Question #200771

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


1
Expert's answer
2021-05-31T16:37:22-0400

xˉ=84.50σ=14.50n=30H0:μ=90H1:μ<90\bar{x}=84.50 \\ \sigma=14.50 \\ n=30 \\ H_0: \mu= 90 \\ H_1: \mu<90

The test statistic:

Z=xˉμσ/nZ=84.509014.50/30=2.08Z=\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<ZcritZ<Z_{crit} .

Z=2.08<Zcrit=1.645Z=-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.


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