Prompt | Accounting homework help - Vintage-paper.com

Prompt B: The population has a mean of 5 and a standard deviation . A sample of 100 istaken.According to Prompt B, What is the standard deviation of the sampling distribution?a) .014 b) .14 c) 14 d) 1.4Prompt B: The population has a mean of 5 and a standard deviation. A sample of 100 istaken.According to Prompt B, What is the probability that your sample has a mean less than 4.9?a) .7625 b) .8164 c) .2675 d) .1234Given a normal distribution with a mean of 22 and standard deviation of 25, what is theprobability that x is greater than 34.3?a) .0069b) .3114 c) 0d) .9931A hypothesis test is done in which the alternative hypothesis is that more than 10% of a population is left-handed. The p-valuefor the test is calculated to be 0.25, so the researcher fails to reject the null hypothesis. Which statement is correct? a) We cannot conclude that more than 10% of the population is left-handed.b) We can conclude that exactly 25% of the population is left-handed.c) We can conclude that more than 25% of the population is left-handed.d) We can conclude that more than 10% of the population is left-handed.Given a normal distribution with a mean of 22 and standard deviation of 5, what values represent the middle 95%?a) 17, 27 b) 22c) 8,37 d) 12, 32DDT is an insecticide that accumulates up the food chain. Predator birds can be contaminatedwith quite high levels of the chemical by eating many lightly contaminated prey. It is believed thatthis causes eggshells to be thinner and weaker than normal and makes the eggs more prone to breakage. A studywas conducted of eggs laid by contaminated birds. A “normal” eggshell has a mean thickness of 0.2 mm, but the researchers wanted to investigate the claim that the eggshells were thinner than normal. What would be an appropriate null and alternative hypothesis?Decreasing the confidence level, while holding the sample size the same, will do what to thelength of your confidence interval?a) cannot be determined from the given information b) make it smallerc) make it biggerd) it will stay the samePrompt C: A sample of 16 college students with part-time jobs were asked how much money they made in the past two months. The result was a mean of $645 with a standard deviation of $31. We will construct a 95% confidence interval for the mean using a T-distribution.According to Prompt C, What is the critical value for this confidence interval?a) 2.120 b) 2.602 c) 2.131 d) 1.96Prompt C: A sample of 16 college students with part-time jobs were asked how much money they made in the past two months. The result was a mean of $645 with a standard deviation of $31. We will construct a 95% confidence interval for the mean using a T-distribution.According to Prompt C, What is the margin of error for this confidence interval? a) $15.19b) $31c) $66.06 d) $16.52Prompt C: A sample of 16 college students with part-time jobs were asked how much money they made in the past two months. The result was a mean of $645 with a standard deviation of $31. We will construct a 95% confidence interval for the mean using a T-distribution.According to Prompt C, What is the 95% confidence interval for the average amount of money earned in 2 months?a) (321.7, 365.8) b) (531.2, 612.9) c) (876.2, 981.5) d) (628.5, 661.5)Prompt E: Susie and Joan complain that they make very little money at their jobs; they are both radio announcers. The population of radio announcers averages $27,000 per year with a standard deviation of $5,300. Susie makes $25,000 per year and Joan makes $32,500.According to Prompt E, what is the z-score for Joan’s salary?a) .47b) -.94c) .94d) -.47Prompt E: Susie and Joan complain that they make very little money at their jobs; they are both radio announcers. The population of radio announcers averages $27,000 per year with a standard deviation of $5,300. Susie makes $25,000 per year and Joan makes $32,500.According to Prompt E, what salary corresponds with a z-score of 1.5?a) $22,200b) $-19,550c) $32,800d) $35,450Prompt E: Susie and Joan complain that they make very little money at their jobs; they are both radio announcers. The population of radio announcers averages $27,000 per year with a standard deviation of $5,300. Susie makes $25,000 per year and Joan makes $32,500.According to Prompt E, What salary corresponds with 75th percentile?a) $31,075b) $22,800 c) $25,450 d) $32,800True or False. The more samples you take the more your sampling distribution becomes morenormal centered around the true proportion/mean. a) Trueb) Have a great summerc) Falsed) Stats is the best class ever

Prompt | Accounting homework help

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