Sampling Distributions & Confidence Intervals Explained

Final assignment Submitted: By Apurva Mohapatra Sampling Distribution and Confidence Interval - Segment 6 PROBLEM 1 A childcare agency was interested in examining the monthly amount that family pay for childcare outside the home per child. A random sample of 64 families was selected and the average and standard deviation were computed to be $675 and $40 respectively. a) Find a 99% confidence interval for the monthly average amount spent per child. CI 99%=2.57 SRS (Simple Random Sample) =64 Average =675 S.D.= 40 Standard error = S.D./ √(SRS) = 40/√(64) = 40/8 = 5 Confidence interval (C.I.) for upper bound = average +(Z-score* 5) = 675+(2.57*5) = 675+12.85 =687.85 Confidence interval(C.I.) for lower bound = average -(Z-score* 5) = 675-(2.57*5) =675-12.85 =662.15 We are 99% confident that the average number of mixers sold is 687.85and 662.15. b) A social worker claims that the average monthly amount spent for childcare outside the home is $700 per child. You suspect it is less. Based on the sample data, can you reject the social worker’s claim? The value of $700 is not contained in the confidence interval computed above. This means the data provides strong evidence to reject, the social works claim, family spent less than $700 per child for childcare outside the home. Page 1 of 8

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PROBLEM 2 The mean time between failures (in hours) for a Telektronic Company radio used in light aircraft is 420h. After 15 new radios were modified in an attempt to improve reliability, tests were conducted to measure the times between failures. For the 15 new radios, the average time between failures was 434.73h and the standard deviation was 18.01h. Compute a 95% confidence interval. Does it appear that the modification improved the reliability? CI 95%=1.96 SRS (Simple Random Sample) = 15 Average = 420 Average (after 15 radio were added)= 434.73 S.D.= 18.01 Standard error = S.D./ √(SRS) = 18.01/√(15) = 18.01/ 3.8729833462 =4.65 Confidence interval (C.I.) for upper bound = average +(Z-score* 4.65) = 434.73+(1.96*4.65) =434.73+ 9.114 =443.844 Confidence interval(C.I.) for lower bound = average -(Z-score* 4.65) = 434.73-(1.96*4.65) =434.73-9.114 = 425.61 It indicates that the average time within failures is the value between 425.61 and 443.84. These vales are significantly higher than the 420. PROBLEM 3 Fifty new computer chips were tested for speed in a certain application. The average speed, in Mhz, for the new chips was 495.6 and the standard deviation was 19.4. a) Compute a 95% confidence interval for the average speed using the sample of 50 computer chips. CI 95%=1.96 SRS (Simple Random Sample) =50 Average =495.6 S.D.= 19.4 Page 3 of 8

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Standard error = S.D./ √(SRS) = 19.4/√(50) = 19.4/7.071 = 2.74 Confidence interval (C.I.) for upper bound = average +(Z-score* 2.74) = 495.6+(1.96*2.74) = 495.6+5.37 = 500.97 Confidence interval(C.I.) for lower bound = average -(Z-score* 2.74) = 495.6-(1.96*2.74) = 495.6-5.37 = 490.23 b) An older version of the chips had a speed of 481.2 Mhz. Can you conclude that the mean speed for the new chips is greater than the average speed of 481.2 Mhz of the old chips? We can conclude the mean speed of the new chips is significantly greater than the speed of 481.2 Mhz of the old chips. Since 481.2 Mhz is smaller than the values in the 95% confidence interval. c) A second sample of 100 new chips were selected and tested. The observed average computer speed was 500.3 mhz with standard deviation equal to 20.3. Can you conclude that the results from the first study are consistent with the second larger study? CI 95%=1.96 SRS (Simple Random Sample) =100 Average =500.3 S.D.= 20.3 Standard error = S.D./ √(SRS) = 20.3 /√(100) = 20.3/10= 2.03 Confidence interval (C.I.) for upper bound = average +(Z-score* 2.03) = 500.3+(1.96*2.03) = 500.3+3.9788 = 504.2788 Confidence interval(C.I.) for lower bound Page 4 of 8

