You can flesh out the argument .....

Since the population variance is not known, the test statistic to use is which may be compared with Student's t-distribution with (n - 1) degrees of freedom.

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The critical value is -2.262 at the significance level. Therefore ......

Alternatively, a 95% confidence interval for is given by where in this problem , and n = 10.

95% confidence interval: . Therefore .....

The critical value is got using the t-distribution with 10 - 1 = 9 degrees of freedom. The t-distribution is is used because the population standard deviation is not known and the sample size is small (n < 50).