0.2 Practice tests (1-4) and final exams  (Page 16/36)

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83 . $\Sigma X\sim N\left(n{\mu }_{x},\left(\sqrt{n}\right)\left({\sigma }_{x}\right)\right)$ so $\Sigma X\sim N\left(4000,28.3\right)$

84 .The probability is 0.50, because 5,000 is the mean of the sampling distribution of sums of size 40 from this population. Sums of random variables computed from a sample of sufficient size are normally distributed, and in a normal distribution, half the values lie below the mean.

85 . Using the empirical rule, you would expect 95 percent of the values to be within two standard deviations of the mean. Using the formula for the standard deviation is for a sample sum: $\left(\sqrt{n}\right)\left({\sigma }_{x}\right)=\left(\sqrt{40}\right)\left(7\right)=44.3$ so you would expect 95 percent of the values to be between 5,000 + (2)(44.3) and 5,000 – (2)(44.3), or between 4,911.4 and 588.6.

86 . $\mu -\left(\sqrt{n}\right)\left({\sigma }_{x}\right)=5000-\left(\sqrt{40}\right)\left(7\right)=4955.7$

87 . $5000+\left(2.2\right)\left(\sqrt{40}\right)\left(7\right)=5097.4$

7.3: using the central limit theorem

88 . The law of large numbers says that as sample size increases, the sample mean tends to get nearer and nearer to the population mean.

89 . You would expect the mean from a sample of size 100 to be nearer to the population mean, because the law of large numbers says that as sample size increases, the sample mean tends to approach the population mea.

90 . X ~ N (0.10, 0.20)

91 . $\overline{X}\sim N\left({\mu }_{x},\frac{{\sigma }_{x}}{\sqrt{n}}\right)$ and the standard deviation of a uniform distribution is $\frac{b-a}{\sqrt{12}}$ . In this example, the standard deviation of the distribution is $\frac{b-a}{\sqrt{12}}=\frac{0.10}{\sqrt{12}}=0.03$
so $\overline{X}\sim N\left(0.15,0.003\right)$

92 .

8.1: confidence interval, single population mean, population standard deviation known, normal

Use the following information to answer the next seven exercises. You draw a sample of size 30 from a normally distributed population with a standard deviation of four.

1 . What is the standard error of the sample mean in this scenario, rounded to two decimal places?

2 . What is the distribution of the sample mean?

3 . If you want to construct a two-sided 95% confidence interval, how much probability will be in each tail of the distribution?

4 . What is the appropriate z -score and error bound or margin of error ( EBM ) for a 95% confidence interval for this data?

5 . Rounding to two decimal places, what is the 95% confidence interval if the sample mean is 41?

6 . What is the 90% confidence interval if the sample mean is 41? Round to two decimal places

7 . Suppose the sample size in this study had been 50, rather than 30. What would the 95% confidence interval be if the sample mean is 41? Round your answer to two decimal places.

8 . For any given data set and sampling situation, which would you expect to be wider: a 95% confidence interval or a 99% confidence interval?

8.2: confidence interval, single population mean, standard deviation unknown, student’s t

9 . Comparing graphs of the standard normal distribution ( z -distribution) and a t -distribution with 15 degrees of freedom ( df ), how do they differ?

10 . Comparing graphs of the standard normal distribution ( z -distribution) and a t -distribution with 15 degrees of freedom ( df ), how are they similar?

Use the following information to answer the next five exercises. Body temperature is known to be distributed normally among healthy adults. Because you do not know the population standard deviation, you use the t-distribution to study body temperature. You collect data from a random sample of 20 healthy adults and find that your sample temperatures have a mean of 98.4 and a sample standard deviation of 0.3 (both in degrees Fahrenheit).

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