Question

In: Statistics and Probability

19) -In a crash test at five miles per hour, the mean bumper repair cost for...

19) -In a crash test at five miles per hour, the mean bumper repair cost for 14  midsize cars was $547 with a standard deviation of $85. In a similar test  of 23 small cars, the mean bumper repair cost was $347 with a standard   deviation of $185. At a = 0.05, can you conclude that the mean bumper repair cost is the same for midsize cars and small cars? a) List the p-value b) accept or reject.

- Last year the average daily change in the Dow Jones Industrial Index was 7.3 points. A random sample of 15 trading days this year showed the average change to be 5.6 points with standard deviation 10.2. Test the claim that the average change in the index is less this year than it was last year. a) List the p-value b) Have we proven that the change is significant? (yes or no) c) what type of error could be made with the type of conclusion drawn from the results of this study?

Solutions

Expert Solution

There is no sufficient evidence to conclude that the mean bumper repair cost is same for midsize cars and small cars.

(b)The change is not significant. There is no sufficient evidence to tell that the change is significant.

(c) We might be making type 2 error because type 2 error is accepting Ho when H1 is true and we are accepting Ho.


Related Solutions

company crashed four cars in succession at 5 miles per hour the cost of repair for...
company crashed four cars in succession at 5 miles per hour the cost of repair for each of the four crashes was $425 $455 400 $16 $232 compute the range sample variance and variance standard deviation cost of repair
1-The mean speed of vehicles along a stretch of highway is 75 miles per hour with...
1-The mean speed of vehicles along a stretch of highway is 75 miles per hour with a standard deviation of 3.8 miles per hour. Your current speed along this stretch of highway is 62 miles per hour. What is the z-score for your speed? z- score =_______ (Round to two decimal places) 2- For a statistics test the mean is 63 and the standard deviation is 7.0, and for a biology test the mean is 23 and the standard deviation...
an insurance company crashed four cars in succession at 5 miles per hour. the cost of...
an insurance company crashed four cars in succession at 5 miles per hour. the cost of repair of the four crashes was $429 $459 $318. Compute the range, sample variance, and sample standard deviation cost of repairs
an insurance company crashed four cars in succession at 5 miles per hour. the cost of...
an insurance company crashed four cars in succession at 5 miles per hour. the cost of repair for each of the four crashes was $428, $463, $413, $230. Compute the range, sample variance and the sample standard devition cost of repair
In a random sample of five microwave​ ovens, the mean repair cost was ​$65.00 and the...
In a random sample of five microwave​ ovens, the mean repair cost was ​$65.00 and the standard deviation was ​$13.50. Assume the population is normally distributed and use a​ t-distribution to construct a 90​% confidence interval for the population mean mu. What is the margin of error of mu​? Interpret the results. The 90​% confidence interval for the population mean mu is ​( nothing​, nothing​). ​(Round to two decimal places as​ needed.)
In a random sample of five microwave​ ovens, the mean repair cost was ​$60.00 and the...
In a random sample of five microwave​ ovens, the mean repair cost was ​$60.00 and the standard deviation was ​$14.00. Assume the variable is normally distributed and use a​ t-distribution to construct a 95​% confidence interval for the population mean u. What is the margin of error of u​? The 95​% confidence interval for the population mean u is l(_, _ ) ​(Round to two decimal places as​ needed.
In a random sample of five mobile​ devices, the mean repair cost was ​$90.00 and the...
In a random sample of five mobile​ devices, the mean repair cost was ​$90.00 and the standard deviation was ​$11.00. Assume the population is normally distributed and use a​ t-distribution to find the margin of error and construct a 95​% confidence interval for the population mean. Interpret the results. The 95​% confidence interval for the population mean ​(76.34,103.66). ​(Round to two decimal places as​ needed.) The margin of error is ​$____?? I NEED THIS ANSWER, OR EXPLAINED HOW TO FIGURE...
In a random sample of five microwave​ ovens, the mean repair cost was $60.00 and the...
In a random sample of five microwave​ ovens, the mean repair cost was $60.00 and the standard deviation was $11.50 Assume the population is normally distributed and use a​ t-distribution to construct a 90​% confidence interval for the population mean μ. What is the margin of error of μ​? Interpret the results. (a)The confidence interval for the population mean is _, _   ​(Round to one decimal place as​ needed.) (b)The margin of error is _ ​(Round to two decimal places...
A plant manager considers that operational cost per hour of five alternative machines. The cost per...
A plant manager considers that operational cost per hour of five alternative machines. The cost per hour is sensitive to three potential weather conditions: cold, mild, and warm. The following table represents the operational cost per hour for each alternative-state of nature combination. Alternatives Cold Mild Warm Machine 1 50 40 45 Machine 2 45 42 47 Machine 3 40 35 54 Machine 4 60 25 48 Machine 5 45 40 45 Probability 30% 50% 20% The EVPI is .
In a crash test of 26 minivans of a Japanese manufacturer, collision repair costs are found...
In a crash test of 26 minivans of a Japanese manufacturer, collision repair costs are found to have a distribution that is roughly bell shaped, with a mean of $1850 and a standard deviation of $340. Construct a 95% confidence interval for the mean repair cost in all such vehicle collisions by filling in the following blanks (Round confidence limits to the nearest dollar). (Please provide clear explanation and steps and calculations) a) df = b) Critical value = c)...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT