Scores on the sat college entrance test in a recent year wer

Assignment Help Applied Statistics

Reference no: EM13811108

Question 1:

A study of voting chose 663 registered voters at random shortly after an election. Of these 72% said they had voted in the election. Election records show that only 56% of all registered voters voted in the election. The boldface number (that is, 72%) is a

Question 1 option:

parameter

sampling distribution

statistic

Question 2:

Question 2 options:

parameter

sampling distribution

statistic

Question 3:

Annual returns on the more than 5000 common stocks available to investors vary a lot. In a recent year, the mean return was 8.3% and the standard deviation of returns was 28.5%. The law of large numbers says that

Question 3 options:

if you invest in a large number of stocks chosen at random, your average return will have approximately a Normal distribution.

you can get an average return higher than the mean 8.3% by investing in a large number of stocks.

as you invest in more and more stocks chosen at random, your average return on these stocks will get closer and closer to 8.3%.

Question 4:

Scores on the SAT college entrance test in a recent year were roughly Normal with mean 1026 and standard deviation 209. You choose an SRS of 100 students and find their average SAT score, x-bar. If you repeatedly take SRS of 100 students and calculate x-bar for each sample, then the mean of the sampling distribution you create will be closest to

Question 4 options:

102.6

1026

10.26

Question 5:

Scores on the SAT college entrance test in a recent year were roughly Normal with mean 1026 and standard deviation 209. You choose an SRS of 100 students and find their average SAT score, x-bar. If you repeatedly take SRS of 100 students and calculate x-bar for each sample, then the standard deviation of the sampling distribution you create will be closest to

Question 5 options:

20.9

209

6.92

Question 6:

Scores on the SAT college entrance test in a recent year were roughly Normal with mean 1026 and standard deviation 209. You choose an SRS of 100 students and find their average SAT score, x-bar. If you repeatedly take SRS of 100 students and calculate x-bar for each sample, then you will create a sampling distribution for the average scores from samples of 100.

Using this sampling distribution you create, what is the probability that you randomly select one sample and the average score, x-bar, for this sample is greater than 1050?

Question 6 options:

0.4404

0.8749

0.1251

Question 7:

The number of hours a light bulb burns before failing varies from bulb to bulb. The distribution of burnout times is strongly skewed to the right. The central limit theorem says that

Question 7 options:

the sampling distribution of average burnout time of a large number of bulbs has a distribution that is close to Normal.

as we look at more and more bulbs, their average burnout time gets closer and closer to the mean for all bulbs of this type.

the average burnout time of a large number of bulbs has a distribution of the same shape (strongly skewed) as the distribution for the individual bulbs.

Question 8:

A machine manufactures parts whose diameters vary according to the Normal distribution with mean 40.150 mm and standard deviation 0.003mm. An inspector measures a random sample of 4 parts. The probability that the average (x-bar) diameter of these 4 parts is less than 40.148 mm is about

Question 8 options:

0.9082

0.0918

0.2510

Question 9:

What is the difference between a statistic and a parameter?

Question 9 options:

A statistic is a value that comes from the sample and a parameter is a value that comes from the population.

A statistic is a value that comes from the population and a parameter is a value that comes from the sample.

A statistic is a value that is usually unknown and a parameter is a value that is calculated from sample data.

Question 10:

The sampling distribution of a statistic is

Question 10 options:

the extent to which the sample results differ systematically from the truth.

the probability that we obtain the statistic in repeated random samples.

the distribution of values taken by a statistic in all possible samples of the same size from the same population.

Question 1:

To give a 99% confidence interval for a population mean mu, you would use a critical value of

Question 1 option:

z* = 2.576

z* = 1.645

z* = 1.96

Question 2:

Sulfur compounds cause "off-odors" in wine, so winemakers want to know the odor threshold, the lowest concentration of a compound that the human nose can detect. The odor threshold for dimethyl sulfide (DMS) in trained wine tasters is about 25 micrograms per liter of wine. The untrained noses of consumers may be less sensitive, however. Here are the DMS odor thresholds for 10 untrained students:

31 31 43 36 23 34 32 30 20 24

Assume that the standard deviation of the odor threshold for untrained noses is known to be sigma = 7 micrograms per liter.

Compute a 95% confidence interval for the true mean odor threshold for all consumers.

