Furthermore, after running a 95% confidence interval on this data, we can say that we are 95% confidence interval on this data and we are 95% certain that the avarage (mean) annual income falls at 47,210 on the higher end of the tail, which is less than 50,000. Since the p-value of .002 is less than the alpha value of .05, we can say beyond a reasonable doubt that the manager's alternative hypothesis is correct and we will reject the null hypothesis because we have sufficient evidence for rejection.…
11. We wish to develop a confidence interval for the population mean. The population follows the normal distribution, the standard deviation of the population is 3, and we have a sample of 10 observations. We decide to use the 90 percent level of confidence. The appropriate value of to represent the level of confidence is…
Null hypothesis: the bottles of the brand of soda produced do not contain less than the advertised sixteen (16) ounces of product.…
In this bottle case study the null hypothesis is there are sixteen ounces in each bottle. The alternate hypothesis is there are not sixteen ounces in each bottle. In order to understand the outcome of the test of hypothesis we must analyze the data set and our projections in the previous to answer them. Analyzing the data from above, we have a mean of 15.2 and a standard deviation of 0.73. Adding one standard deviation to the mean, we still do not get to sixteen ounces, which compute to be 15.93 ounces. Clearly from the calculation that the null hypothesis is incorrect and has been disproven, meaning that the alternate hypothesis is in fact the correct…
A manager at AJ Davis department store has speculated on certain customer statistics in hopes to get a better understanding of his target market, thus improving revenue though marketing and sales initiatives. His first assumption is that the average annual income of 50 randomly selected customers is less than $50,000. After calculating the data this assumption proved to be true, as we can say with 95% certainty that the income is less than $47,510. The next assumption is that more than 40% of his customers live in urban areas. After computing the data, we cannot be certain this is accurate. The data falls too close for me to be confident in this assumption; the 95% confidence interval is too wide. The third assumption is that the average number of years the customers have lived at their current home is less than 13 years. Again, this was an accurate assumption as we can say with a high degree of confidence than most customers have lived at their current location for less than 13 years as the mean was found to be 12.38 years. Finally, the manager assumed the average mean credit balance for suburban customers is greater than $4300. This was found to be an accurate assumption. The 95% lower confidence limit was found to be greater than $4300.…
Conclusion: Reject the null hypothesis, since the observed significance (p-value) is less than the significance level 0.05. The sample provides enough evidence to support the claim that there is significant difference in the proportions of red and brown candies.…
a. Construct a 90% confidence interval for the population mean m 5 +- 1.88 b. Construct a 95% confidence interval for the population mean m 5+- 2.39 c. Construct a 99% confidence interval for the population mean m 5+- 3.75 d. Assume that the sample mean and sample standard deviation s remain exactly the same as those you just calculated but are based on a sample of n=25 observations rather than n=6 observations. Repeat parts a–c. What is the effect of increasing the sample size on the width of the confidence intervals? a)5+- 78.5 +-94.5 +-1.28 b) the sample size of the width will decrease 5.37 For the binomial sample information summarized in each part, indicate whether the sample size is large enough to use the methods of this chapter to construct a confidence interval for p.…
A physician who specializes in weight control has three different diets she recommend. As an experiment, she randomly selected 15 patients and then assigned 5 to each diet. After three weeks, the following weight losses in pounds were noted. At the 0.05 significance level, can she conclude that there is a difference in the mean amount of weight loss among the three…
4. If the limits of the confidence interval of the difference between the means of two normally distributed populations were 0.5 and 2.5 at the 95% confidence level, then we can conclude that we are 95% certain that there is no significant difference between the two population means. false…
What is the 99% confidence interval for the population mean? A) [17.42, 20.78] B) [17.48, 20.72] C) [14.23, 23.98] D) [0.44, 3.80] Answer: A Use the following to answer questions 84-86: A survey of 25 grocery stores revealed that the average price of a gallon of milk was $2.98 with a standard error of $0.10. 84. What is the 95% confidence interval to estimate the true cost of a gallon of milk? A) $2.81 to $3.15 B) $2.94 to $3.02 C) $2.77 to $3.19 D) $2.95 to $3.01 Answer: C 85. What is the 98% confidence interval to estimate the true cost of a gallon of milk? A) $2.73 to $3.23 B) $2.85 to $3.11 C) $2.94 to $3.02 D) $2.95 to $ 3.01 Answer: A 90. A pharmaceutical company wanted to estimate the population mean of monthly sales for their 250 sales people. Forty sales people were randomly selected. Their mean monthly sales were $10,000 with a standard deviation of $1000. Construct a 95% confidence interval for the population mean. A) [9,690.1, 10,309.9] B) [9,715.5, 10,284.5] C) [8,040, 11,960] D) [8,000, 12,000] Answer: B Use the following to answer questions 91-92: A survey of an urban university (population of 25,450) showed that 750 of 1100 students sampled attended a home football game during the season. What inferences can be made about student attendance at football games? 91. Using the 99% level of confidence, what is the confidence interval? A) [0.767, 0.814] B) [0.0.6550, 0.7050] C) [0.6659, 0.6941] D) [0.0.6795, 0.6805] Answer: C 92. Using the 90% level of…
We are 95% confident the proportion is in this interval… if the sample mean or proportion is in the confidence interval the null hypothesis could be inferred. Else, the alternative hypothesis would.…
at 5% level of significance we have strong enough evidence to reject the Ho that the true mean weight of salmon is no different than 40 in favourof Ha that the true mean weight is significantly greater than 40.…
The calculations determined that in both cases the claims are rejected. In the case of the claim that 20% of skittles are red, the class proportion of 20.4% red skittles and is found to be unlikely to be correct and therefore is rejected because the calculated p-value is less than the significance level. Also, the claim that the mean number of skittles per bag is 55 is tested against the class mean of 59.98. This is also determined to be unlikely correct and is rejected as well because the calculated test statistic falls within the critical…
SSER - sum of squares of error of reduced model SSEF - sum of squares of error of full model…
CASE 2: DRINK-AT-HOME, INC. Drink-At-Home, Inc. (DAH, Inc.), develops, processes, and markets mixes to be used in nonalcoholic cocktails and mixed drinks for home consumption. Mrs. Lee, who is in charge of research and development at DAH, Inc., this morning notified Mr. Dick Jones, the president, that exciting developments in the research and development section indicate that a new beverage, an instant pina colada, should be possible because of a new way to process and preserve coconut. Mrs. Lee is recommending a major program to develop the pina colada. She estimates that expenditure on the development may be as much as $100,000 and that as much as a year's work may be required. In the discussion with Mr. Jones, she indicated that she thought the possibility of her outstanding people successfully developing such a drink now that she'd done all the really important work was in the neighborhood of 90 percent. She also felt that the likelihood of a competing company developing a similar product in 12 months was 80 percent. Mr. Jones is strictly a bottom line guy and is concerned about the sales volume of such a beverage. Consequently, Mr. Jones talked to Mr. Besnette, his market research manager, whose specialty is new product evaluation, and was advised that a market existed for an instant pina colada, but was some-what dependent on acceptance by both grocery stores and retail liquor stores. Mr. Besnette also indicated that the sales reports indicate that other firms are considering a line of tropical drinks. If other firms should develop a competing beverage the market would, of course, be split among them. Mr. Jones pressed Mr. Besnette to make future sales estimates for various possibilities and to indicate the present (discounted value of future profits) value. Mr. Besnette provided Table 1. Mr. Besnette's figures did not include (1) cost of research and development, (2) cost of new production equipment, or (3) cost of introducing the pina colada. The cost of…