1) A floodlight is stated to last an average of 65 hours. A theater manager believes that the average life is less than that stated by the manufacturer. He takes a random sample of 21 bulbs and determines that the sample average is 62.5 hours, with a standard deviation of 3 hours. Using a .01 level of significance, is there enough evidence to conclude that the manufacturer’s claim is false?
2) According to Fortune, the average annual pay for a biomedical engineer in 1995 was $72,500, and the average annual pay for a chemical engineer was $73,970. If each statistic is based on two independent random samples of 120 engineer each, and the standard deviations were $5,000 for the biomedical engineers and $6,500 for the chemical engineers, test for the existence of a difference in the population average pay for the two professions at the 0.05 level of significance.
3) A group of five statistics students received the following grades on a test: 81, 74, 62, 92, and 76. An intervention plan is used and the same students increased their scores on a retest to: 85, 79, 75, 95, and 84. At the 0.05 significance level, is the intervention plan helpful in increasing grades?
4) One group was taught an assembly procedure using the usual sequence of steps. Another separate group was taught using an experimental technique. The amount of time (in seconds) required to assemble the unit for two samples are:
Usual Method: 41, 36, 42, 39, 36, 48, 49, 37, 39, 40
Experimental Method: 21, 27, 36, 20, 19, 21, 39, 24, 22, 37
At a.05 level of significance, test the statement that the experimental method requires less time to assemble the unit.
5) Tedko Associates is a marketing research firm that specializes in comparative shopping. Tedko is hired by General Motors to compare the selling prices of Pontiac Sunbird with Chevy Cavalier. Posing as a potential customer, a representative of Tedko visited 8 Pontiac dealerships and 6 Chevrolet dealerships and obtained quotes on comparable cars. The standard deviation for the selling prices of 8 Pontiac Sunbirds is $350 and on six Cavaliers, $290. At the 0.01 significance level do Pontiac Sunbirds exhibit greater variance in quotes than Cavaliers?
6) A consumer organization wants to know if there is a difference in the price of a particular toy at three different types of stores. The price of the toy was checked in a sample of five discount toy stores, five variety stores, and five department stores. The results are shown below. Conduct the test by using the five-step procedure and completing ANOVA table. If you reject Ho, span confidence intervals to show which means are different. Use 0.05 level of significance.
Discount Store Variety Department
12 15 19
13 17 17
14 14 16
12 18 20
15 17 19
Anova: Single Factor
SUMMARY
Groups Count Sum Average Variance
Discount Store 5 66 13.2 1.7
Variety 5 81 16.2 2.7
Department 5 91 18.2 2.7
ANOVA
Source of Variation SS df MS F
Treatments (Stores) 63.333
Error 28.4
Total 91.733
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Reply:Need help with a take-home midterm????
To answer question 1. There is not enough evidence to conclude that the manufacturer’s claim is false. Simply a 95% Confidence interval would be smaller that a 99% Confidence interval. The 95% Confidence interval if we could assume a normal distribution would be the observed 62.5 hours plus or minus 1.96 times the standard deviation of 3. Since we have a small sample of 21 the t-distribution would be appropriate, but we know that the calculation with a t-distribution will be even wider. So a interval of adding/subtracting 1.96*3 to 62.5 will be narrower than appropriate, but 65 is still within the limits of this smaller interval.
This answer won't get you a pass in any class I teach, but after years of experience this is my thought process on such a problem. This type of understanding a problem frequently allows totally avoiding any calculation outside of multiplying by two.
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