chap10A

# Intheappendixtableforthetdistributionwithn18degrees

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Unformatted text preview: n data set, for a chosen significance level α , the decision rule is to reject the null hypothesis in favour of the alternative if the test statistic: t= x−a s n is such that: t > tc or t < − tc where t c is the critical value that satisfies: P(t ( n − 1 ) > t c ) = α 2 That is, the null hypothesis is rejected if: t > tc ↑ absolute value of the t-statistic A p‐value for the test can be calculated as: p‐value = 2 ⋅ P(t ( n − 1 ) > t ) The p‐value for the two‐tailed test is double the p‐value from a one‐tailed test. 19 Chapter 10 Example: Exercise 10.24, page 352 of the textbook A data set has 9 observations on weight (in ounces) of bottles of shampoo. The manufacturing process should give a weight of 20 ounces. Test the null hypothesis: H0 : μ = 20 the process is operating correctly against the alternative: H1 : μ ≠ 20 the process is not operating correctly From the data set, the sample statistics are: n = 9, x = 20.356 (ounces) and s = 0.6126 The t‐test statistic is: 20 t= x − a 20.356 − 20 = = 1.74 s 0.6126 9 n Chapter 10 The calculation of a p‐value for this 2‐sided test is shown in the graph. f(t) PDF of t ( 8 ) p-value / 2 -1.74 p-value / 2 0 1.74 t With Microsoft Excel find this probability with the function: TDIST( 1.74 , 8, 2) ↑ Two-tailed test This returns the answer: p‐value = 0.12. Note that thep‐value is the s...
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## This note was uploaded on 02/06/2014 for the course ECON ECON 325 taught by Professor Whistler during the Spring '10 term at UBC.

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