6. Given below are the birth weights of babies born to mothers who took special vitamin supplements while pregnant:
3.23 4.57 3.93 4.33 3.39 3.68 4.68 3.52 3.84 3.02 4.29 2.47 4.13 4.47 3.22 3.43 2.54 3.40
a. Make a 99% confidence interval for the mean weight of babies whose mothers take vitamin supplements.
b. Do a hypothesis test to determine if these babies weight is more than the mean weight for the population of all babies which is 3.39 kg using α = .005
c. In a short paragraph, describe the relationship between your answer to part (a) and your answer to part (b).
6. Given below are the birth weights of babies born to mothers who took special vitamin...
30 The following table contains the birth weights (in kilograms) of male babies born to 3.37 3.82 4.33.92 4.24 4.45 3.89 4.10 4.23 4.27 3.81 3.88 4.31 3.97 4.33 4.07 mothers on a special vitamin supplement. The standard deviation of birth weights for male babies is 0.470 kg. Does the vitamin supplement appear to affect the variation among birth weights? 30 The following table contains the birth weights (in kilograms) of male babies born to 3.37 3.82 4.33.92 4.24 4.45...
The median wage for economics degree holders is determined by the following equation: log( wage) = Be + B educ + B, exper+ B temure + B.age+ B married + u where educ is the level of education measured in years, exper is the job-market experience in years, tenure is the time spend with the current company in years, age is the age in years and married is a dummy variable indicating if a person is married. 935 reg Iwage...
1. Two manufacturing processes are being compared to try to reduce the number of defective products made. During 8 shifts for each process, the following results were observed: Line A Line B n 181 | 187 Based on a 5% significance level, did line B have a larger average than line A? *Use the tables I gave you in the handouts for the critical values *Use the appropriate test statistic value, NOT the p-value method *Use and show the 5...
Gain (V/V) R Setting Totals Averages Sample 1 Sample 2 Sample 3 4 ап 7.8 8.1 7.9 3 5.2 6.0 4.3 = 359.3 i=1 j=1 2 4.4 6.9 3.8 1 2.0 1.7 0.8 This is actual data from one of Joe Tritschler's audio engineering experiments. Use Analysis of Variance (ANOVA) to test the null hypothesis that the treatment means are equal at the a = 0.05 level of significance. Fill in the ANOVA table. Source of Variation Sum of Squares...
Gain (V/V) R Setting Totals Averages Sample 1 Sample 2 Sample 3 4 ап 7.8 8.1 7.9 3 5.2 6.0 4.3 = 359.3 i=1 j=1 2 4.4 6.9 3.8 1 2.0 1.7 0.8 This is actual data from one of Joe Tritschler's audio engineering experiments. Use Analysis of Variance (ANOVA) to test the null hypothesis that the treatment means are equal at the a = 0.05 level of significance. Fill in the ANOVA table. Source of Variation Sum of Squares...