Practice Questions: Statistics W, Fall 9 Solutions Question.. The standard deviation of Z is 89... P(=6) =..3. is definitely inside of a 95% confidence interval for..4. (a) YES (b) YES (c) NO (d) NO Questions : An opinion poll asks a random sample of adults whether they favor banning the ownership of handguns by private citizens. A commentator believes that more than half of all adults favor such a ban. State the null and alternative hypothesis to investigate this claim. H : p.5 : p.5 H a Questions 3: You want to design a study to estimate the proportion of constituents in a certain congressional district who support a proposal up for debate in congress. You intend on using a 99% confidence interval and you would like the margin of error of the interval to be.3 or less. What is the sample size required to achieve this goal? To guarantee that the margin of error is.3 or less, choose the worst case scenario - p ˆ. 5 * z n m pˆ(.576 pˆ).3 843.3 We need to sample n=844 people.
Questions 4: A set of cards consists of four red cards and six black cards. The cards are shuffled thoroughly and a card is chosen at random. After recording the color of the card, it is put back into the deck. (a) If a person repeatedly keeps drawing cards until she gets a red card, how long would you expect it to take? (b) What is the probability that the first red card comes on the third draw? (c) Suppose she draws six cards. What is the probability that she gets less than red cards? (d) Suppose she draws 6 cards. Approximate the probability that she gets more than red cards. This process is an example of repeated Bernoulli trials with probability of success (getting a red card) equal to.4. (a) Let = number of draws until first success (red card). is Geometric(.4) E ( ).5 p (b) P ( 3) (.6).4. 44 (c) Let = number of red cards in 6 draws is Binomial(6,.4) P ( ) P( ) P( 6 ) (.4) (.6) 6 6 (.4) (.6) 5.3 (d) When np and n ( p) np 6(.4) 4 and standard deviation: np ( p) 6(.4)(.6) 44. Hence, is approximately N 4,., is approximately Normal with mean: 4 P ( ) P( Z ) P( Z.67).955
Questions 5: Suppose a simple random sample of size is made from a normally distributed population with mean and standard deviation. Using the obtained data we were able to calculate y 98. 7 and s. Compute a 9% Confidence Interval for. Compute a 9% Confidence Interval for. y t * s n df = 9. t *. 833 98.7.833 98.7.59 Questions 6: A medical study was done to evaluate the effectiveness of a cholesterol lowering drug. The treatment group had 8 subjects and the control group had 8 subjects. The cholesterol levels of the subjects were measured both before and after the treatment. The results were that 6 of the subjects in the treatment group had a reduced cholesterol level, while 5 of the control subjects had a reduced cholesterol level. Can you conclude that the drug is effective at the 5% significance level? Can we reject the claim of no effect at the 5% significance level? H : p p and H a : p p 6 5 p ˆ.775 and p ˆ. 65 n 8 n 8 p ˆ n n 6 8 5.7 8 6 z ( pˆ pˆ pˆ( pˆ) n n ).775.65.7(.7) 8 8.7 Pvalue P( Z.7).9 Reject H at the 5% level.
Questions 7: In an experiment on what influences opinion, students were asked whether they agree or disagree with the following statement: I hold that a little rebellion, now and then, is a good thing, and as necessary in the political world as storms are in the physical. One group was told that Thomas Jefferson said this, and a second group was told that Lenin said it. The data were Agree Disagree Jefferson 54 6 Lenin 44 6 Test the hypothesis that the source of the quotation does not effect the distribution of whether students agree with the statement. You can use the 5% level of significance. Two-way table for the data: Agree Disagree Total Jefferson 54 6 6 Lenin 44 6 6 Total 98 H : The source of the quotation does not effect the distribution of whether students agree with the statement H a : The source does effect the distribution. Calculate the expected count for each cell: 6 98 Cell (,): 49 6 98 Cell (,): 49 6 Cell (,): 6 Cell (,): All expected counts are greater than 5. (54 49) 49 (44 49) 49 (6 ) (6 ) 5.566 r=, c=, df = (-)(-) =. Use the table with degree of freedom. P( 5.566) is between. and.. We can reject the null hypothesis at the 5% level.
Question 8: A study examined the relationship between ozone level (in parts per million) and the population (in millions) in 6 U.S. cities. A regression analysis with ozone level as the response variable, y, and population as the explanatory variable, x, gave the following results: R-squared = 84.4% s e = 5.454 Variable Coefficient SE(Coefficient) Intercept 8.89.395 Population 6.65.9 Test the hypothesis that there is no linear relationship between the ozone level and population at the 5% level of significance. H H : : a Test the hypothesis at the 5% level of significance. b 6.65 t 3.48 SE( b ).9 Compare with the t-model with 4 degrees of freedom. Reject if t>.45. Reject H. :