Mean Variance Observations Hypothesized Mean Difference df t Stat P(T<=t) one-tail t Critical one-tail P(T<=t) two-tail t Critical two-tail Ob. Decision rule: reject Ho if t > 2.898 Oc. Decision rule: reject Ho if t < -2.898 C. Store A 30.868 89.721 Oa. Decision rule: reject Ho if t > 2.567 or t < -2.567 Od. Decision rule: reject Ho if t > 2.567 Oe. 10 0 17 0.716 0.2418 2.567 0.4836 2.898 Decision rule: reject Ho if t > 2.898 or t < -2,898 Store B 28.232 54.934 12

Assume that a randomly selected subject is given a bone density test. Those test scores are normally distributed with a mean of 0 and a standard deviation of 1. Draw a graph and find the probability of a bone density test score greater than-1.54. 2 Sketch the region. Choose the correct graph below. OA. F2 Wiew an example Get more help. W S X -1.54 5373255.docx H amand # 3 SO F3 E Q D Q C C DSA 4 Q "A R OB. F *NO REF SAMPLE....pdf % OR LO 5 V & -1.54 T G 6 A B D MacBook Air Y H TEMPLATE Sel.....docx & 7 F7 U O C. N * 8 DII 1 M ( 9 K resume..docx DD O < O L F10 O D. P command -1.54 1.54 Clear all : ; { + [ option Check answer ? Show All } ] 11 :T X dele

n a sample of 13 randomly selected high school​ seniors, the mean score on a standardized test was 1191 and the standard deviation was 167.0 Further research suggests that the population mean score on this test for high school seniors is 1020 Does the​ t-value for the original sample fall between minus−t 0.99t0.99 and t 0.99t0.99​? Assume that the population of test scores for high school seniors is normally distributed.       Question content area bottom Part 1 The​ t-value of tequals=enter your response here ▼   does does not fall between minus−t 0.99t0.99 and t 0.99t0.99 because t 0.99t0.99equals=enter your response here.

Confidence interval of the population mean is given by: X+2 x where X is the mean, s is the standard deviation and n is the sample size. Create a function called "summary" that lists mean, %15 trimmed mean and the 95% bounds of confidence interval for a random sample. Use the function to calculate the statistics for the given sample where x= (1200, NA, 1205, 1203, 1296,1304) 1. 2.

H0 : =150.00 kPa H1 :>150.00 kPa b) α= 0.01; sample variance : 1300 (freq. table ) sample deviation : 36.0555 (freq. table ) Calculation of Data  Mean of the results=  = 6700/50 =134 Mid Range of the results= (217.32+46.32)/2= 131.82 Range = (217.32-46.32) = 171 Modal Class= Class 121-145 with the highest frequency of 15 Median of the sorted set of data = 132.25 Mode = there are not clear modes In the samples.  Sample variance S o= 1300 Sample deviation: S =36.55

In a test of the effectiveness of garlic for lowering cholesterol, 43 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured before and after the treatment. The changes (before - after) in their levels of LDL cholesterol (in mg/dL) have a mean of 3.3 and a standard deviation of 17.5. Construct a 95% confidence interval estimate of the mean net change in LDL cholesterol after the garlic treatment. What does the confidence interval suggest about the effectiveness of garlic in reducing LDL cholesterol? Click here to view a t distribution table. Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. What is the confidence interval estimate of the population mean µ? mg/dL < µ< mg/dL (Round to two decimal places as needed.)

3 5 y = f(x) 1.732 2.236 2.645 3.316 Use inverse interpolation to find the value of x for y=3 constructing divided differences table for the data given above.

A company would like to know if efficiency can be improved through a training course for employees. A sample of employees is taken. The time taken for an employee to complete a task before training and after training is observed. Mean Variance Observations Hypothesized Mean Difference df t Stat P(T<=t) one-tail t Critical one-tail P(T<=t) two-tail t Critical two-tail Confidence Level Before n = 33.1 16.81 30 0 29 -2.282 0.051 -1.699 0.025 -2.045 95% After 31.7 12.25 30 p= Ex: 1.234: -3 -2 t = Ex: 1.234 Ex: 9 Degrees of freedom: df = Point estimate for before training: before Ex: 1.234 Point estimate for after training: after -1 0 1 2 3

H0 , test statistic, and p value    please

Data of variable-A is collected as shown to the right. Is the mean of Variable-A less than 80? State null hypothesis and choose level of significance, a = 0.05. (Use Excel ttest function, provide clear logic and reasoning.) Variable-A Variable-B 78.8 81.9 76.9 78.0 67.8 69.7 68.9 74.7 70.5 70.0 80.0 68.2 63.0 88.7 78.7 86.7 85.4 69.5 87.0 75.1 82.4 71.8 78.2 58.1 76.9 74.8 60.2 93.2 82.6 84.6 67.2 66.5 71.7 77.2 68.9 74.0 69.3 77.7 89.1 78.6 76.3 68.3 74.0 89.0 69.4 82.1 76.3 78.0 66.2 80.3 73.9 82.4 79.1 83.1 80.4 66.4 66.2 67.6 80.9 75.3