{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Ch. 11

# The Practice of Statistics: TI-83/89 Graphing Calculator Enhanced

This preview shows pages 1–4. Sign up to view the full content.

Inference for Distributions 11.1 (a) s / f i = 9 . 3 / f i = 1.7898. (b) Since s / f i = 0.01, s = (0.01) (fi) = 0.0173. 11.2 (a) 2.015. (b) 2.518. 11.3 (a) 2.145. (b) 0.688. 11.5 (c) The t, curve is a bit shorter at the peak and slightly higher in the tails (see TI-83 plot below). (d) The t, curve has moved toward coincidence with the standard normal curve. (e) The t,, curve cannot be distinguished from the standard normal curve. As the degrees of freedom / increase, the t (df) curve approaches the standard normal density graph. 11.6 (a) 0.228. (b), ( 4 7 and (d) Absolute df P(t > 2) difference (e) As the degrees of freedom increases, the area to the right of 2 under the t,, distribution gets closer to the area under the standard normal curve to the right of 2. 11.7 (a) 14. (b) 1.82 is between 1.761 ( p = 0.05) and 2.145 ( D = 0.025). (c) The P-value is between 0.025 and 0.05 (in fact, P = 0.0451). (d) t = 1.82 is significant at a = 0.05 but not at a = 0.01.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
174 Chapter 1 1 L d 11.8 (a) 24. (b) 1.12 is between 1.059 (0 = 0.15) and 1.318 (0 = 0.10). (c) The P-value is between 0.30 and 0.20 (in fact, P = 0.2738). (d) t = 1.12 is not significant at either a = 0.10 or at a = 0.05. 11.9 (a) Since the sample size is small (n < 15), the distribution of the CSB vitamin C data should be close to normal. We can check this using a stemplot and a normal probability plot. Since there are no outliers and the normal plot is reasonably linear, the assumption of nor- mality seems justified despite the small number of observations. We also must assume that the eight observations represent an SRS from the population of all possible amounts of vitamin C in samples of CSB. Since the eight observations were taken from a production run, this seems like a reasonable assumption provided that the observations were taken at regular intervals. (b) We will use the t-procedure. 15 = 22.50, s = 7.19, df = n - 1 = 7. Using Table C with df = 7, we find t' = 2.365. The 95% confidence interval is therefore 22.50 + (2.365) (7.19/d) = 22.5 + 6.0,or(16.5,28.5). L (c) Letting p = the mean vitamin C content per 100 g, we wish to test Ho: p = 40 vs. Ha: p i 40. The t test statistic is t = = -6.88 and the corresponding P-value (from soft- ware) is 0.0002. Clearly, this result is incompatible with a process mean of p = 40. We reject Ho and conclude that the vitamin C content for this run does not conform to specifications (specifically, it is lielow the specifications). 11.10 (a) The stemplot shown below has stems in 1000s, split 5 ways. The data are right-skewed with a high outlier of 2433 (and possibly 1933). The normal plot shows these two outliers, but otherwise it is not strikingly different from a line. (b) ?? = 926, s = 427.2, standard error = 69.3 (all in mg). (c) Use of the t-procedure is justified here because the sample size is large (n = 38 > 30) and thus the distribution of ?? will be approximately normal by the central limit theorem. Using Table C 2500 < Q) 3 2000 : -v .s 1500 E .2 1000: 0 w s 500: o 0 00 0 0 0 o - ~ I a , ~ I ~ I ~ I ~ -3 -2 -1 0 1 2 3 z score
Inference for Distributions /' with 30 degrees of freedom, we have t* = 2.042. The approximate 95% confidence interval is then 926 + (2.042) (69.3), or 784.5 to 1067.5 mg; MINITAB reports 785.6 to 1066.5 mg.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 16

Ch. 11 - Inference for Distributions 11.1(a s f i = 9 3 f i...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online