Question

Example:
A simple sample of heights of 6400 English men has a mean of 170 cm and a
S.D. of 6.4 cm, while a simple sample of heights of 1600 Americans has a mean
of 172 cm and a S.D. of 6.3 cm. Do the data indicate that Americans are, on the
average, taller than the Englishmen?​

Answer 1

Answer:

87..............................

Answer 2

Answer:

According to the data, Americans are, on the average, taller than the Englishmen.

Step-by-step explanation:

Solution:

Given sample of heights of Englishmen, n₁ = 6400

Mean of heights of Englishmen, $\overline{x_1}=170 cm$

Standard deviation of Englishmen heights, $\sigma_1=6.4~ cm$

Sample of heights of Americans, n₂ =1600

Mean of heights of Americans, $\overline{x_2}=172 cm$

Standard deviation of Americans heights, $\sigma_2=6.3~ cm$

Null Hypothesis: When $H_0:\mu_1=\mu_2$,  then there is no significant difference between heights of Englishmen and Americans.

For one tailed test stating the alternative hypothesis: When $H_0:\mu_1 < \mu_2$, Americans are taller than Englishmen.

The test statistic is given by

$z=\dfrac{\overline{x_1}-\overline{x_2}}{\sqrt{\dfrac{\sigma_1^{2} }{n_{1} }+\dfrac{\sigma_2^{2} }{n_{2} } } }$

Substituting the given values,

$z=\dfrac{170-172}{\sqrt{\dfrac{(6.4)^{2} }{6400}+\dfrac{(6.3)^{2} }{1600}} }=\dfrac{-2}{\sqrt{\dfrac{40.96 }{6400}+\dfrac{39.69}{1600}} }$

 $=\dfrac{-2}{\sqrt{0.0064+0.0248} }=\dfrac{-2}{\sqrt{0.0312} }$

 $=\dfrac{-2}{0.17665} }\approx-11.322$

And hence, $|z|=11.322$

The table value of z at 1% level is 2.33.

And at 5% significance level, the table value of z is 1.96.

Since the calculated value of z is greater than the table value,  the Null Hypothesis can be rejected.

Therefore, it can be concluded that the Americans are, on the average, taller than the Englishmen.