Express the confidence interval 31.7%
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Express the confidence interval 31.7%<p<44.9%31.7%<p<44.9% in the form of ˆp±Ep^±E.
______% ±±_________ %
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- 6.6 You estimated a regression with the following output. Source | SS df MS Number of obs = 209 -------------+---------------------------------- F(1, 207) = 92470.91 Model | 622579639 1 622579639 Prob > F = 0.0000 Residual | 1393670.56 207 6732.70804 R-squared = 0.9978 -------------+---------------------------------- Adj R-squared = 0.9978 Total | 623973310 208 2999871.68 Root MSE = 82.053 ------------------------------------------------------------------------------ Y | Coef. Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------- X | 34.2537 .1126432 304.09 0.000 34.03163 34.47578 _cons | 8.806569 10.31802 0.85 0.394 -11.5353 29.14844…Approximately one-third of Canadians have some form of privately-covered long-term disability insurance. If random groups of 165 Canadians are sampled, find the range of values that the middle 95% of the sample proportions of those with long-term disability health insurance are likely to fall. = ? < p̂ < ? Round to four decimal places if necessary Enter 0 if normal approximation cannot be usedScores on a dental anxiety scale range from 0 (no anxiety) to 20 (extreme anxiety). The scores are normally distributed with a mean of 11 and a standard deviation of 4. Find the z-score for the given score on this dental anxiety scale. 16 Z16 (Type an integer or a decimal.)
- X has PDF f(x) = #/2 if0Fawns between 1 and 5 months old have a body weight that is approximately normally distributed with mean ? = 28.4 kilograms and standard deviation ? = 4.1 kilograms. Let x be the weight of a fawn in kilograms. Convert the following z intervals to x intervals. (Round your answers to one decimal place.) (a) −2.17 < z___ < x(b) z < 1.28x < ___(c) −1.99 < z < 1.44___ < x < ___Question 23 > Suppose f(10) = 7 and g(10) = 3. %3D 1. (f – 9)(10) %3D 2. (fg)(10) = () 3. (10) A decimal is not an acceptable answer. 4. |(10) A decimal is not an acceptable answer. Submit QuestionGroup I Given N1 = 20 E(X- 8) = 10 E(X-8)? = 152 50. Group II N2 = 30 (Y- 10) = - 15 E(Y-10)2 = 300 %3D %3D %3D Find arithmetic mean and standard deviation of each group and the combined arithmeti nean, standard deviation and the overall coefficient of variation.Find the area of the shaded region. The graph depicts the standard normal distribution of bone density scores with mean 0 and standard deviation 1. ← 2 W S The area of the shaded region is (Round to four decimal places as needed.) आईड SA H nd D X 2Demo Outline....docx 3 E 8.0 F3 D C $ 4 a F4 R F V chapt21_lecture....pdf % 5 e " T G 1 ^ 6 B MacBook Air F6 Y H & 7 chapt22_lecture.....pdf N U √₁ Vi J * 8 DII FB I M 1, (0,0) ( 9 K chapt23_lecture....pdf DD FO More O < I H ) O F10 command P J L V Time Remaining:00:42:40 Next - : ; I 4 F11 { [ option + = 11 z = -0.92 ? 1 I Show All F12 1 X X Q C delete LPls help with this homework. Consider the estimator s2* = ae'e where a is a constant. For which value of a do we obtain the lowest mean squared error (under normality)?Let x = red blood cell (RBC) count in millions per cubic millimeter of whole blood. For healthy females, x has an approximately normal distribution with mean ? = 3.6 and standard deviation ? = 0.6. (a) Convert the x interval, 4.5 < x, to a z interval. (Round your answer to two decimal places.) ____< z(b) Convert the x interval, x < 4.2, to a z interval. (Round your answer to two decimal places.)z < ____(c) Convert the x interval, 4.0 < x < 5.5, to a z interval. (Round your answers to two decimal places.) _____< z < ______(d) Convert the z interval, z < −1.44, to an x interval. (Round your answer to one decimal place.)x < ____(e) Convert the z interval, 1.28 < z, to an x interval. (Round your answer to one decimal place.) ____< x(f) Convert the z interval, −2.25 < z < −1.00, to an x interval. (Round your answers to one decimal place.)_____ < x < ____(g) If a female had an RBC count of 5.9 or higher, would that be considered unusually…Test a claim that the mean amount of lead in the air in U.S. cities is less than 0.038 microgram per cubic meter. It was found that the mean amount of lead in the air for the random sample of 56 U.S. cities is 0.038 microgram per cubic meter and the standard deviation is 0.068 microgram per cubic meter. At alphaαequals=0.01, can the claim be supported? Complete parts (a) through (e) below. Assume the population is normally distributed. d) Decide whether to reject or fail to reject the null hypothesis. ▼ Fail to reject Reject Upper H 0H0 because the standardized test statistic ▼ is is not in the rejection region. (e) Interpret the decision in the context of the original claim. There ▼ is not is enough evidence at the nothing% level of significance to ▼ reject support the claim that the mean amount of lead in the air in U.S. cities is ▼ equal greater than or equal less than or equal not equal greater than less than nothing…How do I solve
