statistics

[jaimetab]

The average household spends $90 a day. Assume a sample of 25 households showed a sample mean daily expense of $84.50 with a sample standard deviation of $14.50. Using a 5% significance level, can you conclude that the mean daily expense are significantly different than the US average? Usin the classical approach to the hypothesis testing.

 

[galactus]

Since the sample size is less than 30, we can use a t-test with 24 degrees of freedom.

\(\displaystyle t=\frac{(84.5-90)\sqrt{25}}{14.5}=-1.8966\)

Looking in the t-table for a two-tailed test with .05 significance level, we get \(\displaystyle \pm2.0639\)

\(\displaystyle H_{0}:{\mu}=90\)

\(\displaystyle H_{a}:{\mu}\neq 90 \;\ \text{claim}\)

The p-value is .07, which is greater than the alpha level.

Also, the test statistic is not in the rejection region.

So, we DO NOT reject the null hypothesis.

Since we do not reject, we can not conclude there is a significant difference.

 

[jaimetab]

Thanks for your help