Ch. 8/9 Review

A 95% confidence interval for p, the proportion of all shoppers at a large grocery store who purchase cookies, was found to be (0.236, 0.282).

The point estimate and margin or error for this interval are:

• Point estimate = 0.259; Margin of Error = 0.046

• Point estimate = 0.259; Margin of Error = 0.023

• Point estimate = 0.046; Margin of Error = 0.023

• Point estimate = 0.023; Margin of Error = 0.046

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CCSS.HSS.IC.B.4

A marketing company discovered the following problems with a recent poll:

I. Some people refused to answer questions.

II. People without telephones could not be in the sample.

III. Some people never answered the phone in several calls.

Which of these sources is included in the ±2% margin of error announced for the poll?

• Only source I

• Only source II

• Only source III

• All three sources of error

• None of these sources of error

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CCSS.HSS.IC.B.4

Which of the following statements are true?

I. Both Normal distributions and t distributions are unimodal and symmetric.

II. A t distribution has more area in its tails than a Normal distribution.

III. A t distribution is used when we use the sample standard deviation in place of the population standard deviation.

• I only

• I and II only

• I and III only

• II and III only

• I, II, and III

A local planning board in Elm County took a random sample of residents and asked them, “Are you in favor of a $200 annual increase in property taxes to pay for improvements in the local roads and sidewalks?” The 95% confidence interval for the proportion of residents who responded “Yes” was calculated to be (0.49, 0.59). Which statement is correct?

• About 95% of all people in this survey reported a proportion between 0.49 and 0.59.

• About 95% of the time, the true proportion of Elm County residents who approve of the tax increase is between 0.49 and 0.59.

• There’s a 95% chance that the interval contains the proportion of the sample of Elm County residents that approve of the tax increase.

• About 5% of all random samples like ours fail in producing a confidence interval that captures the true proportion of Elm County residents who approve of the tax increase.

• If the true proportion of Elm County residents who approve of the tax increase is actually 0.59, then the interval (0.49, 0.59) is not likely to emerge from a random sample.

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CCSS.HSS.IC.B.4

When the sample size is small, why is it necessary to inspect a graph of the sample data when calculating a confidence interval for a population mean?

• To verify that the population can be considered approximately Normal.

• To verify that the Central Limit Theorem applies.

• So we can use a z critical value instead of a t critical value

• So that the statistic is an unbiased estimator

• So the formula for the standard error is approximately correct

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CCSS.HSS.IC.B.4

Which of the following are ways to decrease the margin of error for a confidence interval?

• decrease the sample size

• decrease the confidence level

• increase the population size

• increase the point estimate

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CCSS.HSS.IC.B.4

A marketing consultant wants to estimate the proportion of all shoppers at a certain mall who make at least one purchase. He stands at a mall exit for two hours on a weekday afternoon and flips a coin each time a shopper leaves. If the coin comes up heads, he asks them if they have made any purchases during this visit. After two hours, he has 132 responses, 104 of whom made a purchase. Which condition for constructing a confidence interval for a proportion has he failed to satisfy?

• The data is a random sample from the population of interest.

• The number of successes is at least 10

• The number of failures is at least 10

• The sample size is at least 30

• The sample is less than 10% of the population

After once again losing a football game to the archrival, a college’s alumni association conducted a survey to see if alumni were in favor of firing the coach. An SRS of 100 alumni from the population of all living alumni was taken, and 64 of the alumni in the sample were in favor of firing the coach. Suppose you wish to see if a majority of living alumni are in favor of firing the coach. The appropriate test statistic is

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CCSS.HSS.IC.B.4

The principal of a large high school wants to know if students spend more than 1 hour doing homework per night, on average. To investigate, he surveys a random sample of 100 students and use the results to test the following hypotheses: µ = 1 versus : µ > 1 where µ is the mean amount of homework done per night for all the students at the school.

In the sample, mean time spent on homework per night was 1.23 hours, resulting in a P-value of 0.18. Interpret the P-value in context.

• Assuming that the true average amount of time spent doing homework per night is 1 hour, there is an 18% probability that the null hypothesis is true.

• Assuming that the true average amount of time spent doing homework per night is 1 hour, there is an 18% probability that the sample mean would be 1.23 or higher by chance alone.

• Assuming that the true average amount of time spent doing homework per night is more than 1 hour, there is an 18% probability that the sample mean would be 1.23 or higher by chance alone.

• Only 18% of the students do homework for more than an hour per night on average.

• There is an 18% chance that the average amount of time spent on homework per night is greater than 1 hour.

We want to test H_o: µ = 1.5 vs. H_a: µ ≠ 1.5 at the alpha = 0.05. A 95% confidence interval for µ calculated from a random sample is 1.55 to 3.65. Based on this finding we

• fail to reject H_o.

• reject H_o.

• cannot make any decision at all because the value of the test statistic is not available.

• cannot make any decision at all because the distribution of the population is unknown.

• cannot make any decision at all because (1.55, 3.65) is only a 95% confidence interval for µ.

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CCSS.HSS.IC.B.4