= average -(Z-score* 2.03) = 500.3-(1.96*2.03) = 500.3-3.9788 = 496.3212 The second study confirms the results in (a) at the 95% confidence interval contains values larger than 481.2Mhz. the range is defined for the 2 955 confidence interval overlaps and provide similar results. Tests of Significance - Segment 8 Problem 1 A special cable has a breaking strength of 800 pounds. A researcher selects a sample of 80 cables and finds that the average breaking strength is 793 pounds, with standard deviation equal to 12 pounds. Can one reject the claim that the breaking strength is 800 pounds? Step 1: Find H 0 & H a Step 2: Z-score= avg -Ho/(S.D./√SRS) Step 3: Probability Step 4: Conclusion a. Write down the test hypotheses, the test statistic and the p-value. H 0 = 800 H a ≠800 b. Should the null hypothesis be rejected at the significance level of 0.01? Z-score= avg -Ho/(S.D./√SRS) = 793-800/(12/√80) = 793-800/(12/8.94) = 7/ 1.342 =-5.21 The Z value is less than 0.01. The P value is almost zero, we can reject the null hypothesis, at 1% level. The data support the alternative hypothesis the breaking strength is not 800 pounds. Page 5 of 8

Problem 2 A bank wonders whether omitting the annual credit card fee for customers who charge at least $3,000 in a year would increase the amount charged on their credit card. The bank makes an offer to an SRS of 500 existing credit card customers. It then compares how much these customers charge this year with the amount they charges last year. The mean increase is $565, and the standard deviation is $267. a. Is there significant evidence at the 1% level that the mean amount charged increases under the no-fee offer? State the null Ho and alternative hypothesis Ha and carry out a statistical test. ( HINT: the hypothesis test evaluates if the average increase in charges is significantly larger than zero ) H 0 = 0 H a>0 SRS= 500 Avg = $565 S.d.= 267 Z-score= avg -Ho/(S.D./√SRS) = 565/(267/√500) = 565/(267/22.36) = 565/11.941 =47.317 The P value is equal to zero, the test is highly significant data shows strong evidence that there is a significant increase in charges under the no fee offer. b. Give a 95% confidence interval for the mean amount of the increase SRS= 500 Avg = $565 S.d.= 267 CI 95%=1.96 Standard error = S.D./ √(SRS) = 267/√500 = 267/22.36 = 11.941 Confidence interval (C.I.) for upper bound = average +(Z-score* 11.941 ) = 565 +(1.96* 11.941 ) = 565 +23.40436 = 588.404 Page 6 of 8

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Confidence interval(C.I.) for lower bound = average -(Z-score* 11.941 ) = 565 -(1.96* 11.941 ) = 565 -23.40436 = 541.596 c. A critic points out that the customers would probably have charged more this year than last even without the new offer because the economy is more prosperous and interest rates are lower. Briefly describe the design of an experiment to study the effect of the no-fee offer that would avoid this criticism. To test the effectiveness of the no-fee offer a randomized controlled experiment can be conducted where randomly selected customersare assigned to two groups. The controlled group that are not offered the no-fee option and the treatment group that is offered the no-fee option. The creadit card charges will be recorded from the two groups for a certain period of time. Problem 3 A random sample of 10 one-bedroom apartments from your local newspaper has these monthly rents (dollars): 500, 650, 600, 505, 450, 550, 515, 495, 650, 395. Do these data give good reason to believe that the mean rent of all advertised apartments is greater than $500 per month? a. State hypotheses. H 0 = 500 H a> 500 b. Find the mean and standard deviation using SPSS or Excel. Avg = (500+650+ 600+505+ 450+ 550+515+ 495+ 650+395)/10=5,310/10= 531 SD = 82.7915723 c. Compute the p-value and draw your conclusions. Z-score= avg -Ho/(S.D./√SRS) =531-500/(82.79/√10) =31//(82.79/3.162) =31/26.1804967485 =1.184 1-NORMDIST (1.18,0,1,TRUE)=1- 0.881 = 0.119 Mean is 0 because its normal distribution and S.D. is 1. Page 7 of 8

Since the p-value is larger than 0.05 data does not provide enough evidence that the average rent is greater than $500 Page 8 of 8

Final IT_ApurvaMohapatra

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