If you choose 40 random employees from the corporation, the standard error would equal 6/Square root of 40 = .95 days. The 12 days in this department corresponds to (12-8.2)/.95 = 4 standard errors above the corporation average of 8.2. This is much higher than two or three standard errors, and it appears to be beyond chance variation. Chapter 9 Exercise 3 The p- value tells you how likely it would be to get results at least as extreme as this if there was no difference in the taste and only chance variation was operating. In this problem, p-value of 0.02 means that, if there is no difference in taste, then there is only 2% chance that 70% or more people would declare one drink better than the
The alternative hypothesis H1 is that the mean annual income μ is less than $50,000. H1:μ < $50,000 Significance level chosen is 5% or α = 0.05 Here, the population standard deviation is unknown. Hence, we use a t statistic Therefore the test statistic used is t = X-μS/n follows a t distribution with n-1 degrees of freedom From the t table corresponding to 0.05 probability, the critical value tα =1.6766. Hence the critical region is t < -1.6766. Alternatively, we reject the null hypothesis, if the p value is less than the significance level Substituting the value we get t = 43.74-5014.6396/50 = -3.02 The p value corresponding to t = -3.02 and 49 d.f.
Why or why not? | Not unusual, because it is within 2 standard deviations of the mean | 8d. Fifty adults are randomly selected. What is the likelihood that the mean of their body temperatures is 97.98°F or lower? | 0.62 percent | 9a.
Since we cannot deny that the annual income average is $50,000, we have no choice but to keep it as a consideration moving forward. B. The true population proportion of customers who live in an urban area exceeds 40%. Scenario B is very similar to scenario A because we were able to indicate a weak p-value to indicate that the population proportion of urban customers could be exactly 40%. With hypothesis testing we were able to test the probability of
3. The rate variance is calculated by the difference between the $16.90 actual labor rate vs the $16 budgeted rate, then we multiply the difference by the 9000 actual labor hours which gives us an $8,100 unfavorable rate variance. To figure out the efficiency variance we multiply the $16 budgeted rate by the difference between the 9,000 actual labor hours and the 10,000 budgeted hours, giving us a $16,000 favorable efficiency variance. As a result of the difference between the rate and efficiency variances we end up having a $7,900 favorable flexible budget variance. 4.
Chapter 6 Sampling Distributions True/False 1. If we have a sample size of 100 and the estimate of the population proportion is .10, we can estimate the sampling distribution of [pic]with a normal distribution. Answer: True Difficulty: Easy 2. A sample size of 500 is sufficiently large enough to conclude that the sampling distribution of [pic] is a normal distribution, when the estimate of the population proportion is .995. Answer: False Difficulty: Medium 3.
The total risk score is 4.14, the greatest relative or standardized difference between pretest and 3 month outcomes. This t ratio has a statistical significance of 0.05 - the least acceptable value for statistical significance. Also the larger the t ratio, the smaller the observed p value and increased odds of being able to reject the null hypothesis. 3. Which t-ratio listed in Table 3 represents the smallest relative difference between the pretest and 3 months?
The R^2=86.52%, while R=0.80. From the Analysis of Variance, we saw the P Value= 0.000 the F Value= 85.6452. When we tested with the significance level of 5% we concluded the P value was less than, therefore we concluded to reject the null hypothesis for this level. We also performed the 95% Confidence level to be ($0.009and $0.015) for B1. In addition, we can estimated that a customer with a $4000 credit balance to have an income in between (41.7665, 46.6130) in $1000 using the 95% CI confidence levels to calculate the income level.
In the 2010 census, the population of Ellettsville was 6378. Out of this number 95.8 percent of the people are White, 1.2 percent African American, and the other 3 percent of people are Native American, Asian, and Hispanic or made up of two or more races. Of these people there are 1944 households with 1345 families consisting of 3 or more people. 41 percent of these households have children under the age of 18 and 52 percent of the households have married couples in them. Also found in the census was that the average household income in the town of Ellettsville is 37, 275 dollars.
Mr. Cogan also states that in 2008 an estimated 30 percent of soldiers took their own lives while on deployment and that another 35 percent committed suicide after returning home. Mr. Cogan will also go on to say that the Veterans Affairs (VA) has only treated around 400,000 of the 1.7 million men and women who have served and that the numbers might be far worse than the VA actually knows about.