Answer to Question #200771 in Statistics and Probability for T MABOTJA

Question

Suppose, the average U. S. household spends $90 per day. You recently took a random sample of 30 households in Huntsville and the results revealed a mean of $84.50. Suppose the SD 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 has decreased

Answer 1

$\bar{x}=84.50 \\ \sigma=14.50 \\ n=30 \\ H_0: \mu= 90 \\ H_1: \mu<90$

The test statistic:

$Z=\frac{\bar{x}-\mu}{\sigma / \sqrt{n}} \\ Z = \frac{84.50-90}{14.50/ \sqrt{30}}=-2.08$

The level of significance α=0.05

Since the test is left tailed. Thus, the critical value using the standard normal table is -1.645.

Reject H₀ if $Z<Z_{crit}$ .

$Z=-2.08<Z_{crit}=-1.645$

Thus, reject the H₀.

Hence, it can be concluded that the average amount spent per day by U.S. households has decreased.