A teacher at a school claims that the students in her class are above average intelligence. A random sample of 30 students IQ scores have a mean score of 112. Is there sufficient evidence to support the teacher's claim? The mean population IQ is 100 with a standard deviation of 15.

The alpha level = 0.05

The critical value = 1.645

I don't know what formula to use to find the test statistics.

  • $\begingroup$ you use the z-test because the s.d. is known $\endgroup$
    – AmR
    Jul 7 '18 at 18:33
  • $\begingroup$ How do you know that it's known? $\endgroup$
    – Hx3
    Jul 8 '18 at 0:16
  • $\begingroup$ I get confused when to use the z and t test $\endgroup$
    – Hx3
    Jul 8 '18 at 0:17
  • $\begingroup$ well you use z when the s.d. is known and t when it's unknown. It will be stated $\endgroup$
    – AmR
    Jul 8 '18 at 0:19
  • $\begingroup$ Okay, thanks! @AmR $\endgroup$
    – Hx3
    Jul 8 '18 at 0:24

1) State the null and alternate hypothesis:

$H_0:μ = 100$
$H_1:μ > 100$

2) Find the alpha level. There was no alpha level given so by default we use $0.05$.

3) Find the reject region (critical value). By using the z-table, the area of $0.05$ is equal to the z-score of $1.645$.

4) Find the test statistic.


= $\frac{112.5-100}{15/\sqrt{30}}$

= 4.56

5) Since the test statistics of 4.56 is greater than the critical value of 1.645, we reject the null hypothesis.

  • $\begingroup$ How does a random sample of 30 students infer that the students in her class have above average IQs? I know its only academic but to me it's not a very well thought out question. $\endgroup$
    – Phil H
    Jul 7 '18 at 18:58
  • $\begingroup$ Yeah i really didn't like the way it was worded. $\endgroup$
    – AmR
    Jul 7 '18 at 19:40
  • $\begingroup$ I up voted your answer for the math procedure and for the fact that you steered clear of making a summary conclusion. $\endgroup$
    – Phil H
    Jul 7 '18 at 19:58
  • $\begingroup$ My teacher uses this example for all the classes she teaches $\endgroup$
    – Hx3
    Jul 8 '18 at 0:15
  • $\begingroup$ My older brother had her 4 years ago and he remembers having that question $\endgroup$
    – Hx3
    Jul 8 '18 at 0:20

Since the standard deviation of 15 is known, you would use the formula: $Z=\frac{\bar{x}-μ}{σ/\sqrt{n}}$. Where $\bar{x}$ = 112, $μ = 100$, $σ = 15$, and $n = 30$.

If the test statistic is less than the critical value, you fail to reject the null. If test statistic is greater than the critical value, you reject the null.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.