A. Identify the Parameter of Interest. Μ = the Mean Oven Temperature

Chapters 11 and 12

1. We wish to see if the dial indicating the oven temperature for a certain model oven is properly calibrated. Four ovens of this model are selected at random. The dial on each is set to 300ºF, and, after one hour, the actual temperature of each is measured. The temperatures measured are 305º, 310º, 300º, and 305º. Assume that the actual temperatures for this model when the dial is set to 300º are normally distributed.

a. Identify the parameter of interest. µ = the mean oven temperature b. Identify the appropriate procedure to find out if the ovens are heating to the right temperature. What are the

H0 :m = 300 null and alternative hypothesis? 1-sample t-test, Ha :m 300 c. Perform the procedure. What are your conclusions? (check conditions – given that it is normal, SRS, 40< pop, each oven is independent of each other.) x - m 305- 300 t =0 = = -2.449 3 s n 4.08 4

P( t3 >2.449) = p = .09 p > a =.05. Therefore we fail to reject the null hypothesis. There is not significant evidence to conclude that mean oven temperature is different than 300 degrees.

2. Bags of a certain brand of tortilla chips claim to have a net weight of 14 ounces. Net weights actually vary slightly from bag to bag and are normally distributed with mean m. A representative of a consumer advocate group wishes to see if there is any evidence that the mean net weight is less than advertised. To do this, he selects sixteen bags of this brand at random and determines the net weight of each. He finds the sample mean to be x =13.82 and the sample standard deviation to be s = 0.24.

a. Identify the parameter of interest. The parameter of interest is m , the mean weight of bags of tortilla chips. b. Identify the appropriate procedure to determine if a bag of tortilla chips contains the right amount. What are the null and alternative hypothesis? Perform a one-sample t-test Conditions met

H0 :m = 14

Ha :m < 14 c. Perform the procedure. What are your conclusions? x - m 13.2- 14 t =0 = = -3 15 s n 0.24 16

P( t15 < -13.333) = p = .004 p < a =.05. There is significant evidence to conclude that there are not enough tortilla chips in each bag. (Reject the null in favor of the alternative.

3. The water diet requires one to drink two cups of water every half hour from when one gets up until one goes to bed, but otherwise allows one to eat whatever one likes. Four adult volunteers agree to test the diet. They are weighed prior to beginning the diet and after six weeks on the diet. The weights (in pounds) are

Person 1 2 3 4 Weight before the diet 180 125 240 150 Weight after six weeks 170 130 215 152

For the population of all adults, assume that the weight loss after six weeks on the diet is normally distributed. We want to determine if the diet leads to weight loss. a. Identify the parameter of interest.

Parameter of interest is mean loss of weight after six weeks on the diet, mL= m B - m D

Let mB represent subject’s weight before the diet

Let mD represent subject’s weight after the diet b. Identify the appropriate procedure to determine if the diet leads to weight loss. Matched pairs – t-test (one sample)

H0 :mL = 0

Ha:m L > 0 c. Perform the procedure. What are your conclusions?

x - m 7- 0 t =0 = =1.0265 3 s n 13.638 4

P( t3 >1.0265) = p = .1901 At an alpha level of a = .05 (or even a = 0.10 ) there is not significant evidence to conclude that the diet leads to weight loss. Our p-value of p =0.1901 >a = 0.05 , so we can not reject the null hypothesis.

4. Do students tend to improve their SAT Mathematics (SAT-M) score the second time they take the test? A random sample of four students who took the test twice received the following scores.

Student 1 2 3 4 First score 450 520 720 600 Second score 440 600 720 630

Assume that the change in SAT-score for the population of all students taking the test twice is normally distributed. We want to know how much taking the test twice helped.

a. Identify the parameter of interest.

The parameter of interest is the mean difference in scores, mD , on SAT test taken twice where

mD= m2 nd - m 1 st b. Identify the appropriate procedure to determine how much taking the test twice helped their score Since we are trying to find out how much taking the test twice helped, we construct a matched pairs t-interval (one sample). The problem doesn’t specify, but we will construct a 95% confidence interval. c. Perform the procedure. What are your conclusions?

We are 95% confident that the mean difference (increase) in SAT scores was between -39.31 and 89.309. 5. A newspaper conducted a statewide survey concerning the 1998 race for state senator. The newspaper took a random sample (assume it is an SRS) of 1200 registered voters and found that 620 would vote for the Republican candidate. Pollsters want to determine if the Republican candidate is going to win.

a. Identify the parameter of interest. The parameter of interest is p, the proportion of people who will vote for the Republican candidate. b. Identify the appropriate procedure to find if the Republican candidate is going to win. What are the null and alternative hypothesis? We will conduct a one-proportion z-test

H0 : p .50

Ha : p > .50 c. Perform the procedure. What are your conclusions? pˆ - p (.51667- .5) z =0 = =1.1547 p(1- p) ( 0.5 0.5) n 1200 P( z >1.1547) = .1241

Our p-value of .1241 > a = .05 , so there is not enough evidence for us to conclude that the Republican candidate will win the election. There is not enough evidence to reject the null hypothesis.

