SOLUTION: ) A coffee company claims that their coffee can contains 300g of coffee. Suppose the sample of 36 coffee cans provide a sample mean of = sh 292. Is = 292 small enough to cause us

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Question 1206540: ) A coffee company claims that their coffee can contains 300g of coffee. Suppose the sample of 36 coffee cans provide a sample mean of = sh 292. Is = 292 small enough to cause us to reject H0. Using = 292, = 18 and n = 36.
Found 2 solutions by Theo, Boreal:
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
claim is that the coffee can contains 300 grams of cofee.
sample of 36 cans provides a sample mean mean of 292 with a standard deviation of 18 and a sample size of 36.

standard error = standard deviation divided by square root of sample size = 18 / 6 = 3.

t-score is used because the standard deviation was taken from the sample.

t-score with 35 degrees of freedom formula = (292 - 300) / 3 = 2.6667.

area to the left of that t-score with 35 degrees of freedom = .005759.

here's what the t-test calculator graph looks like.






one tailed critical t-score at 99% confidence interval requires a t-score of less than or equal to -2.4377.

since -2.6667 is greater than that, the sample results indicate those results are highly unlikely to come from random variation in sample means from different sample of size 36.

the conclusion is that the coffee cans really do contain less grams of coffee than claimed.

the test t-=score of -2.67 is greater than the critical t-score - 2.44.
the test p-value of .00576 is less than the critical p-value .01.








Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
it's not stated here clearly but s=18 gm.
Ho: mean is 300
Ha: mean not 300
alpha =0.05 p[reject Ho| Ho true]
test statistic is a t df=35 t=(x mean-300)/s/sqrt(n)
critical value at 0.05 level for a two way test is +/-2.030
t=(292-300)/(18/sqrt(36)
=-2.667, which is beyond the critical value, so we reject Ho
p=0.012.