Assume measurements for Ph levels in soil follow exactly a normal relative frequency distribution with population mean µ = 5 and population standard deviation o = 1.4. Use Empirical rule to determine percentage of Ph levels in interval 3.6 to 7.8.
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Q: Overproduction of uric acid in the body can be an indication of cell breakdown. This may be an…
A: Hey, since there are multiple subparts posted, we will answer first three question. If you want any…
Q: Assume measurements for Ph levels in soil follow exactly a normal relative frequency distribution…
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Q: Overproduction of uric acid in the body can be an indication of cell breakdown. This may be an…
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Q: Total plasma volume is important in determining the required plasma component in blood replacement…
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Q: Overproduction of uric acid in the body can be an indication of cell breakdown. This may be an…
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Q: Overproduction of uric acid in the body can be an indication of cell breakdown. This may be an…
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Q: Assume measurements for Ph levels in soil follow exactly a normal relative frequency distribution…
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Q: Overproduction of uric acid in the body can be an indication of cell breakdown. This may be an…
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- I want to summarize what is the most and least resistant to salinity?Unfortunately, arsenic occurs naturally in some ground water.t A mean arsenic level of u = 8.0 parts per billion (ppb) is considered safe for agricultural use. A well in Texas is used to water cotton crops. This well is tested on a regular basis for arsenic. A random sample of 36 tests gave a sample mean of x = 7.1 ppb arsenic, with s = 2.2 ppb. Does this information indicate that the mean level of arsenic in this well is less than 8 ppb? Use a = 0.01. A USE SALT (a) What is the level of significance? State the null and alternate hypotheses. O Ho: H= 8 ppb; H,: H > 8 ppb O Ho: H 8 ppb; H: H = 8 ppb (b) What sampling distribution will you use? Explain the rationale for your choice of sampling distribution. O The standard normal, since the sample size is large and a is unknown. O The Student's t, since the sample size is large and a is known. O The standard normal, since the sample size is large and a is known. O The Student's t, since the sample size is large and a is unknown. What is…Unfortunately, arsenic occurs naturally in some ground water.t A mean arsenic level of u = 8.0 parts per billion (ppb) is considered safe for agricultural use. A well in Texas is used to water cotton crops. This well is tested on a regular basis for arsenic. A random sample of 36 tests gave a sample mean of x = 7.1 ppb arsenic, with s = 2.2 ppb. Does this information indicate that the mean level of arsenic in this well is less than 8 ppb? Use a = 0.01. A USE SALT (a) What is the level of significance? State the null and alternate hypotheses. O Họ: u = 8 ppb; H,: u > 8 ppb O Ho: H 8 ppb; H,: u = 8 ppb (b) What sampling distribution will you use? Explain the rationale for your choice of sampling distribution. The standard normal, since the sample size is large and a is unknown. O The Student's t, since the sample size is large and a is known. O The standard normal, since the sample size is large and a is known. The Student's t, since the sample size is large and a is unknown. What is the…
- Unfortunately, arsenic occurs naturally in some ground water.t A mean arsenic level of u = 8.0 parts per billion (ppb) is considered safe for agricultural use. A well in Texas is used to water cotton crops. This well is tested on a regular basis for arsenic. A random sample of 36 tests gave a sample mean of x = 6.7 ppb arsenic, with s = 3.0 ppb. Does this information indicate that the mean level of arsenic in this well is less than 8 ppb? Use a = 0.01. n USE SALT (a) What is the level of significance? State the null and alternate hypotheses. O Ho: H = 8 ppb; H,: u > 8 ppb O Ho: H = 8 ppb; H,: H + 8 ppb O Ho: H 8 ppb; H,: u = 8 ppb O Ho: H = 8 ppb; H,: µ 0.100 O 0.050 < P-value < 0.100 O 0.010 < P-value < 0.050 O 0.005 < P-value < 0.010 P-value < 0.005 Sketch the sampling distribution and show the area corresponding to the P-value. MacBook Pro escAir samples for worker's exposure to oil mist were found to indicate the following levels in Table 1. Calculate TWA. Table 1 Sample Measured Amount Time of Sample 3 mg/m 28 mg/m Sample 1 2 hours Sample 2 6 hours (a) 6.57 mg/m³ (b) 6.75 mg/m' (c) 54 mg/m (d) 31 mg/mTotal plasma volume is important in determining the required plasma component in blood replacement therapy for a person undergoing surgery. Plasma volume is influenced by the overall health and physical activity of an individual. Suppose that a random sample of 50 male firefighters are tested and that they have a plasma volume sample mean of x = 37.5 ml/kg (milliliters plasma per kilogram body weight). Assume that ? = 7.20 ml/kg for the distribution of blood plasma. I need help with part B and D
