Answer to Question #251710 in Statistics and Probability for Gel

Question

the average length of time for students to register for summer classes at a certain college has been 50 minutes. A new registration procedure using modern computing machines is being tried. If a random sample of 35 students had an average registration time of 42 minutes with a standard deviation of 11.9 minutes under the new system, test the hypothesis that the population mean is now less than 50 minutes, using 0.01 level of significance. Assume the population of times to be normal.

Answer 1

Sinve the population is normally distributed and standart deviation is known, we can use Z-statistic

Null hypothesis: $u = 50$

Alternative hypothesis: $u < 50$

Test statistic: $Z={\frac {(u-u{\scriptscriptstyle 0})*\sqrt{n}} {σ}}$, where u - sample mean, $u{\scriptscriptstyle 0}$ - claimed mean, n - sample size, σ - standard deviation

In our case: $Z={\frac {(42-50)*\sqrt{35}} {11.9}} = -3.98$

Due to the form of the alternative hypothesis, left-tailed test is appropriate

The critical value can be found as $P(N(0,1)<Z{\scriptscriptstyle {cr}}) = \alpha$, where $\alpha$ - level of significance, $Z{\scriptscriptstyle {cr}}$ - critical value

In our case: $P(N(0,1)<Z{\scriptscriptstyle {cr}}) = 0.01 \to Z{\scriptscriptstyle {cr}}$ = - 2.33

Since $Z$ < $Z{\scriptscriptstyle {cr}}$, we reject the null hypothesis.

There are statistically significant evidence that the population mean is now less than 50