Answer:
a
The null hypothesis is
[tex]H_o : \mu = 21[/tex]
The Alternative hypothesis is
[tex]H_a : \mu< 21[/tex]
b
[tex]\sigma_{\= x} = 0.8944[/tex]
c
[tex]t = -2.236[/tex]
d
Yes the mean population is significantly less than 21.
Step-by-step explanation:
From the question we are given
a set of data
20 18 17 22 18
The confidence level is 90%
The sample size is n = 5
Generally the mean of the sample is mathematically evaluated as
[tex]\= x = \frac{20 + 18 + 17 + 22 + 18}{5}[/tex]
[tex]\= x = 19[/tex]
The standard deviation is evaluated as
[tex]\sigma = \sqrt{ \frac{\sum (x_i - \= x)^2}{n} }[/tex]
[tex]\sigma = \sqrt{ \frac{ ( 20- 19 )^2 + ( 18- 19 )^2 +( 17- 19 )^2 +( 22- 19 )^2 +( 18- 19 )^2 }{5} }[/tex]
[tex]\sigma = 2[/tex]
Now the confidence level is given as 90 % hence the level of significance can be evaluated as
[tex]\alpha = 100 - 90[/tex]
[tex]\alpha = 10[/tex]%
[tex]\alpha =0.10[/tex]
Now the null hypothesis is
[tex]H_o : \mu = 21[/tex]
the Alternative hypothesis is
[tex]H_a : \mu< 21[/tex]
The standard error of mean is mathematically evaluated as
[tex]\sigma_{\= x} = \frac{\sigma}{ \sqrt{n} }[/tex]
substituting values
[tex]\sigma_{\= x} = \frac{2}{ \sqrt{5 } }[/tex]
[tex]\sigma_{\= x} = 0.8944[/tex]
The test statistic is evaluated as
[tex]t = \frac{\= x - \mu }{ \frac{\sigma }{\sqrt{n} } }[/tex]
substituting values
[tex]t = \frac{ 19 - 21 }{ 0.8944 }[/tex]
[tex]t = -2.236[/tex]
The critical value of the level of significance is obtained from the critical value table for z values as
[tex]z_{0.10} = 1.28[/tex]
Looking at the obtained value we see that [tex]z_{0.10}[/tex] is greater than the test statistics value so the null hypothesis is rejected
