Question

The average height of 18-year-old males is 70 inches. A random sample of sixteen 18-year old males is selected. The average heights and standard deviation of heights came out to be 68.8 and 8 inches, respectively. At level of significance can we conclude that the average height has changed? a = .01

a) Set up test hypothesis

b) Write test statistics and sketch.

c) Apply classical or P-value method

d) Draw conclusion and interpret.

Answer 1

Solution:

Given:

Sample Size = n = 16

Sample mean = $\bar{x} = 68.8$ in.

Sample standard deviation = s = 8 in.

Population Mean = $\mu = 70$ in.

level of significance = 0.01

We have to test if  the average height has changed.

Part a) Set up test hypothesis

$H_{0}: \mu = 70$

$H_{1}: \mu \neq 70$

Part b) Write test statistics and sketch.

$t=\frac{ \bar{x}-\mu }{s /\sqrt{n}}$

$t=\frac{ 68.8 - 70 }{ 8 /\sqrt{16 }}$

$t=\frac{ -1.2 }{ 8 / 4}$

$t=\frac{ -1.2 }{ 2 }$

t-0.600

Part c) Apply classical or P-value method

Using classical method: find t critical values:

df = n - 1 = 16 -1 = 15

Two tail area = level of significance = 0.01

t Table cum. prob one-tail two-tails t 50 t 995 75 0.25 ss 0.01 975 85 90 95 0.50 0.20 0.15 0.10 0.05 0.025 0.005 0.50 0.30 1.00 0.40 0.20 0.10 0.05 0.02 0.01 df 6366 9.925 5.841 4.504 4.032 3707 3.499 3. 55 3.250 3.169 3.106 3.0 5 3.012 2.977 3.078 1.886 1.638 1.533 1 0.000 1.000 1.376 1.963 6.314 12.71 31.82 0.000 0.000 0.816 1.061 1.386 2.920 4.303 6.965 0.765 0.978 1.250 2.353 3.182 4.541 4 0.000 0.741 0.941 1.190 2.132 2.776 3.747 0.920 0.000 0.727 1.156 1.476 1.440 2.015 2.571 3.365 2.447 3.143 6 0.000 0.718 0.906 1.134 1.943 1.415 7 0.000 0.711 0.896 1.119 1.895 2.365 2.998 1.860 8 0.000 0.706 0.889 1.108 1.397 2.306 2.262 2.896 0.703 0.883 0.000 1.100 1.383 1.833 2.821 0.700 1.093 1.812 1.796 10 0.000 0.879 1.372 2.228 2.764 1.088 1.083 1.363 11 0.000 0.697 0.876 2.201 2.718 12 0.000 0.695 0.873 1.356 1.782 2.179 2.681 2.160 13 0.000 0.694 0.870 1.079 1.350 1.771 2.650 2.624 14 0.000 0.692 0.868 1.076 1.345 1.761 2.145 2 131 066 15 1.074. 1.341 1.753 2.602 2.947 U.O00 U.09T

t critical values = $\pm 2.947$

Part d) Draw conclusion and interpret.

Decision Rule:

Reject null hypothesis H0, if absolute t test statistic value > t critical value = 2.947 , otherwise we fail to reject H0

since absolute t test statistic value = 0.600 < t critical value = 2.947 , we fail to reject H0.

Thus at 0.01 level of significance we do not have sufficient evidence to conclude that the average height has changed.