8th-ed Chapter 08, Section 8.3, Intelligent Tutoring Problem 047 A sample of 10 observations taken from a normally distributed population yielded the following data. 46 54 30 48 46 39 47 36 41 56 Note: Use Table V in Appendix C to compute the probabilities. (a) What is the point estimate of p? (b) Find t for 95% confidence level. (c) Make a 95% confidence interval for u. (d) What is the margin of error of estimate for p in part (c)? Recall the following definitions from section 8.4 of the text. The confidence level is denoted by (1 - a)100%, where a is called the significance level. a is the sum of the probabilities of left and right tail corresponding to a given t value with degrees of freedom (df) = n - 1 from the t-distribution. That is, a = P(IT| > t). By symmetry, a/2 = P(T > t). Notice that we can find this probability or find t using the Table V of appendix C. The (1 - a)100%, confidence interval for p, if o is unknown and population normal or n 2 30 is given by It tsz, where the value oft is obtained from the t distribution table forn - 1 degrees of freedom (table V in appendix C or by calculator) and Sz = E is the standard deviation of X when s is the standard deviation of the sample of size n. Vn The quantity tsz is called the Margin of Error and is denoted by E.

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A sample of 10 observations taken from a normally distributed population yielded the following data.

46 54 30 48 46 39 47 36 41 56

(a) Find the mean, x¯ for this data.

x¯=

(b) Find the standard deviation, s for this data.

Round the answer to three decimal places.

s =

(c) What is the point estimate of µ?

µ =

(d) Find α for 95% confidence level.

α =

(e) Find α/2 for 95% confidence level.

α/2 =

(f) Find the degrees of freedom, df for the t distribution.

df =

(g) Find t for 95% confidence level.

t =
Round the answer to three decimal places.

8th-ed Chapter 08, Section 8.3, Intelligent Tutoring Problem 047
A sample of 10 observations taken from a normally distributed population yielded the following data.
46 54 30 48 46 39 47 36 41 56
Note: Use Table V in Appendix C to compute the probabilities.
(a) What is the point estimate of p?
(b) Find t for 95% confidence level.
(c) Make a 95% confidence interval for u.
(d) What is the margin of error of estimate for p in part (c)?
Recall the following definitions from section 8.4 of the text.
The confidence level is denoted by (1 - a)100%, where a is called the significance level. a is the sum of the probabilities of left and right tail corresponding to a given t value with degrees of freedom (df) = n - 1 from the
t-distribution. That is, a = P(IT| > t). By symmetry, a/2 = P(T > t). Notice that we can find this probability or find t using the Table V of appendix C.
The (1 - a)100%, confidence interval for p, if o is unknown and population
normal or n 2 30 is given by It tsz, where the value oft is obtained from the t distribution table forn - 1 degrees of freedom (table V in
appendix C or by calculator) and Sz =
E is the standard deviation of X when s is the standard deviation of the sample of size n.
Vn
The quantity tsz is called the Margin of Error and is denoted by E.

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

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