A study of commuting times reports the travel times to work of a random sample of 20 employed adults in New York State. The mean isx = 31.25 minutes and the standard deviation is s = 21.88 minutes. What is the standard error of the sample mean, x?

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a. 1.094b. 4.893c. 6.988d. 21.88

Question 2

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True or False: The standard Normal distribution is more spread out than a t distribution with 6 degrees of freedom.

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a. Trueb. False

Question 3

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In order to estimate with a 90% confidence interval using the formula

x t*[s/(n)]

whatvalue should we use for t* if we plan to take a sample of size n = 17?

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a. 1.645b. 1.740c. 1.746d. 6.314

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We are testing the hypotheses H₀: = 6 versus H_a: < 6. Suppose our calculated standardized test statistic is t = -3.59 with n = 33. What is the appropriate P-value for this one-sided test?

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a. P-value < 0.0001b. P-value < 0.0005c. 0.0005 < P-value < 0.0010d. 0.05 < P-value < 0.10

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Suppose we are testing the hypotheses H₀: = 174 versus H_a: 174. and that our calculated standardized test statistic is t = 2.44 with n = 9. What is the appropriate P-value for this two-sided test?

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a. 0.0146b. 0.020 < P-value < 0.025c. 0.040 < P-value < 0.050d. 0.9927

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The following situation applies to questions 6 10: Hallux abducto valgus (call it HAV) is a deformation of the big toe that often requires surgery. Doctors used X-rays to measure the angle (in degrees) of deformity in 38 consecutive patients under the age of 21 who came to a medical center for surgery to correct HAV. The angle is a measure of the seriousness of the deformity. Here are the data:

28	32	25	34	38	26	25	18	30	26	28	13	20
21	17	16	21	23	14	32	25	21	22	20	18	26
16	30	30	20	40	25	26	28	31	38	32	21

It is reasonable to regard these patients as a simple random sample of young patients who require HAV surgery. Carry out SOLVE and CONCLUDE steps of a 95% confidence interval for the mean HAV angle in the population of all such patients.

Begin with the first part of the SOLVE step by creating a stemplot of the data to check the Normality condition. Which of the following best describes your plot?

Select one:

a. The plot is extremely left skewed.b. The plot is extremely right skewed.c. The plot is unimodal (single-peaked), with no outliers.d. The plot is unimodal (single-peaked) symmetric with an outlier at the lower end of the distribution.

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CHECK OF CONDITIONS:

Data collection check: This is an observational study, so we need to check random selection of data. The problem states that it is safe to regard the patients as a simple random sample, so the data collection condition is met.

From the data, we compute x= 25.421053 degrees and s = 7.474756 degrees. In order to create a 95% confidence interval for , which formula should be used?

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a. xt*[/(n)]b. xt*[s/(n)]

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When calculating the 95% confidence interval for , what value of t* should be used according to the t table C?

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a. 1.645b. 1.697c. 1.960d. 2.021e. 2.042

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Calculate a 95% confidence interval for the mean HAV angle in the of all young patients, based on the 38 consecutive patients. Regardless of your answer above, use t* = 2.026. Note that values for xand s are given above.

Select one:

a. (23.04, 27.80)b. (22.96, 27.88)c. (10.77, 40.07)d. (10.16, 40.69)

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Referring to the information given in the above questions, suppose the 95% confidence interval for was calculated to be (22.5 degrees, 27.5 degrees) (which it is not). Which of the following is a correct interpretation of this confidence interval?

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a. We are 95% confident that the HAV angles of all patients under the age of 21 are in the interval of 22.5 to 27.5 degrees.b. We are 95% confident that the sample mean HAV angle of all patients under the age of 21 are in the interval of 22.5 to 27.5 degrees.c. There is a 95% chance that the mean HAV angle of all patients under the age of 21 is somewhere between 22.5 and 27.5 degrees.d. We are 95% confident that the mean HAV angle for all patients under the age of 21 is somewhere between 22.5 and 27.5 degrees.

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The following situation applies to Questions 11-15:

A tire store advertises that the average price of a new set of their tires is only $150. One of their recent customers believes their advertised average is too low, that the true mean price for a set of their tires exceeds $150. He plans to carry out a test of hypothesis at = 0.05. What are the correct hypotheses this customer is interested in testing?

Select one:

a. H₀: = 150 versus H_a: < 150b. H₀:= 150versusH_a: > 150c. H₀:= 150versusH_a: 150d. H₀: x= 150versusH_a: x< 150e. H₀:x= 150versusH_a:x > 150f. H₀:x= 150versusH_a:x 150

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Assuming the conditions were met, using of computed valuesx= 156.9 and s = 11.8 from the 8 data values. What is the value for the t test statistic for this test?

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a. 0.58b. 1.55c. 1.65d. 2.44

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Suppose the value for the t test statistic ist = 1.79, what is the P-value for a one-sided test?

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a. 0.02 < P-value < 0.025b. 0.025 < P-value < 0.05c. 0.05 < P-value < 0.10d. 0.10 < P-value < 0.20

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Suppose the P-value of the test is found to be 0.0715 using statistical software. What is the appropriate conclusion at = 0.05?

Select one:

a. Reject the null hypothesis. We have enough evidence to conclude that the mean price of a set of tires is significantly greater than $150.b. Reject the null hypothesis. We do not have sufficient evidence to conclude that the mean price of a set of tires is significantly greater than $150.c. Fail to reject the null hypothesis. We do not have sufficient evidence to conclude that the mean price of a set of tires is significantly greater than $150.d. Fail to reject the null hypothesis. We have sufficient evidence to conclude that the mean price of a set of tires is significantly greater than $150.