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Case -I A total of 2058 students take a difficult test. Each student has an independent 0.6205 probability of passing the test. a.What is the probability that at least 1300 students will pass the test. b.What is the probability that between 1250 and 1300 students both numbers inclusive will pass the test? ANSWER

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Page 1: Statistics All

Case -I

A total of 2058 students take a difficult test. Each student has an independent 0.6205 probability of passing the test.

a. What is the probability that at least 1300 students will pass the test.

b. What is the probability that between 1250 and 1300 students both numbers inclusive will pass the test?

ANSWER

Page 2: Statistics All

Case II

A sample of dry battery cells tested to find the length of life produced the following result.

Mean= 12 hoursStd deviation- 3 hoursa. More than 15 hoursb. Less than 6 hours

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Case III

• An auditor takes a random sample of n=36 from the population of 1000 accounts receivables . The mean value of the accounts receivables for the population is Rs.260 with standard deviation of Rs. 45. what is the probability that the sample mean will be less than Rs. 250?

Page 4: Statistics All

Case IV

A marketing research analyst selects a random sample of 100 customers who purchased a particular item from central store. Hundred customers spend an average of Rs.250 with standard deviation of Rs. 70 for 95% customers estimate the mean purchase amount?

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Case V

An auditor takes a random sample of n=25. The mean value of the accounts receivables for the population is Rs.260 with standard deviation for the sample is Rs. 45. what is the probability that the sample mean will be less than Rs. 250?

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Case VI

• A manager at a departmental store wants to know the average purchase by the consumer over last one year. A random sample of 25 bills gives the sample mean Rs. 357 and sample standard deviation Rs.140 give 95% confidence interval for the average purchase of the year for the population

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Case VII

Suppose we are testing claim by Tata Motors that Tata Nano car gives an average running of 20 kmpl. We know from previous testing that standard deviation is 3kmpl now from the new production lot we test drive 100 cars and find average 19.5kmpl. Could we reasonably say that Tata’s claim is acceptable at 5% significance level?

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Case VIII(normal distribution or z test)

The statistical quality control department takes a random sample of 200 cans of 5kg packs of cooking oil. It gave an average weight of 4.98kg. Based on earlier experience of the machine used the standard deviation of the population mean is 0.21kg. Do we accept the hypothesis that the net average weight is 5kg per can at 10% significant level?

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Case IX t-distribution

The average breaking strength of steel rods is specified to be 18.5 thousands lbs. for this a sample of 14 rods was tested. The mean and standard deviation obtained were 17.85 and 1.955. conclude that the breaking strength is 18.5 thousands lbs. significant level is .1?

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Case X binomial distribution

Your company has launched a new product. Your company is a reputed company with 50% market share of similar range of products. Your competitor also enters with their new product equivalent to your new product. Based on earlier experience you initially estimated that your market share of the new products would be 50%. You carry out random sampling of 15 customers who have purchased your product and found out that only 4 customers have actually purchased the product. Test if you are likely to have half of the market share at 10% confidence level?

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Case XI approximation of binomial distribution

Your company has launched a new product. Your company is a reputed company with 50% market share of similar range of products. Your competitor also enters with their new product equivalent to your new product. Based on earlier experience you initially estimated that your market share of the new products would be 50%. You carry out random sampling of 25 customers who have purchased your product and found out that only 7 customers have actually purchased the product. Test if you are likely to have half of the market share at 10% confidence level?

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Case XII proportion and hypothesis testing

Marketing manager believes that his new marketing campaign has increased company’s market share from earlier known market share of 30%. A sample of 50 customer indicate that only 12 purchased company’s product. What should the manager conclude at 5% significance level?

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Case XIII(chi square test – two tailed)

The surface hardness of composite material sheet is known to have variance of 0.40. for the received, the sample variance is 0.22 for random sample of 9 sheets. Is it right to conclude that this shipment has a variance different from 0.40 at 5% significance level?

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Case XIV

The following data refers to the yearly earnings of 20 individuals before and after training program under vocational guidance scheme. Is there any improvement in the earning after the training? Test at significance level of 0.05

The data is given in MS EXCEL