### CFA Practice Question

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### CFA Practice Question

A floor manager of a large department store is studying the buying habits of the store's customers. Suppose he assumes that the monthly income of these customers is normally distributed with a standard deviation of 500. If he draws a random sample of size N = 100 and obtains a sample mean of YBAR = 800, a 0.95 confidence interval for the true population mean is 702 < MU < 898. Do you think that it would be quite unreasonable for the true population mean to be \$600? Explain.

Based on the above confidence interval, we would reject the hypothesis that MU = 600 (at ALPHA = .05).

User Comment
topdan Any more detailed explanation please ?
1. H(o): x <= 600 vs H(a): x > 600
2. rejection point 1.96
3. if z > 1.96, reject H(o)
4. z = (800 - 600)/(500/sqrt(100)) = 4
mchu It is in the notes.

Confidence intervals can be used to test hypotheses. Note that the right side of the equation is the left endpoint of the confidence interval. Essentially, if the confidence interval contains the value of the unknown population parameter as hypothesized under H0, then H0 would not be rejected in a two-sided hypothesis test with corresponding &#945;, whereas if the confidence interval does not contain the value of the unknown population parameter as hypothesized under H0, then H0 would be rejected in a two-sided hypothesis test with corresponding &#945;.

Since 600 is not in the confidence interval, the H0 would be rejected.
chamad Plese note this is 2 tailed rest
H0 : Income = 600
Ha : income != 600
The confidence intervl is (702 < MU < 898)
So obviously 600 is outside this interval
Reject H0