approximately 120 drives for a daily average total of 360 drives per day. The company runs quality checks called PDQ tests on one of their drives every hour of production. The test takes up to 20 minutes. Their historical “in control” process is defined with a mean of 7.0 and a standard deviation of .3. Four-D performs their own PDQ tests on the drives that DataStor ships them. They sample ten drives at random. If any of the drives have a PDQ score of 6.2 or below‚ then the entire shipment will be rejected
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MHR 705 Problem Set 4 Top Dollar Sales (TDS) is a national firm that sells automobile and life insurance. TDS employs 500 insurance agents. Each agent works somewhat independently to contact and service clients. However‚ TDS is organized into 100 different geographically regions. The average sales for a region is $1‚000‚000‚ with a standard deviation of $100‚000. The manager of each geographical region has the autonomy to establish a compensation plan. The average annual compensation
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sample mean to be 499 ml. The specification of the machine says that generally the standard deviation is 1 ml. He does a hypothesis testing at 10% level of significance. Q2) A tyre company wants to test the stress of tyres. The tyres should withstand a minimum load of 80‚000kgs but excess load would burst the tyres. From the past experience‚ it is known that the standard deviation of the load is 4000kgs. A sample of 100 tyres was selected and tests were carried out. The result showed that mean stress
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one’s own means. This is also called bootstrap funding. Example Let’s understand this with an interested example of a candy factory. Knowing that the candy bars have mean weight. Since it is not feasible to weigh each candy that is produced in the factory‚ we use sampling techniques and randomly choose 100 candies. We calculate the mean of these 100 candies‚ and say that the population mean falls under a margin of error. Suppose that after a couple of months‚ we want to know the mean candy weight
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PuttingPeople2Work has a growing business placing out-of-work MBAs. They claim they can place a client in a job in their field in less than 36 weeks. You are given the following data from a sample. Sample size: 100 Population standard deviation: 5 Sample mean: 34.2 Formulate a hypothesis test to evaluate the claim. (Points : 10) Ho: µ = 36; Ha: µ ≠ 36 Ho: µ ≥ 36; Ha: µ < 36 Ho: µ ≤ 34.2; Ha: µ > 34.2 Ho: µ > 36; Ha: µ ≤ 36 Ans. b. H0 must always have equal sign‚ < 36 weeks 2. (TCO
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Elements of a Test of Hypothesis 1. Null Hypothesis (H0 ) - A statement about the values of population parameters which we accept until proven false. 2. Alternative or Research Hypothesis (Ha )- A statement that contradicts the null hypothesis. It represents researcher’s claim about the population parameters. This will be accepted only when data provides sufficient evidence to establish its truth. 3. Test Statistic - A sample statistic (often a formula) that is used to decide whether to reject H0
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Population & Sample Means 2. Expected Values 3. Population & Sample Variances 4. Population & Sample Covariances 5. Population & Sample Correlation Coefficients 6. Estimators Statistic – a numerical characteristic of a sample Statistical inference – drawing conclusion about a population based on information contained in a sample Random sample – every sample of the same size in the population has the same chance of being selected 1 2 POPULATION MEAN AND SAMPLE MEAN Set of all possible
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the median‚ mean‚ minimum‚ and maximum values should be assessed and examined to determine if this hypothesis is accurate. Team B will be using the mean home prices in group one and two to determine if there is a significant difference in home prices for homes less than 15 miles from the city compared to those equal to or greater than 15 miles from the center of the city. Based on the possible testable outcomes‚ Team B will use the Null Hypothesis and Alternate hypothesis with the mean prices in group
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largest 2 3 7 8.5 9 3. A) Compute the mean‚ median‚ first quartile‚ and third quartile. Order Array: 5.25 5.29 5.32 5.32 5.34 5.36 5.40 5.40 5.40 5.41 5.42 5.42 5.44 5.44 5.44 5.45 5.45 5.46 5.47 5.47 5.49 5.50 5.50 5.50 5.51 5.52 5.53 5.53 5.53 5.53 5.54 5.54 5.55 5.55 5.56 5.56 5.57 5.57 5.57 5.58 5.58 5.58 5.61 5.61 5.62 5.63 5.65 5.67 5.67 5.77 Mean = 275.07 / 50 = 5.5014 Mode = 5.53 (four times) Median
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results of textbook‚ MP3 and ID card were cleared up to Appendix 1 and Appendix 2. The table below shows that the experimental results of different gender of teenagers from 15 to 18 have different estimated capacity. As we can see clearly from the mean about textbook‚ males was 20.24cm and females was 18.4cm‚ which shows males had 0.46cm of the deviation and females had 2.3cm of the deviation. And from the result of MP3 and ID card‚ the males were 7.11cm and 4.61cm‚ however‚ females were 7.5cm and
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