SOLUTION: A ski shop manager claims that the average of the sales for her/his shop is $1800 a day during the winter months. Ten winter days are selected at random, and the mean of the sales

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Question 463673: A ski shop manager claims that the average of the sales for her/his shop is $1800 a day during the winter months. Ten winter days are selected at random, and the mean of the sales is $1830. The standard deviation of the population is $200. Can one reject the claim at α = 0.05? Find the 95% confidence interval of the mean. Does the confidence interval interpretation agree with the hypothesis test results? Explain. Assume that the variable is normally distributed
Answer by edjones(8007) About Me  (Show Source):
You can put this solution on YOUR website!
H[o]=1800
mu=1800
n=10, m=1830, s=200
.
t=%28m-mu%29%2F%28s%2Fsqrt%28n%29%29=%281830-1800%29%2F%28200%2Fsqrt%2810%29%29=.474
t[.025][9]=2.262 (two tailed)
Since .474<2.262 then we accept the null hypothesis. The ski shop manager is not refuted.
.
Ed