1. A Type II error occurs in hypothesis testing when we
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[removed] |
reject the alternative hypothesis and the alternative hypothesis is true. |
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[removed] |
fail to reject the null hypothesis and the null hypothesis is not true. |
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[removed] |
fail to reject the alternative hypothesis and the alternative hypothesis is not true. |
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[removed] |
reject the null hypothesis and the null hypothesis is true. |
5 points
1. The average natural gas bill for a random sample of 21 homes in the 19808 zip code during the month of March was $311.90 with a sample standard deviation of $51.60. The 90% confidence interval around this sample mean is ________.
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[removed] |
($285.98, $337.82) |
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[removed] |
($292.46, $331.34) |
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[removed] |
($266.82, $356.98) |
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[removed] |
($278.09, $345.71) |
5 points
1. The National Center for Education Statistics would like to test the hypothesis that the proportion of Bachelor’s degrees that were earned by women equals 0.60. A random sample of 140 college graduates with Bachelor degrees found that 75 were women. The National Center for Education Statistics would like to set α = 0.10. The conclusion for this hypothesis test would be that because the absolute value of the test statistic is
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[removed] |
less than the absolute value of the critical value, we cannot conclude that the proportion of Bachelor’s degrees that were earned by women does not equals 0.60. |
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[removed] |
less than the absolute value of the critical value, we can conclude that the proportion of Bachelor’s degrees that were earned by women does not equals 0.60. |
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[removed] |
more than the absolute value of the critical value, we can conclude that the proportion of Bachelor’s degrees that were earned by women equals 0.60. |
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[removed] |
more than the absolute value of the critical value, we can conclude that the proportion of Bachelor’s degrees that were earned by women does not equals 0.60. |
5 points
1. CoStar would like to test if the vacancy rate for “warehouse” stores is more than 8%. The correct hypothesis statement to test for this vacancy rate would be
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[removed] |
H0: π ≤ 0.08; H1: π > 0.08.
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[removed] |
H0: π = 0.08; H1: π < 0.08. |
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[removed] |
H0: π = 0.08; H1: π ≠ 0.08. |
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[removed] |
H0: π ≠ 0.08; H1: π = 0.08. |
5 points
1. “Gamers” are described as people who regularly play video games. A random sample of 120 people was selected and 66 were found to be male. The 90% confidence interval for the proportion based on this sample is ________.
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[removed] |
(0.390, 0.710) |
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[removed] |
(0.527, 0.573) |
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(0.475, 0.625) |
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[removed] |
(0.413, 0.687) |
5 points
1. The National Center for Education Statistics would like to test the hypothesis that the proportion of Bachelor’s degrees that were earned by women equals 0.60. A random sample of 140 college graduates with Bachelor degrees found that 75 were women. The National Center for Education Statistics would like to set α = 0.10. The test statistic for this hypothesis test would be ________.
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[removed] |
zp = -1.21 |
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[removed] |
zp = 2.17 |
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[removed] |
zp = -1.55 |
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[removed] |
zp = 1.36 |
5 points
1. A random sample of 130 mortgages in the state of Mississippi was randomly selected. From this sample, 14 were found to be delinquent on their current payment. Using this sample, the point estimate for the proportion of delinquent mortgages in Mississippi is ________.
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[removed] |
0.127 |
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[removed] |
0.119 |
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[removed] |
0.084 |
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[removed] |
0.108 |
5 points
1. Many hotel chains that offer free Wi-Fi service to their customers have experienced increasing demand for Internet bandwidth and increasing costs. Marriott International would like to test the hypothesis that the proportion of customers that are carrying two Wi-Fi devices exceeds 0.40. A random sample of 130 Marriott customers found that 63 have two Wi-Fi devices. Marriott International would like to set α = 0.05. The conclusion for this hypothesis test would be that because the test statistic is
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[removed] |
less than the critical value, we can conclude that the proportion of Marriott customers that carry two Wi-Fi devices is less than 0.40. |
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[removed] |
more than the critical value, we can conclude that the proportion of Marriott customers that carry two Wi-Fi devices is less than 0.40. |
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[removed] |
more than the critical value, we can conclude that the proportion of Marriott customers that carry two Wi-Fi devices is greater than 0.40. |
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[removed] |
less than the critical value, we can conclude that the proportion of Marriott customers that carry two Wi-Fi devices is greater than 0.40. |
5 points
1. Dr. Jordan operates a dental practice with 315 patients and she would like to estimate the proportion of her patients that have dental insurance from their employers. A random sample of 110 patients found that 40 had insurance coverage. The 95% confidence interval to estimate the proportion of patients with dental insurance is ________.
