Что такое диаграмма рассеяния и для чего она используется?

A manufacturer must test that his bolts are 1.00 cm long when they come off the assembly line. He must recalibrate his machines if the bolts are too long or too short. After sampling 64 randomly selected bolts off the assembly line, he calculates the sample mean to be 1.05 cm . He knows that the population standard deviation is 0.18 cm . Assuming a level of significance of 0.05 , is there sufficient evidence to show that the manufacturer needs to recalibrate the machines? Step 3 of 3: Draw a conclusion and interpret the decision. Answer 2 Points We reject the null hypothesis and conclude that there is sufficient evidence at a 0.05 level of significance that the mean length of the bolts is different from 1.00 cm and the machines need to be recalibrated. We reject the null hypothesis and conclude that there is insufficient evidence at a 0.05 level of significance that the mean length of the bolts is different from 1.00 cm and the machines need to be recalibrated. We fail to reject the null hypothesis and conclude that there is sufficient evidence at a 0.05 level of significance that the mean length of the bolts is different from 1.00 cm and the machines need to be recalibrated. We fail to reject the null hypothesis and conclude that there is insufficient evidence at a 0.05 level of significance that the mean length of the bolts is different from 1.00 cm and the machines need to be recalibrated.

A manufacturer must test that his bolts are 2.00 cm long when they come off the assembly line. He must recalibrate his machines if the bolts are too long or too short. After sampling 169 randomly selected bolts off the assembly fine, he calculates the sample mean to be 2.05 cm . He knows that the population standard deviation is 0.29 cm . Assuming a level of significance of 0.01 , is there sufficient evidence to show that the manufacturer needs to recalibrate the machines? Step 3 of 3 : Draw a conclusion and interpret the decision. Answer 2 Points We reject the null hypothesis and conclude that there is insufficient evidence at a 0.01 level of significance that the mean length of the bolts is different from 2.00 cm and the machines need to be recalibrated. We fail to reject the null hypothesis and conclude that there is insufficient evidence at a 0.01 level of significance that the mean length of the bolts is 2.00 cm and the machines need to be recalibrated. We fail to reject the null hypothesis and conclude that there is sufficient evidence at a 0.01 level of significance that the mean length of the bolts is different from 2.00 cm and the machines need to be recalibrated. We reject the null hypothesis and conclude that there is sufficient evidence at a 0.01 level of significance that the mean length of the bolts is different from 2.00 cm and the machines need to be recalibrated.

A manufacturer must test that his bolts are 1.00 cm long when they come off the assembly line. He must recalibrate his machines if the bolts are too long or too short. After sampling 64 randomly selected bolts off the assembly line, he calculates the sample mean to be 1.05 cm . He knows that the population standard deviation is 0.18 cm . Assuming a level of significance of 0.05 , is there sufficient evidence to show that the manufacturer needs to recalibrate the machines? Step 2 of 3 : Compute the value of the test statistic. Round your answer to two decimal places. Answer 2 Points

A newsletter publisher believes that less than \( 58 \% \) of their readers own a Rolls Royce. For marketing purposes, a potential advertiser wants to confirm this claim. After performing a test at the 0.05 level of significance, the advertiser decides to reject the null hypothesis. What is the conclusion regarding the publisher's claim? Answer 2 Points O There is sufficient evidence at the 0.05 level of significance that the percentage is less than \( 58 \% \). There is not sufficient evidence at the 0.05 level of significance to say that the percentage is less than \( 58 \% \).

Frise 15.3 Find the lower quartile, \( Q_{1} \), the upper quartile, \( Q_{3} \), and data sets: Note: You should first rank the data. \( \begin{array}{ll}\text { (i) }\{2,5,7,3,3,2,8\} & \text { (ii) } \\ \text { (ii) }\{5,8,6,4,5,3,4,12\} & \text { (iv) }\{8,7,8,7,6,5,6,5,4,3,4,3\} \\ \text { Leaf }\end{array} \)