1. ## hypothesis testing

Hello all--I am having trouble with this question--I feel I understand the concept, but I keep getting a test (z) statistic above 12, which seems way too high. Help??

The average number of employees in medium sized US businesses (population) is 110. A business administrator at a certain business claims that the average number is different from 110. A random of 40 business were chosen. The sample mean score was 130 and population standard deviation is 10. Is there sufficient evidence to support the business administrator’s claim? Take alpha =.05 and use (+/-) 1.96 as the critical value.

1 Determine the test statistic for the test. Round to the nearest hundredth.
2 Should the null hypothesis be accepted or rejected?
3 Is the test statistically significant? Provide your rationale for the conclusion.

2. ## Re: hypothesis testing

Originally Posted by mathizfun
Hello all--I am having trouble with this question--I feel I understand the concept, but I keep getting a test (z) statistic above 12, which seems way too high. Help??

The average number of employees in medium sized US businesses (population) is 110. A business administrator at a certain business claims that the average number is different from 110. A random of 40 business were chosen. The sample mean score was 130 and population standard deviation is 10. Is there sufficient evidence to support the business administrator’s claim? Take alpha =.05 and use (+/-) 1.96 as the critical value.

1 Determine the test statistic for the test. Round to the nearest hundredth.
2 Should the null hypothesis be accepted or rejected?
3 Is the test statistically significant? Provide your rationale for the conclusion.
$\displaystyle H_0: \mu=110$
$\displaystyle H_1: \mu\neq110$

Note that I would use $\displaystyle H_1: \mu>110$, but it appears your problem wants a two-tailed test.

The test statistic is $\displaystyle Z=\dfrac{\bar{x}-\mu_0}{\sigma} = \dfrac{130-110}{10} = 2$.

Here is why you got $\displaystyle Z>12$: You divided the standard deviation by the square root of the sample size. You should only divide the standard deviation by the square root of the sample size to get a standard error term for the t-test, which uses degrees of freedom. You can try this problem out using the t test and appropriate degrees of freedom and the test statistic T will be very similar to Z because the sample size is greater than 30. When $\displaystyle N>30$, the Z test is a sufficient approximation of the T-Test. Additionally, it is usually safe to assume $\displaystyle \sigma = s$ for large sample sizes.