1. A 95% confidence interval for the mean IQ of college students aged 20-25 years old was computed as (103, 110), based on a sample of 130 observations of IQ. The correct interpretation of this interval is: (d) (a) 95% of the individuals in the population should have IQs between 103 and 110. (b) About 95% of future samples will have confidence intervals that contain the true mean. (c) 95% of individuals in the sample have IQs between 103 and 110. (d) There is a 95% probability that the future sample mean IQ will be between 103 and 110. 2. In a random sample of 25 students at college, the average IQ. was . = 103. Assume that the IQ of a student in college follows a Normal distribution with unknown mean and standard deviation o = 15. A 90% confidence interval for u is required to construct (1) What is the critical value z*? z*=1.645 (2) What is the stand deviation of X? o/vn =3 (3) A 90% confidence interval for u is (c) (a) (78, 128). (b) (97, 109). (c) (98, 108). (d) (102, 104) 3. A shipment of 1000 small lab mice arrives at the animal care facility with a nominal weight of 10 g per mouse. A sample of 20 mice is selected and weighed. The average weight of the sample is 10.2 g. From past shipments, it is known that the standard deviation of weight among mice is 2 g. Find a 95% confidence interval for the mean weight of all the mice in the shipment. (b) (a) 10.2 + 0.45 g (b) 10.2 + 0.88g (c) 10 + 0.88 g (d) 10 + 0.45 g4. A shipment of 1000 small lab mice arrives at the animal care facility with a nominal weight of 10 g per mouse. A sample of 20 mice is selected and weighed. The average weight of the sample is 9.8 g. From past shipments, it is known that the standard deviation of weight among mice with a nominal weight of 10 g is 2 g. The technician wishes to test the hypothesis that the average weight of the mice matches the nominal value versus that the mice are underweight. The null and alternative hypotheses are (a) OH, : M=10; H, : # 30, assuming that the standard deviation is unchanged when taking vitamin C. Based on these data, find the P-value of the appropriate test. (1) What is the value of the test statistic? X -HO Z= = 0. 83 o (2) The P-value is (a) O p = 0.2033. The observed x is not a good evidence against Ho. O p = 0.2033. The observed x is a good evidence for the company's claim. O p = 0.4066. The observed > is not a good evidence against Ho. 6. State Type I and Type II error7. From a sample of 400 births, the standard deviation of the gestation time was 10 days, and the mean was 250 days. The standard error for the sample mean was (d) O 0.98. O 0.025. O 12.5. O 0.5. 8. The value of t* for a 95% confidence interval when there were 15 pieces of data is O 2.145. 1.960. 1.761 O 2.131. (a) 9. The gestation time for humans is approximately Normally distributed with mean p. The average gestation time of a random sample of 25 babies is found to be * = 245 days, and the standard deviation of the 25 gestation times is found to be s = 10 days. A 90% confidence interval for u is (a) O 245 + 3.42 O 245 + 3.29. O 245 + 4.13. O 245 + 3.9210. Critical values. Using Table C or technology, what critical value t* would you use for a confidence interval for the mean of the population in each of the following situations? A 95% confidence interval based on n = 10 observations t*=2.26211. Is it significant? The one-sample t statistic for testing HOW = 0 H.: # > 0 from a sample of n = 15 observations has the value t = 1.82. (a) What are the degrees of freedom for this statistic? (b) Give the two critical values * from Table C that bracket t. What are the one- sided P-values for these two entries? Alternatively, use technology to obtain the exact one-sided P-value for this test. (c) Is the value t = 1.82 significant at the 5% level? Is it significant at the 1% level? solution: (a) df=n-1=14 (b) 1.761 and 2.145 (c) 0.025