Answer:
a
[tex]z = - 2.12[/tex]
b
[tex]p- value = 0.017003[/tex]
c
There is sufficient evidence to show that the population mean is less than 20
d
[tex]z_{0.05} =- 1.645[/tex]
e
The decision rule is
Reject the null hypothesis
f
The conclusion is
There is sufficient evidence to show that the population mean is less than 20.
Step-by-step explanation:
From the question we are told that
The null hypothesis is [tex]H_o: \mu = 20[/tex]
The alternative hypothesis is Ha: µ < 20
The sample size is n = 50
The sample mean is [tex]\= x = 19.4[/tex]
The level of significance is [tex]\alpha = 0.05[/tex]
The population standard deviation is [tex]\sigma = 2[/tex]
Generally the test statistics is mathematically represented as
[tex]z = \frac{\= x - \mu }{ \frac{\sigma }{ \sqrt{n} } }[/tex]
=> [tex]z = \frac{19.4 - 20 }{ \frac{2}{ \sqrt{50 } } }[/tex]
=> [tex]z = - 2.12[/tex]
From the z table the area under the normal curve to the left corresponding to -2.12 is
[tex](P< -2.12) = 0.017003[/tex]
Generally the p-value is
[tex]p- value = 0.017003[/tex]
From values obtained we see that [tex]p-value < \alpha[/tex] hence
The decision rule is
Reject the null hypothesis
The conclusion is
There is sufficient evidence to show that the population mean is less than 20.
Generally form the normal distribution table the critical value at a level of significance of [tex]\alpha = 0.05[/tex] is
[tex]z_{0.05} =- 1.645[/tex]
Generally given that [tex]z < z_{0.05}[/tex]
The decision rule is
Reject the null hypothesis
The conclusion is
There is sufficient evidence to show that the population mean is less than 20.
