Exam 12: Inference About a Population

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The test statistic to test hypotheses about the population variance has a chi-squared distribution with n - 1 degrees of freedom when the population random variable has a(n) ____________________ distribution.

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Concert Tickets A simple random sample of 100 concert tickets was drawn from a normal population.The mean and standard deviation of the sample were $120 and $25, respectively. -{Concert Tickets Narrative} Explain how to use the confidence interval to test the hypotheses at α\alpha = 0.10.

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Two uses for the population variance are to measure risk and consistency.

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The degrees of freedom for the test statistic for μ\mu when σ\sigma is unknown is:

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For a 99% confidence interval of the population mean based on a sample of n = 25 with s = 0.05, the critical value of t is:

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Single Mothers' Ages A random sample of 10 single mothers was drawn from a Obstetrics Clinic.Their ages are 22, 17, 27, 20, 23, 19, 24, 18, 19, and 24 years. -{Single Mothers' Ages Narrative} Test to determine if we can infer at the 5% significance level that the population mean is not equal to 20.

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Generally speaking, mass marketing has given way to target marketing, which focuses on satisfying the demands of a particular segment of the entire market.

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Hourly Fees A random sample of 15 hourly fees for car washers (including tips) was drawn from a normal population.The sample mean and sample standard deviation were Hourly Fees A random sample of 15 hourly fees for car washers (including tips) was drawn from a normal population.The sample mean and sample standard deviation were   = $14.9 and s = $6.75. -{Hourly Fees Narrative} Can we infer at the 5% significance level that the population mean is greater than 12, assuming that you know the population standard deviation is equal to 6.75? = $14.9 and s = $6.75. -{Hourly Fees Narrative} Can we infer at the 5% significance level that the population mean is greater than 12, assuming that you know the population standard deviation is equal to 6.75?

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The t-distribution is used to develop a confidence interval estimate of the population mean when the population standard deviation is unknown.

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Mystic Pizza Mystic Pizza in Mystic, Connecticut, advertises that they deliver your pizza within 15 minutes of placing an order or it is free.A sample of 25 customers is selected at random.The average delivery time in the sample was 13 minutes with a sample standard deviation of 4 minutes. -{Mystic Pizza Narrative} What is the required condition of the technique used in the previous question?

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The test statistic for p is approximately normal when np and n(1 - p) are both ____________________.

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Mass marketing is especially effective for ____________________ goods such as gasoline, which are difficult to differentiate from the competition except through price and availability.

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The t-distribution assumes that the population is normally distributed.

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The chi-squared test and estimator of the population variance ____________________(are/are not) valid if the population is slightly to moderately nonnormal.

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In a hypothesis test for the population variance, the hypotheses are H0: σ\sigma 2 = 100 vs.H1: σ\sigma 2 \neq 100.If the sample size is 15 and the test is being carried out at the 10% level of significance, the rejection region is:

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The statistic The statistic   when the sampled population is normal is Student t-distributed with n degrees of freedom. when the sampled population is normal is Student t-distributed with n degrees of freedom.

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Mass Marketing refers to the mass production and marketing by a company of a single product for the entire market.

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Sinus Drug A company claims that 10% of the users of a certain sinus drug experience drowsiness.In clinical studies of this sinus drug, 81 of the 900 subjects experienced drowsiness -{Sinus Drug Narrative} Compute the p-value of the test.

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If the lower and upper confidence limits of the population mean μ\mu , using a sample of size 100, are 225 and 280, respectively, then the lower and upper confidence limits of the total of a population of size 3000 are respectively

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A left-tailed area in the chi-squared distribution equals 0.95.For 6 degrees of freedom the critical value equals 12.592.

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