Question 2 options:

(26.06 micrograms/liter, 34.74 micrograms/liter)

(29.03 micrograms/liter, 31.77 micrograms/liter)

(26.76 micrograms/liter, 34.04 micrograms/liter)

Question 3:

A sample of 31 seventh-grade girls in a Midwest school district took an IQ test and their scores were reported. Their scores were used to compute a 99% confidence interval for the average IQ score of all seventh-grade girls in the school district. The 99% confidence interval computed was (92, 115).

Interpret this interval.

Question 3 options:

99% of all seventh-grade girls scored between 92 and 115 on the IQ test.

99% of all samples of 31 seventh-grade girls would produce a mean IQ test score between 92 and 115.

We are 99% confident that the true mean IQ score for all seventh-grade girls in this school district lies somewhere between 92 and 115.

Question 4:

A confidence level, C, gives

Question 4 options:

the probability that the interval will capture the true parameter value in repeated samples.

a guarantee that the interval computed captures the true paramter value.

the probability that the statistic is a good estimate of the parameter.

Question 5:

The mean score of adult men on a psychological test that measures "masculine stereotypes" is 4.88. A researcher studying hotel managers suspects that successful managers score higher than adult men in general. A random sample of 48 managers of large hotels has mean x-bar = 5.91. The null and alternative hypotheses that would be used to test this claim are

Question 5 options:

H_0: mu = 4.88

H_a: mu > 4.88

H_0: mu = 5.91

H_a: mu does not equal 5.91

H_0: mu = 4.88

H_a: mu < 4.88

H_0: mu > 4.88

H_a: mu = 4.88

Question 6:

The mean score of adult men on a psychological test that measures "masculine stereotypes" is 4.88. A researcher studying hotel managers suspects that successful managers score higher than adult men in general. A random sample of 48 managers of large hotels has mean x-bar = 5.91. Assume the population standard deviation is sigma = 3.2.

Using the null and alternative hypotheses that you set up in problem 5, the value of the test statistic for this hypothesis test is

Question 6 options:

z = 2.23

z = 0.32

z = 1.54

Question 7:

The mean score of adult men on a psychological test that measures "masculine stereotypes" is 4.88. A researcher studying hotel managers suspects that successful managers score higher than adult men in general. A random sample of 48 managers of large hotels has mean x-bar = 5.91. Assume the population standard deviation is sigma = 3.2.

Given the null and alternative hypotheses that you set up in problem 5, what conclusion would you make based on alpha = 0.05?

Question 7 options:

Since the p-value is greater than the alpha level, we would fail to reject the null hypothesis, indicating that this sample of men does not provide evidence that managers score higher than adult men, in general.

Since the p-value is less than the alpha level, we would reject the null hypothesis, indicating that this sample of men does provide evidence that managers score higher than adult men, in general.

Since the test statistic value is greater than the alpha level, we cannot make a conclusion for this test based on this sample. We would need to take many more samples of 48 men before we can make a conclusion.

Question 8:

If a z statistic (test statistic) has value z = -1.30, then the two-sided P-value is

Question 8 options:

0.9032

0.1936

0.0968

Question 9:

A researcher performs a hypothesis test to test the claim that for a particular manufacturer, the mean weight of cereal in its 18 ounce boxes is less than 18 ounces. The computer display is shown below. Using a significance level of 0.05, what is your conclusion from these data?

N	MEAN	STDEV	SE MEAN	T	P VALUE
weight	21	17.883	0.212	0.0463	-2.53	0.01

Question 9 options:

Because the p-value of 0.01 is smaller than the significance level of 0.05, we reject the null hypothesis. There is sufficient evidence to suggest that the mean weight is less than 18 ounces.

Because the p-value of 0.01 is smaller than the signifcance level of 0.05, we reject the null hypothesis. There is sufficient evidence to suggest that the mean weight is equal to 18 ounces.

Because the p-value of 0.01 is smaller than the significance level of 0.05, we fail to reject the null hypothesis. There is insufficient evidence to suggest that the mean weight is less than 18 ounces.

Since x-bar = 17.883, which is clearly less than 18, we should reject the null hypothesis.

