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Pages 8 (2008 words)
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…
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.
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