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Problem
The average height of an American was reported to be 70. Many believe that college students are shorter than average. A random sample of students produced the following data:

$n=140$, $mean = 65.39$, $s = 4.64$

Find the test statistics

Steps Taken to Solve
I decided that this type of problem was a 1 sample T-test. These are my steps to the Hypothesis Test.

  1. Null Hypothesis = No difference, so $M = M_o = 0$
    Alternative Hypothesis = $M < 50$
  2. $t = \frac{50 - 0}{\frac{4.64}{\sqrt{140}}} = 166.7467$

Is this approach correct so far?

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  • $\begingroup$ If the average is really $50$ (inches) then the population may include babies and young children. $\endgroup$
    – Henry
    Aug 24, 2016 at 21:59
  • $\begingroup$ @henry just for fun $\endgroup$ Aug 24, 2016 at 22:00
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    $\begingroup$ Note that your alternative hypothesis is the population has mean $M \lt 50$ but your sample has a mean greater than $50$. That should give you a check on your answer $\endgroup$
    – Henry
    Aug 24, 2016 at 22:01
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    $\begingroup$ @Henry Oh wow I just noticed the typo! I corrected the original question. $\endgroup$ Aug 24, 2016 at 22:03
  • $\begingroup$ Still seems to be confusion between 70 and 50. More editing needed? And what is $M_0$? $\endgroup$
    – BruceET
    Aug 24, 2016 at 23:34

1 Answer 1

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The null hypothesis is $H_0: \mu = 70$ and the alternative is $H_a: \mu < 70.$ Roughly speaking, the question is whether the sample mean $\bar X = 65.39$ is enough below $70$ that we should agree with the view that college students are 'shorter than average'.

The test statistic is $T = \frac{\bar X - 70}{S/\sqrt{n}} = -11.76.$

a = 65.39;  s = 4.64;  n = 140;  mu.0 = 70
t = (a - mu.0)/(s/sqrt(n));  t
[1] -11.75566

This is an extremely negative value of the T statistic. The P-value is the probability under $H_0$ of value as or more extreme (in the direction of the alternative) than the observed value -11.76. The distribution of the T statistic under the null hypothesis has Student's t distribution with $n - 1 = 139$ degrees of freedom. That is essentially $0,$ found in R as:

> pt(-11.76, 139)
[1] 6.78294e-23

It is almost impossible for 'college students' randomly sampled from a normal population with $\mu = 70$ to give dats with $\bar X = 65.39$ and $S = 4.46.$ The null hypothesis is rejected at any reasonable level of significance.

Note: Student's t distribution with df = 139 is very nearly standard normal.

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