A statistics practitioner wishes to test the following hypothesis: H0 : ì = 600 against H1 : ì < 600

A sample of 50 observations yielded the statistics: mean x = 585 and standard deviation sx = 45. The
test statistic of a test to determine whether there is enough evidence at the 10% significance level to reject
the null hypothesis is
(1) 2.3570
(2) 0.3333
(3) 16.6667
(4) 23.570
(5) −2.3570

Judy, Sarah, Sam, Cindy, Joseph or whomever. First, we like for you to use the same screen name. It makes it easier for us to keep up with the questions.

Second, you are unlikely to get repsonses when you show no work.

4. 23.570

To test the hypothesis H0: ì = 600 against H1: ì < 600, we can use a one-sample t-test. The test statistic for a one-sample t-test is given by:

t = (x - ì) / (sx / sqrt(n))

Where:
- x is the sample mean
- ì is the hypothesized mean under the null hypothesis (600 in this case)
- sx is the standard deviation of the sample
- n is the sample size

In this question, we are given the sample mean x = 585, the standard deviation sx = 45, and the sample size n = 50.

Plugging these values into the formula, we get:

t = (585 - 600) / (45 / sqrt(50))

Simplifying further:

t = (-15) / (45 / sqrt(50))

t = (-15) / (45 / 7.0711) [approximating sqrt(50) as 7.0711]

t ≈ -0.3333

The test statistic t is approximately -0.3333.

To determine if there is enough evidence to reject the null hypothesis at the 10% significance level, we compare the test statistic with the critical value from the t-distribution.

Since the alternative hypothesis is ì < 600, we are performing a left-tailed test. Therefore, we need to find the critical value that corresponds to a 10% significance level and (n - 1) degrees of freedom.

The degrees of freedom for a one-sample t-test is (n - 1), which in this case is (50 - 1) = 49.

Looking up the critical value for a left-tailed test with 49 degrees of freedom and a 10% significance level in a t-table or using statistical software, we find that the critical value is approximately -1.660.

Since the test statistic (-0.3333) is greater than the critical value (-1.660), we do not have enough evidence to reject the null hypothesis at the 10% significance level. Therefore, the answer is (2) 0.3333.