Data on salaries in the public school system are published annually by a teacher
ID: 2947081 • Letter: D
Question
Data on salaries in the public school system are published annually by a teachers' association. The mean annual salary of (public) classroom teachers is $51.3 thousand. Assume a standard deviation of $7.6 thousand. Complete parts (a) through (e) below a. Determine the sampling distribution of the sample mean for samples of size 64. The mean of the sample mean is H S 51,300. (Type an integer or a decimal. Do not round.) The standard deviation of the sample mean is ?x-$ 950. (Type an integer or a decimal. Do not round.) b. Determine the sampling distribution of the sample mean for samples of size 256 The mean of the sample mean is S 51300 Type an integer or a decimal. Do not round.) The standard deviation of the sample mean is ?--$ 475 . (Type an integer or a decimal. Do not round.) c. Do you need to assume that classroom teacher salaries are normally distributed to answer parts (a) and (b)? Explain your answer. A. Yes, because x is only normally distributed if x is normally distributed. B. Yes, because the sample sizes are not sufficiently large so that x will be approximately normally distributed, regardless of the distribution of>. CNo, because the sample sizes are sufficiently large so that x will be approximately normally distributed, regardless of the distribution of x D. No, because if x is normally distributed, then x must be normally distributed d. What is the probability that the sampling error made in estimating the population mean salary of all classroom teachers by the mean salary of a sample of 64 classroom teachers will be at most $1000? (Round to three decimal places as needed.)Explanation / Answer
Solution:- Given that mean = 51300, sd = 7600 , n = 64
P (the difference between sample mean and population mean is at most 1,000) is
d. P((x-mean) <= 1000)= P((x- mean)/(sd/sqrt(n)) <= (1000/(7600/sqrt(64))
= P(Z <= 1.0526)
= 0.853 (rounded)
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Given that mean = 51300, sd = 7600 , n = 256
P (the difference between sample mean and population mean is at most 1,000) is
d. P((x-mean) <= 1000) = P((x- mean)/(sd/sqrt(n)) <= (1000/(7600/sqrt(256))
= P(Z <= 2.1053)
= 0.9826
= 0.983 (rounded)
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