Reference no: EM13237044
1. Suppose the monthly percentage changes in the TSE 300 index from 1980 to 2000 are normally distributed with a standard deviation of 2.05 percentage points. We wish to estimate the mean value of monthly percentage changes during this period. To do so, we randomly selected a sample of 20 such changes, the values of which are as follows:
3.92
|
1.69
|
1.79
|
-1.00
|
-2.05
|
1.41
|
2.65
|
-1.09
|
3.92
|
1.48
|
2.00
|
0.85
|
0.31
|
1.46
|
3.92
|
.094
|
0.46
|
2.91
|
0.23
|
0.71
|
Find a 90 percent confidence interval estimate of the population mean.
Find a 95 percent confidence interval estimate of the population mean.
2. Using the same sample data as in question 7, suppose that we know that the population was normally distributed, but we do not know what the population standard deviation was. Find a 90 percent confidence interval estimate of the population mean.
3. A research firm conducted a survey to determine the mean amount smokers spend on cigarettes during a week. A sample of 49 smokers found a sample average of $54 and a sample standard deviation of $10. What is the 95 percent confidence interval estimate of the population mean.
Interval Estimators: Population Proportion
4. In a survey conducted by Ipsos-Reid in August 2009, 63 percent of the 1000 Canadians surveyed said that they are less likely to buy genetically modified food products. Construct a 95 percent confidence interval estimate for the proportion of Canadians who are less likely to buy genetically modified food products.
5. A market survey was conducted to estimate the proportion of homemakers in Ontario who would recognize the brand name of a cleanser based on the shape and the colour of the container. Of the 1400 homemakers sampled, 420 were able to identify the brand by name. What is the 99 percent confidence interval estimate of the population proportion?
Hypothesis Testing
6. The manufacturer of a steel-belted radial tire claims that the mean mileage the tire can be driven before the tread wears out is 80,000 kilometres. The standard deviation of tread life is 8,000 km. A truck company purchased 49 of these tires and found a mean tread life of only 79,200. On the basis of this data, test the null hypothesis that the mean tread life is at least 80,000 km. against the alternative that the mean life is less than 80,000 km. Use a 5% level of significance.
7. Past records show that the mean life of a certain brand of batteries used in a digital clock is 305 days. The battery was recently modified to last longer. A sample of 20 of the modified batteries had a mean life of 311 days with a sample standard deviation of 12 days. Did the modification increase the mean life of the battery? Assume that the population distribution of battery life is normally distributed and use a 5% level of significance. The null hypothesis is that the mean life is still no greater than 305 days.
8. Global TV news, in a segment on the price of gasoline, reported that the mean price nationwide is $1.29 per litre for regular unleaded gasoline. A random sample of 36 stations in Toronto revealed a mean price of $1.315 per litre and a sample standard deviation of $0.05 per litre. At the 5 percent level of significance, test the null hypothesis that gasoline prices in Toronto are no higher than the national average against the alternative hypothesis that they are.
9. A recent American Medical Association survey of 785 randomly selected university graduates found that 144 of them smoked. The rate of smoking in the general population is 27%. Using a 0.01 significance level, test the claim that the rate of smoking among university graduates is less than the rate in the general population. The null hypothesis is that the rate among university graduates is not less.
10. In a study of air-bag effectiveness, it was found that in 821 crashes of midsize cars equipped with air bags, 46 of the crashes resulted in hospitalization of the drivers. The hospitalization rate for mid-size cares without air bags is 7.8%. Using a 0.01 significance level, test the claim that airbags reduce the hospitalization rate. The null hypothesis is that they do not.
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