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[removed] |
(0.291, 0.437) |
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[removed] |
(0.346, 0.383) |
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[removed] |
(0.327, 0.401) |
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[removed] |
(0.227, 0.501) |
5 points
1. A benefit of point estimates is that they provide information about their accuracy.
[removed] True
[removed] False
5 points
1. The purpose of generating a confidence interval for the mean is to provide an estimate for the value of the population mean.
[removed] True
[removed] False
5 points
1. A random sample of 130 mortgages in the state of Mississippi was randomly selected. From this sample, 14 were found to be delinquent on their current payment. The 98% confidence interval for the proportion based on this sample is ________.
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[removed] |
(0.015, 0.201) |
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[removed] |
(0.029, 0.188) |
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[removed] |
(0.063, 0.153) |
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[removed] |
(0.036, 0.180) |
5 points
1. A survey by American Express Spending reported that the average amount spent on a wedding gift for a close family member is $166. A random sample of 45 people who purchased wedding gifts for a close family member spent an average of $160.50. Assume that the population standard deviation is $38. Use a 95% confidence interval to test the validity of this report and choose the one statement that is correct.
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[removed] |
Because this confidence interval does include $166, the report by American Express Spending is not validated. |
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[removed] |
Because this confidence interval does not include $166, the report by American Express Spending is not validated. |
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[removed] |
Because this confidence interval does include $166, the report by American Express Spending is validated. |
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[removed] |
Because this confidence interval does not include $166, the report by American Express Spending is validated. |
5 points
1. When calculating a confidence interval for the mean using a sample size of less than 30 with an unknown population standard deviation, the ________ is used.
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[removed] |
normal distribution |
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[removed] |
binomial distribution |
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[removed] |
Poisson distribution |
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[removed] |
Student’s t-distribution |
5 points
1. The average natural gas bill for a random sample of 26 homes in the 19808 zip code during the month of March was $305.30 with a sample standard deviation of $46.50. The margin of error for a 98% confidence interval around this sample mean is ________.
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[removed] |
$25.20 |
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[removed] |
$20.15 |
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[removed] |
$18.69 |
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[removed] |
$22.66 |
5 points
1. Proportion data follow the ________ distribution, which can be approximated by the ________ distribution under the following conditions: nπ ≥ 5 and n(1 – π) ≥ 5.
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[removed] |
exponential, binomial |
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[removed] |
Poisson, normal |
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[removed] |
binomial, normal |
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[removed] |
normal, exponential |
5 points
1. If the hypothesis test is used in a quality control setting, the alternative hypothesis takes the stance that the process is not behaving as it should and needs some sort of attention.
[removed] True
[removed] False
5 points
1. The Retail Advertising and Marketing Association would like to estimate the average amount of money that a person spends for Mother’s Day with a 99% confidence interval and a margin of error within plus or minus $12. Assuming the standard deviation for spending on Mother’s Day is $36, the required sample size is ________.
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[removed] |
97 |
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[removed] |
72 |
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[removed] |
108 |
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[removed] |
60 |
5 points
1. The critical z-score for a 98% confidence level is ________.
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[removed] |
±2.575 |
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[removed] |
± 1.96 |
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[removed] |
± 1.28 |
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[removed] |
± 2.33 |
5 points
1. A random sample of 30 business students required an average of 43.4 minutes to complete a statistics exam. Assume that the population standard deviation to complete the exam was 8.7 minutes. The 99% confidence interval around this sample mean is ________.
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[removed] |
(37.1, 49.8) |
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[removed] |
(36.3, 50.5) |
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[removed] |
(40.5, 46.3) |
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[removed] |
(39.3, 47.5) |
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