Question

Assume that the average volume of raindrops globally is normally distributed with mean 0.45 ml and variance 0.01 ml². Michael measures the volume of two raindrop at Rainyville as 0.65 ml and 0.75 ml; and comes to believe that the raindrops at Rainyville are significantly larger than the global average at the 5% level.

If Michael performs the appropriate statistical test, what will be his conclusion?

1. reject the null hypothesis or 2. fail to reject the null hypothesis?

Answer 1

given data are

population mean =0.45 ml

population variance =0.01ml^2

then = =0.1

sample n =2

sample mean =

=(.65+.75)/2

= 1.4/2= 0.70 ml

null hypothesis H0: - =0.45

alternative hypothesis Ha: - 0.45

level of significance =0.05

test statistic: - z=

z=(0.7-0.45)/(0.1/)

= 0.25/0.0707

= 3.5355

decision rule:- at 5% level in 1tail from z table zcri= 1.645

reject H0 if zcal>1.645

decision: - zcal>1.645

so reject the null hypothesis H0

conclusion: - at 5% level their is sufficient evidence to support the claim that " the raindrops at rainyville are significantly larger than the global average "