6. A noted psychic was tested for ESP. The psychic was presented with 200 cards face down and asked to determine if the card was one of five symbols: a star, cross, circle, square, or three wavy lines. The psychic was correct in fifty cases. Assume the 200 trials can be treated as an SRS from the population of all guesses the psychic would make in his lifetime. Suppose you wished to see if there is evidence that the psychic is doing better than just guessing.

a. Identify the parameter of interest. The parameter of interest is p, the proportion of cards that the psychic correctly identifies. b. Identify the appropriate procedure to find out if the psychic is really psychic. What are the null and alternative hypothesis? We will conduct a one-proportion z-test

H0 : p = .20

Ha : p > .20 c. Perform the procedure. What are your conclusions? pˆ - p .25- .20 z =0 = =1.7678 p(1- p) .20( .80) n 200 P( z >1.7678) = .03855 Our p-value of .03855 is less than a 5% significance level, so there is ample evidence to reject the null hypothesis in favor of the alternative and conclude that the psychic is actually psychic.

7. An inspector inspects large truckloads of potatoes to determine the proportion p in the shipment with major defects prior to using the potatoes to make potato chips. Unless there is clear evidence that this proportion is less than 0.10, she will reject the shipment. To find out, she selects an SRS of fifty potatoes from the over 2000 potatoes on the truck. Suppose that only two of the potatoes sampled are found to have major defects.

a. Identify the parameter of interest. The parameter of interest is p, the proportion of potatoes with defects. b. Identify the appropriate procedure to determine if the potato chips are ok. What are the null and alternative hypothesis? We will conduct a one-proportion z-test

H0 : p .10

Ha : p > .10 c. Perform the procedure. What are your conclusions? (Conditions not met np=2<10, but we will proceeded with caution)

pˆ - p .04- .10 z =0 = = -1.414 p(1- p) .10( .90) n 50 P( z > -1.414) = .07865 Our p-value of .07865 is greater than any acceptable significance level (a ) so we cannot reject the null hypothesis, and we conclude that the shipment of potatoes is fine!

8. In a large Midwestern university (the class of entering freshmen being on the order of 6000 or more students), an SRS of 100 entering freshmen in 1993 found that 20 finished in the bottom third of their high school class. Admission standards at the university were tightened in 1995. In 1997 an SRS of 100 entering freshmen found that 10 finished in the bottom third of their high school class. We want to find out how much tightening their standard helped.

Our parameter of interest is pD , the difference in the proportion of students in the bottom third of the class

Let pD = p97 - p 93 b. Identify the appropriate procedure to find out how much tightening their standard helped. We construct a 95% confidence interval, for two-sample proportions c. Perform the procedure. What are your conclusions? * p1(1- p 1) p 2( 1 - p 2 ) ( pˆ1- p ˆ 2 ) � z n1 n 2 骣 (.1)( .9) ( .2)( .8) =0.1 - 0.2� 1.645琪 ( ) 琪桫 100 100 =( -.198, - .002) We are 95% confident that the difference in the proportion of students that were in the bottom third of their HS graduating class is between -19.8% and -0.2%. Because all of the values in the confidence interval are negative, we can conclude that fewer freshman were in the bottom third of their HS graduating class, and tightening their standards must have helped.

9. An SRS of 100 flights of a large airline (call this airline 1) showed that 64 were on time. An SRS of 100 flights of another large airline (call this airline 2) showed that 80 were on time. Is airline 2 more reliable than airline 1?

Our parameter of interest is p1 and p 2 , the proportion of flights on-time for airlines 1 and 2 respectively b. Identify the appropriate procedure to find out if airline 2 is more reliable than airline 1. What are the null and alternative hypothesis? We will conduct a 2-proportion z-test

H0: p 1= p 2

Ha : p1< p 2 c. Perform the procedure. What are your conclusions?

( pˆ- p ˆ ) .64- .80 z =1 2 = = -2.519 p(1- p) p( 1 - p ) ( .64)( .36) ( .80)( .20) 1 1+ 2 2 + n1 n 2 100 100 P( z < -2.519 = .00587) Because our p-value is extremely small (p=.00587) and way below our sig level of .05, we can conclude that airline 2 is in fact more reliable than airline 1. We reject the null in favor of the alternative