. It is slightly larger (fatter) than the Standard Normal (Z) distribution b. It was discovered by a brewmaster at the Guiness brewery in Ireland c. The penalty for using (s) instead of (σ) is reflected in the degrees of freedom d. It results in smaller intervals than when calculating confidence intervals with the same level of confidence as those using the standard normal (Z) distribution. e. As the sample size gets larger, the t-distribution approaches (gets closer to) the corresponding standard normal (Z) distribution value.

d. It results in smaller intervals than when calculating confidence intervals with the same level of confidence as those using the standard normal (Z) distribution.

Which of the following is NOT included when writing a conclusion in the language of the problem (LOP)? (2 pts) a. The Level of Confidence b. The Parameter being estimated c. The Population to which we generalize to d. The Standard Error of the mean e. The Confidence Interval

d. the standard error of the mean

A runner trains by running either 3, 5, or 7 miles during a training run. The discrete distribution of runs is as follows: Miles Probability 3 .60 5 .20 7 .20

expected # of miles on typical run = 4.2 variance of the # of miles on typical run = none probability of less than 5 miles = .60

According to a published report by the M&M/Mars Corporation in 2008, the proportion of Red M&M candies was around 15%. Jill gets a snack bag of M&M's for Halloween. Assuming that there are 9 M&M's in the snack bag, the following binomial distribution where n=9 and p=.15 would apply: P(X=k) P(X<=k) 0 0.2316 0.2316 1 0.3679 0.5995 2 0.2597 0.8591 3 0.1069 0.9661 4 0.0283 0.9944 5 0.0050 0.9994 6 0.0006 1.0000 7 0.0000 1.0000 8 0.0000 1.0000 9 0.0000 1.0000

expected value of red = 1.35 variance of red = 1.148 probability of 4 red in a bag = .0283 probability of 2 or more red = .4005 probability of more than 1 but less than 3 red = .2597 probability of 3 NOT RED = .0006

A Bernoulli trial consists of more than 2 possible outcomes

false

Degrees of Freedom are associated with the Z statistic

false

If the confidence level is .88 (88%), then the level of risk (α) = .06

false

If you knew both the population standard deviation (σ) and the sample standard deviation, you would use the t-distribution to calculate confidence intervals.

false

The one-year postgraduate salary of statistics students is known to have a Normal Distribution. From a sample of 36 statistics students, the average salary was found to be $70,000. It is known that the population standard deviation (σ) of salaries of statisticians one-year post graduation is $8,000. find standard error find 90% confidence interval It has been hypothesized that the true mean one-year postgraduate salary of statistics students is $68,400. Do you find this to be a plausible value given the result of your confidence interval (C.I.)?

standard error of mean(standard deviation of sampling distribution) = $1,333.33 90% confidence interval around the mean = ($67,806.70, $72,193.30) yes, bc the hypothesized value of $68,400 falls in my 90% C.I.

A Standard Normal Distribution is one that has a mean of 0 and a standard deviation of 1.

true

As the sample size gets smaller, the width of a confidence interval gets larger.

true

If the underlying distribution of the variable of interest is already normal, than we can assume that the distribution of sample means is also normal.

true

The Central Limit Theorem states that the sampling distribution of the mean is approximately normal no matter what the underlying distribution is as long as the sample size is large enough (n >= 30)

true

The standard deviation (i.e. "Standard Error") of a sampling distribution where the population standard deviation is unknown is s / root n

true

X is at the exact center of a confidence interval calculated from a sample

true