Question

Suppose, according to a 1990 demographic report from the Brookings Institution, the average U. S. household spends $90 per day. Suppose you recently took a random sample of 30 households in Chicago and the results revealed a mean of $84.50. Suppose the standard deviation of daily expenditure is known to be $14.50. Using a 0.05 level of significance, can it be concluded that the average amount spent per day by U.S. households is different than the Brookings report? Use α = .05

Answer 1

given data are:-

sample size (n) = 30

sample mean () = 84.50

population sd () = 14.50

as here, the population sd is known we will do 1 sample Z test for mean.

hypothesis:-

the test statistic is:-

the p value is :-

[ in any blank cell of excel type =NORMSDIST(-2.078)]

decision:-

p value = 0.0377 < 0.05 (alpha)

so, we reject the null hypothesis.

conclusion:-

there is sufficient evidence to conclude that the average amount spent per day by U.S. households is different than the Brookings report at 0.05 level of significance.

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