An accelerated life test on a large number of type-D alkaline batteries revealed that the mean life for a particular use before they failed is 19.0 hours. The distribution of the lives approximated a normal distribution. The standard deviation of the distribution was 1.2 hours. About 95.44 percent of the batteries failed between what two values?
Answer(s): A
9544/2 = 0.4772. For an area under the curve of 0.4772, the z value is (+/-)2. We know that z = (x-u)/sigma.Therefore, +/-2 = (x-19)/1.2. Rearranging we get x = u +/- z*sigma. So x = 19 +/- 2*1.2 = 19+/-2.4. x is 16.6 and 21.4.
If the p-value is ________ the significance level, the null hypothesis ________.
The lowest significance level at which the null hypothesis can be rejected is called the p-value of the test. Thus, if the p-value is less than the significance level, the null hypothesis can be rejected at that significance level. To fix this in your mind, think about the p-value as the maximum probability that the null hypothesis is true despite observing the value of the test statistic that you have in the sample at hand. For e.g., suppose the p-value for a given sample turns out to be 3%. Then, if you reject the null, the probability that you have made an error in rejecting it is at most 3%. Clearly, if a critic asked you, "Is the probability that you have made an error less than 5%?" your answer would be, "yes." The critic will then conclude that your test is significant at the 5% level and perhaps even at a lower level.
At the end of every year for the next 3 years, you deposit $400 in an account that pays 5% per year, annually compounded. After that, you do not make any more deposits. The amount that you can withdraw after 7 yearsis:
For the first 3 years, your regular deposits constitute an annuity and after that, a simple term deposit. The future value of an n-period annuity that starts paying at the end of the current period equals FV = (C/r)*[(1+r)^n - 1] where C is the payment per period and r is the one-period interest rate. In this example, the annuity is over 3 periods, C = 400 and the per period rate equals 5%. So the future value of the deposits after 3 years equals (400/0.05)*[1.05^3 - 1] = 1,261. Starting in year 3, this amount grows at an annual rate of 5% for a period of 7-3 = 4 years. So in 7 years, the amount in the account equals 1,261*1.05^4 = $1,533.
The coefficient of variation of X is three times that of Y. If X and Y have the same means, the variance of Y is:
The coefficient of variation equals the ratio of the standard deviation to the mean. Hence, if X and Y have the same means, then the standard deviation of X must be 3 times the standard deviation of Y.
A variable, Z, is regressed against variables X and Y and the following equation is estimated: Z = -3.0 + 1.2X - 2.5Y + error. If X = 5 and Y = 3, the estimated value of Z is ________.
Answer(s): D
With the given data, Z = -3 + 1.2*5 - 2.5*3 = -4.5
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