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Statistical Methods - Math Problem Example

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1. The 95% confidence interval for the unknown mean from a Normal distribution is given by 1.96 where is the sample mean, is the population standard deviation and n is the sample size. Here 1.96 is the value of the standard normal curve such that P (Z 1.96) = 0.025…
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Download file to see previous pages 4. Given data: , Sxx = 27100, Syy = 280.1 and Sxy = 2665
Cxx = Sxx - n
= 27100 - 10(51.82) = 267.6
Cyy = Syy - n
= 280.1 - 10(5.12) = 20
Cxy = Sxy - n
= 2665 - 10 (51.8) (5.1) = 23.2
The slope a of the fitted regression line Y on X is a =
Intercept b = = 5.1 - (0.0867) (51.8) = 0.6091
Estimate S for the standard deviation of the model
s = =

5. H0: A = 0 Vs H1: A 0 where A is the slope of the fitted line Y on X.
Test statistic: Under H0, tn - 2 distribution where a is the sample slope parameter, A is the population slope parameter, s is the sample estimate for the standard deviation. The results from Q4 are a = 0.0867, s = 1.4495 and Cxx = 267.6 and A = 0.
Test statistic:
Pr {-2.306 t8 2.306} = 0.95.
The value 0.9786 is contained in this interval and hence we have insufficient evidence to reject the null hypothesis. The p-value is 0.3 which is higher than 0.05. Hence we can conclude that the population slope parameter A = 0.

6. The Fitted line is y = ax* + b where 'a' is the slope and 'b' is the intercept. From previous questions we have the results a = 0.0867 and 0.6091
At x* = 44, the value of the line is y = 0.0867(44) + 0.6091 = 4.424
At x* = 52, the value of the line is y = 0.0867(52) + 0.6091 = 5.118
At x* = 54, the value of the line is y = 0.0867(54) + 0.6091 = 5.291
The 95% confidence interval for mean of Y is given by the formula

where a = 0.0867, b = 0.6091, x* = 44, 52 and 54, = 51.8, n = 10, s = 1.4995, Cxx = 267.6 and t8,0.025 = 2.306
At x* = 44, the confidence interval is

4.4239 1.9784
(2.4455, 6.4023)

At x* = 52, the confidence interval is

5.1175 1.09428
(4.023, 6.21178)
At x* = 54, the confidence...
The value 0.9786 is contained in this interval and hence we have insufficient evidence to reject the null hypothesis. The p-value is 0.3 which is higher than 0.05. Hence we can conclude that the population slope parameter A = 0.
The tabulated value for t32 distribution at upper 5% significance level is 1.694. Since our test statistic value is higher than this value, we reject H0 at 10% significance level. The tabulated value for t32 distribution at upper 2.5% significance level is 2.037. The test statistic is lower than this. Hence we accept H0 at 5% level of significance.
The estimated p-value should be between (0.05, 0.1) excluding the upper and lower limits. The result is statistically significant and we can conclude there is a difference in the mean profit outputs. Since we can only say that the means are not equal and cannot say about which is larger, we recommend carrying out one-sided test and then choosing about which course is best.
This is reasonable as the sum of n-1 values of a sample gives the other value of the statistic and there is dependence between the n terms and so the degrees of freedom of sample size n are n - 1. As we infer about 2 samples it is reasonable to use 2(n-1) as degrees of freedom when the sample sizes and variances are equal. ...Download file to see next pagesRead More
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