- 1. Model 1: OLS, using observations 1-706 Dependent variable: RST Coefficient Std. Error t-ratio p-value const 3586.38 38.9124 92.17 <0.0001 *** TOTWRK −0.150746 0.0167403 −9.005 <0.0001 *** Mean dependent var 3266.356 S.D. dependent var 444.4134 Sum squared resid 1.25e+08 S.E. of regression 421.1357 R-squared 0.103287 Adjusted R-squared 0.102014 F(1, 704) 81.08987 P-value(F) 1.99e-1810538.19 Log-likelihood −5267.096 Akaike criterion 10538.19 Schwarz criterion 10547.31 Hannan-Quinn 10541.71 RSTi =3586.38−0.150746 x TOTWRKi , R2=0.103287,SER=421.1357 (38.9124) (0.0167403) Question? could you please help with this question below. 3) By observing the GRETL output in Part (1) above, provide a Detailed explanation of the coefficient of determination. Based on your analysis, is this a good model? Why or why not?1. Model 1: OLS, using observations 1-706 Dependent variable: RST Coefficient Std. Error t-ratio p-value const 3586.38 38.9124 92.17 <0.0001 *** TOTWRK −0.150746 0.0167403 −9.005 <0.0001 *** Mean dependent var 3266.356 S.D. dependent var 444.4134 Sum squared resid 1.25e+08 S.E. of regression 421.1357 R-squared 0.103287 Adjusted R-squared 0.102014 F(1, 704) 81.08987 P-value(F) 1.99e-1810538.19 Log-likelihood −5267.096 Akaike criterion 10538.19 Schwarz criterion 10547.31 Hannan-Quinn 10541.71 ????=3586.38−0.150746 ? ???????,?2=0.103287,???=421.1357 (38.9124) (0.0167403) Question? could you please help with this question below. 3) By observing the GRETL output in Part (1) above, provide a detailed explanation of the coefficient of determination. Based on your analysis, is this a good model? Why or why not?1. Model 1: OLS, using observations 1-706 Dependent variable: RST Coefficient Std. Error t-ratio p-value const 3586.38 38.9124 92.17 <0.0001 *** TOTWRK −0.150746 0.0167403 −9.005 <0.0001 *** Mean dependent var 3266.356 S.D. dependent var 444.4134 Sum squared resid 1.25e+08 S.E. of regression 421.1357 R-squared 0.103287 Adjusted R-squared 0.102014 F(1, 704) 81.08987 P-value(F) 1.99e-1810538.19 Log-likelihood −5267.096 Akaike criterion 10538.19 Schwarz criterion 10547.31 Hannan-Quinn 10541.71 RSTi =3586.38−0.150746 x TOTWRKi , R2=0.103287,SER=421.1357 (38.9124) (0.0167403) Question? Test the significance of the slope coefficient of the regression in Part (1) above. Use 5% level of significance on: a. Level of significance approach (show your calculations of t-ratio) b. P-value approach (show your calculation of p-value) please show the complete steps as well as the interpretation(s) involved in each of the above…
- 1. Model 1: OLS, using observations 1-706 Dependent variable: RST Coefficient Std. Error t-ratio p-value const 3586.38 38.9124 92.17 <0.0001 *** TOTWRK −0.150746 0.0167403 −9.005 <0.0001 *** Mean dependent var 3266.356 S.D. dependent var 444.4134 Sum squared resid 1.25e+08 S.E. of regression 421.1357 R-squared 0.103287 Adjusted R-squared 0.102014 F(1, 704) 81.08987 P-value(F) 1.99e-1810538.19 Log-likelihood −5267.096 Akaike criterion 10538.19 Schwarz criterion 10547.31 Hannan-Quinn 10541.71 RSTi =3586.38−0.150746 x TOTWRKi , R2=0.103287,SER=421.1357 (38.9124) (0.0167403) Question? A- The researcher claims that the model lacks a fundamental principle, namely, the impact of the worker’s gender on their efficiency. The researcher further claims that Men on average take more resting time than women. To clarify the researcher’s claim, add the binary variable MALE into your model, and write down the estimated results…A photoconductor film is manufactured at a nominal thickness of 25 mils. The product engineer wishes to increase the mean speed of the film, and believes that this can be achieved by reducing the thickness of the film to 20 mils. Eight samples of each film thickness are manufactured in a pilot production process, and the film speed (in microjoules per square inch) is measured. For the 25-mil film, the sample data result is = 1.15 and 81 = 0.11, while for the 20-mil film, the data yield 2 = 1.06 and 82 = 0.09. Note that an increase in film speed vould lower the value of the observation in microjoules per square inch. (a) Do the data support the claim that reducing the film thickness increases the mean speed of the film? Use a = 0.10 and assume that the two population variances are equal and the underlying population of film speed is normally distributed. What is the P-value for this test? Round your answer to three decimal places (e.g. 98.765). The data the claim that reducing the film…A professor of mathematics teaches at a catholic school and at a public school. He wants to know whether attending a catholic school leads to higher grades. He collected data from 7,430 students from both school's types. Considering this information, please answer the following questions: a) How can you interpret the OLS outcome on whether the child attended a catholic school (cathhs) is important to explain students' grade? Source Model Residual Total cathhs lfaminc SS 143098.058 521610.985 motheduc fatheduc 664709.043 df math12 Coefficient Std. err. 17887.2573 8 7,421 70.2885036 7,429 89.4749015 1.623158 .4106669 1.447696 .1431147 .0613988 .7345401 .832485 . 055494 black -5.226717 hispan -1.221689 asian 3.28341 female _cons MS .3915981 .3328612 .4438741 -1.066021 .1950649 16.72931 1.355485 Number of obs F(8, 7421) Prob > F R-squared Adj R-squared Root MSE t P>|t| 3.95 0.000 0.000 10.12 11.96 0.000 15.00 0.000 -13.35 0.000 -3.67 0.000 7.40 0.000 -5.46 0.000 12.34 0.000 = .8181343…