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a. Test-statistic
= (Xbar-Mu)/(Stdev/sqrt(n))
= (29.7-30)/(0.7/sqrt(25))
= -2.14
Answer: -2.14
b.
df = n-1=24
Critical value at significance level of 1% is : -2.7969 , and rounding off to -2.80 ( 2 decimal places)
Answer: -2.80
c. The decision is ?
Since our test-statistic is more than critical test-statistic then we fail to reject the null hypothesis.
Answer: B. Do not reject the null hypothesis.
A manufacturer of detergent claims that the contents of boxes sold weigh on average at least...
Please, determine the appropriate value(s) of -Za, Za, +- Z(a/2) and round to two decimal places as needed. A manufacturer of detergent claims that the contents of boxes sold weigh on average at least 12 ounces. The distribution of weight is known to be normal, with a standard deviation of 0.5 ounces. A random sample of 16 boxes yielded a sample mean weight of 11.23 ounces. Test at the 1% significance level the null hypothesis that the population mean weight...
THANK YOU IN ADVANCE FOR HELPING ME! i appreciate this soo much! this has helped me a lot especially throught times like this. im so grateful for your help 1 2 3 4 (1 point) The time needed for college students to complete a certain paper-and-pencil maze follows a normal distribution with a mean of 30 seconds and a standard deviation of 2.3 seconds. You wish to see if the mean time is changed by vigorous exercise, so you have...
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A cereal company claims that mean weight of cereal boxes is at most 16.1 ounces. Suppose that a plant manager wishes to test whether the true mean weight of cereal boxes is greater than 16.1 ounces. Suppose that for this problem the population standard deviation is 0.4 and the population distribution is normal. The manager obtain a random sample of size 25 and finds a mean of 16.3 ounces. Using p value approach test the claim of company at significance...
1. A process that produces bottles of shampoo, when operating correctly, produces bottles whose contents weigh on average 20 ounces. A random sample of nine bottles from a single production run yielded the following content weights (in ounces):21.4 19.7 19.7 20.6 20.8 20.1 19.7 20.3 20.9Assuming that the population distribution is normal, test at the 5% level that the process is operating correctly, i.e., the mean is 20 ounces.
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