Express the confidence interval 68.8%
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Express the confidence interval 68.8%<p<84% in the form of p±ME.
___% ± ___%
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- Let a population consist of the values 7 cigarettes, 21 cigarettes, and 22 cigarettes smoked in a day. Show that when samples of size 2 are randomly selected with replacement, the samples have mean absolute deviations that do not center about the value of the mean absolute deviation of the population. What does this indicate about a sample mean absolute deviation being used as an estimator of the mean absolute deviation of a population? Calculate the mean absolute deviation for each possible sample of size 2 from the population. Sample Mean Absolute Deviation {7,7} __ {7,21} __ {7,22} __ {21,7} __ {21,21} __ {21,22} __ {22,7} __ {22,21} __ {22 ,22}STAYS THE SAME, INCREASES OR DECREASES PICK 1 OPTION FOR ALL BLANKS.Please show work on paper please
- t3 solve both pleaseLife of Smoke Detectors The average lifetime of smoke detectors that a company manufactures is 5 years, or 60 months, and the standard deviation is 7 months. Find the probability that a random sample of 26 smoke detectors will have a mean lifetime between 58 and 63 months. Assume that the sample is taken from a large population and the correction factor can be ignored. Round the final answer to at least four decimal places and intermediate z-value calculations to two decimal places. P(58+ I/ * 00 A company manufactures light bulbs. The company wants the bulbs to have a mean life span of 997 hours. This average is maintained by periodically testing random samples of 16 light bulbs. If the t-value falls between -th an and t, go then the company will be satisfied that it is manufacturing acceptable light bulbs. For a random sample, the mean life span of the sample is 1002 hours and the standard deviation is 29 hours. Assume that life spans are approximately normally distributed. Is the company making acceptable light bulbs? Explain. and to 90 The company making acceptable light bulbs because the t-value for the sample is t= (Round to two decimal places as needed.) Get more help- Clear all Check answer 3:44 PM 64°F Sunny A 11/23/2021 PrtSc F5. F10 F11 F12 4. 5. 9. 6. K. B. M.SAT scores in one state is normally distributed with a mean of 1481 and a standard deviation of 88. Suppose we randomly pick 49 SAT scores from that state. a) Find the probability that one of the scores in the sample is less than 1457. P(X<1457)P(X<1457) = b) Find the probability that the average of the scores for the sample of 49 scores is less than 1457. P(¯¯¯X<1457)P(X¯<1457) =A statistician wishing to test a hypothesis that students score at most 75% on the final exam in an introductory statistics course decides to randomly select 20 students in the class and have them take the exam early. The average score of the 20 students on the exam is 72% and the standard deviation in the population is known to be = 15%. The statistician calculates the test statistic to be –0.8944. If the statistician chose to do a two-sided alternative, the P-value would be calculated by: a. finding the area to the right of –0.8944 and doubling it. b. finding the area to the left of –0.8944 and doubling it. c. finding the area to the right of the absolute value of –0.8944 and dividing it by two. d. finding the area to the left of –0.8944Test 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…SEE MORE QUESTIONS