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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.

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  • $\begingroup$ 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
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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.

$Z=\frac{\bar{x}-μ}{σ/\sqrt{n}}$

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

= 4.56

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

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  • $\begingroup$ 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
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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.

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