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A manufacturer of cookies and crackers does a small survey of the age at sale of one of its brands. A random sample of 33 retail markets in a particular region is chosen. In each store, the number of days since manufacture of the frontmost box of crackers is determined by a date code on the box. The data (age in days) from the 33 stores showed the sample mean to be 56.869 days, and the sample standard deviation to be 28.97 which almost precisely estimated the population standard deviation. Suppose the manufacturer wants to obtain a 90% confidence interval with a margin of error that depicts 5 days between the lowest and highest points of the interval. Assuming that the sample standard deviation does not change, how large a sample is needed? Show all work.
the probability of an american adult flying a kite at least once in the year 2003 was 0.023. the probability of
PestFree Chemicals has developed a new fungus preventative that may have a significant market among potato growers. Unfortunately, the actual extent.
7% of his eggs are underweight, and that 14% weigh over 72 grams. If a normal model is appropriate, what would the standard deviation of his egg weights be?
Do you think men and women will have different personal distances? Will the larger distances be specified by the men or the women?
Find the probability that a randomly selected student is a male binge drinker, and find the probability that a randomly selected student is a female binge drinker.
Do the sales prices of houses in a given community vary systematically with their sizes (as measured in square feet)?
Lance produces 20 more than 4 times what Jacob can produce in one day. If Lance's total production for the day is 360, how much does Jacob produce in one day
Explain the advantages and disadvantages of each analytic strategy.
in this problem we explore the effect on the standard deviation of adding the same constant to each data value in a
The times it takes workers to complete a task at a certain assembly plant are normally distributed with mean = 10 minutes and standard deviation = 2 minutes.
The manufacturer claims this probability is now 0.80. A sample of size 20 is tested. Determine the odds favoring the new probability for various numbers.
For a t-distribution with ten degrees of freedom, find (t-sub-alpha) for alpha value of 0.05 (this is the value such that the right tail probability is 0.05
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