Question 10:

A can of Coke displays the statement "12 FL OZ." Let mu denote the mean amount of Coke in all cans. People who drink Coke would probably feel cheated if it turned out that the mean amount of Coke in all cans is less than the claimed value. We carry out a test of the hypotheses: H_0: mu = 12, H_a: mu < 12

A random sample of 30 cans of Coke was selected and the amount of fluid was measured. The mean amount of Coke in the sample was 11.90 ounces and the corresponding p-value was 0.20. Assume we want to use alpha = 0.05. What should we conclude?

Question 10 options:

There is strong evidence to suggest that the mean amount of Coke in all cans is less than 12 ounces since the sample mean was 11.90 ounces and the p-value is larger than alpha.

There is insufficient evidence to suggest that the mean amount of Coke in all cans is less than 12 ounces since the p-value is larger than our significance level.

There is evidence to suggest that all cans of Coke contain less than 12 ounces.

The results are statistically significant since x-bar = 11.90 is less than mu = 12.

Question 1:

A certain organization wants to know the opinion of the adult residents, 18 years of age or older, of Johnson City about banning smoking in all public restaurants. Specifically, they want to know what proportion of those adults are in favor of banning smoking in all public restaurants. They organize a survey and select a simple random sample of 200 adult residents and ask them "Do you think that smoking should be banned from all public restaurants?" In the sample, 140 people answered "yes." What is the population of interest in this case?

Question 1 options:

The 200 adults surveyed.

The people that belong to the organization.

All people of Johnson City.

All adult residents (18 years or older) of Johnson City.

Question 2:

Question 2 options:

The 200 adults surveyed

The people that belong to the organization

All people of Johnson City

All adult residents (18 years or older) of Johnson City

Question 3:

Question 3 options:

0.3

0.4

0.6

0.7

Question 4:

Question 4 options:

(0.624, 0.776)

(0.636, 0.764)

(0.641, 0.759)

(0.584, 0.816)

Question 5:

If you were to construct a 99% confidence interval for the true proportion of all adult Johnson City residents that favor banning smoking in restaurants, how would this interval compare to the 95% interval you computed in question 4, if p-hat and n remained the same?

Question 5 options:

The 99% confidence interval would be wider; that is, it would have a larger margin of error.

The 99% confidence interval would be narrower; that is, it would have a smaller margin of error.

There would be no change between the two intervals by increasing the confidence level from 95% to 99%.

Question 6:

From time to time police set up roadblocks to check cars to see if the safety inspection is up to date. At one such roadblock they issued tickets for expired inspection stickers to 22 of 628 cars they stopped. Based on the results at this roadblock, they construct a 95% confidence interval for the true proportion of all drivers with expired inspections and find it to be (2%, 5%). Interpret this interval.

Question 6 options:

95% of the cars stopped have between 2% and 5% expired safety inspection stickers.

We are 95% confident that the true proportion of all automobiles in the region whose safety inspections have expired lies somewhere between 2% and 5%.

We are 95% confident that the sample proportion of all automobiles in the region whose safety inspections have expired lies somewhere between 2% and 5%.

Question 7:

A survey of 865 voters in one state reveals that 408 favor approval of an issue before the legislature. Construct a 95% confidence interval for the percentage of all voters in the state who favor approval.

Question 7 options:

(44.4%, 50.0%)

(43.8%, 50.5%)

(43.1%, 51.2%)

(46.9%, 47.5%)

Question 8:

Use the following information to answer the next three questions.

An opinion poll asks a SRS of 100 college seniors how they view their job prospects. In all, 53 say "Good." Does this poll give us evidence that more than half of all college seniors think their job prospects are good?

The hypotheses used to conduct this test are

Question 8 options:

Ho: p = 0.5

Ha: p > 0.5

Ho: p > 0.5

Ha: p = 0.5

Ho: p = 0.5

Ha: p does not equal 0.5

Ho: p = 0.5

Ha: p < 0.5

Question 9:

The value of the test statistic for this test is about

Question 9 options:

z = 12

z = 6

z = 1.4

z = 0.6

Question 10:

What conclusion would you make for this test if alpha = 0.05?

Question 10 options:

Since the p-value is less than alpha, we would decide to reject the null hypothesis, indicating evidence that the true proportion of seniors that think their job propsects are "Good" is more than 50%.

Since the p-value is greater than alpha, we would decide to fail to reject the null hypothesis, indicating evidence that the true proportion of seniors that think their job propsects are "Good" equals 50%.

Reference no: EM13811108

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