The accompanying table contains the percentage of a certain country's population that uses a certain social media. Complete parts (a) through (c) below. 57 56 43 54 41 36 7 25 43 18 44 5 31 27 58 19 27 36 45 80 58 56 a. Compute the first quartile (Q1), the third quartile (Q3), and the interquartile range. Q1 = enter your response here Q3 = enter your response here Interquartile range = enter your response here
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Q: 56 56 44 54 42 35 6 26 43 17 43 7 30 28 58 19 27 37 44 79 56 57
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The accompanying table contains the percentage of a certain country's population that uses a certain social media. Complete parts (a) through (c) below.
Q1
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=
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enter your response here
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Q3
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=
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enter your response here
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Interquartile range
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=
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enter your response here
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- Suppose you computed r=−0.817 using n=13 data points. Using the critical values table below, determine if the value of r is significant or not. df12345678910CV (+ and -)0.9970.9500.8780.8110.7540.7070.6660.6320.6020.576df11121314151617181920CV (+ and -)0.5550.5320.5140.4970.4820.4680.4560.4440.4330.423df21222324252627282930CV (+ and -)0.4130.4040.3960.3880.3810.3740.3670.3610.3550.349df405060708090100CV (+ and -)0.3040.2730.2500.2320.2170.2050.195 Select the correct answer below: r is significant because it is between the positive and negative critical values. r is not significant because it is between the positive and negative critical values. r is significant because it is not between the positive and negative critical values. r is not significant because it is not between the positive and negative critical values.Calculate the test-statistic, t with the following information.n1=35, ¯x1=2.65, s1=0.58n2=50, ¯x2=2.49, s2=0.86Rounded to 2 decimal places.calculate SN given by Simpson’s Rule for the value of N indicated.
- A survey of 90 recently delivered women on the rolls of a county welfare department revealed that 27 had a history of intrapartum or postpartum infection. What is the critical value of z if we need to conclude that the population proportion with a history of intrapartum or postpartum infection is less than 0.25.An SRS of 100 flights of a large airline (airline 1) showed that 64 were on time. An SRS of 100 flights of another large airline (airline 2) showed that 80 were on time. Let p1 and p2 be the proportion of all flights that are on time for these two airlines. Is there evidence the the on-time rate for airline 1 is smaller than the on-time rate for airline 2? To determine this, you test the following hypotheses: Select one: H0:p1=p2 vs Ha:p1<p2 H0:p1≠p2 vs Ha:p1=p2 H0:p1=p2 vs Ha:p1>p2 H0:p1=p2 vs Ha:p1≠p2A tax accountant would like to test the claim that the proportion of individuals who owe when filing their taxes is greater than 0.20. If the z− test statistic was calculated as z=1.94, does the tax accountant have enough evidence to reject the null hypothesis? Assume α=0.05.
- A pharmcuticle company claims that its new drug reduces systolic blood pressure. The systolic blood pressure (in millimeters of Mercury) for 9 patients before taking the new drug and 2 hours after taking the drug are shown in the table below. is there enough evidence support the company's claim? Let D =(blood pressure before taking new drug)- (blood pressure after taking new drug).use significant levels of a=their 0.05 for the test. Assume that the systolic blood pressure levels are normally distributed for the population of patience both Before & After taking the new drug. 1.State the null and alternative hypothesis for the test. 2. Find the value of the standard deviation of the paired differences. Round to one decimal place. 3.compute the value of the test statistic. Round to three decimal places 4.determine the decision rule for rejecting the null hypothesis Ho. Round the numerical portion to three decimals. 5.make decision for the hypothesis test.Find the z-score that has 73.6% of the distribution's area to its right.Find the z value to the right o the mean so that 54.78% of the area under the distribution curve lies to the left of it.
- State as much information as you can about the P-value for an upper-tailed F test in each of the following situations. (a) df1 = 3, df2 = 15, calculated F = 4.43 P-value > 0.1000.050 < P-value < 0.100 0.010 < P-value < 0.0500.001 < P-value < 0.010P-value < 0.001 (b) df1 = 4, df2 = 18, calculated F = 1.83 P-value > 0.1000.050 < P-value < 0.100 0.010 < P-value < 0.0500.001 < P-value < 0.010P-value < 0.001 (c) df1 = 5, df2 = 20, calculated F = 4.00 P-value > 0.1000.050 < P-value < 0.100 0.010 < P-value < 0.0500.001 < P-value < 0.010P-value < 0.001 (d) df1 = 4, df2 = 35, calculated F = 4.64 P-value > 0.1000.050 < P-value < 0.100 0.010 < P-value < 0.0500.001 < P-value < 0.010P-value < 0.001 You may need to use the appropriate table in the appendix to answer this question.Find the z-score that has 70.54% of the distribution's area to its left. z 0.09 0.08 0.07 0.06 0.05 0.04 0.03 0.02 0.01 0.00 −3.4 0.0002 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 −3.3 0.0003 0.0004 0.0004 0.0004 0.0004 0.0004 0.0004 0.0005 0.0005 0.0005 −3.2 0.0005 0.0005 0.0005 0.0006 0.0006 0.0006 0.0006 0.0006 0.0007 0.0007 −3.1 0.0007 0.0007 0.0008 0.0008 0.0008 0.0008 0.0009 0.0009 0.0009 0.0010 −3.0 0.0010 0.0010 0.0011 0.0011 0.0011 0.0012 0.0012 0.0013 0.0013 0.0013 −2.9 0.0014 0.0014 0.0015 0.0015 0.0016 0.0016 0.0017 0.0018 0.0018 0.0019 −2.8 0.0019 0.0020 0.0021 0.0021 0.0022 0.0023 0.0023 0.0024 0.0025 0.0026 −2.7 0.0026 0.0027 0.0028 0.0029 0.0030 0.0031 0.0032 0.0033 0.0034 0.0035 −2.6 0.0036 0.0037 0.0038…Spacers are typically used in groups of four and tolerances require the total length of the 4 spacers to be between 155 mm and 165 mm. This means that the mean length of spacers in a pack has to be between 38.75 mm and 41.25 mm. The quality control officer for the company is looking at the likelihood of a pack of spacers being outside these tolerances. His interest shifts to the mean length of spacers in a sample of 4, which will be denoted X . i) State the distribution of the mean of a sample of 4 randomly selected spacers (that is, X ) and give the value(s) of its parameter(s). ii) Determine the probability that a randomly selected pack of spacers has a mean length that is inside the range 38.75 mm to 41.25 mm.