Answer :
Answer:
We accept the null hypothesis and reject the alternate hypothesis. There is no evidence to conclude that the population mean is greater than 29. The population mean is less than or equal to 29.
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 29
Sample mean, [tex]\bar{x}[/tex] = 30
Sample size, n = 47
Alpha, α = 0.05
Population standard deviation, σ = 5
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu \leq 29\\H_A: \mu > 29[/tex]
a) This is a one-tailed test because the alternate hypothesis is in greater than direction.
We use One-tailed z test to perform this hypothesis.
b) [tex]z_{stat} > z_{critical}[/tex] , we reject the null hypothesis and accept the alternate hypothesis and if [tex]z_{stat} < z_{critical}[/tex] , we accept the null hypothesis and reject the alternate hypothesis.
c) Formula:
[tex]z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]z_{stat} = \displaystyle\frac{30 - 29}{\frac{5}{\sqrt{47}} } = 1.37[/tex]
d) Now, [tex]z_{critical} \text{ at 0.05 level of significance } = 1.64[/tex]
Since,
[tex]z_{stat} < z_{critical}[/tex]
We accept the null hypothesis and reject the alternate hypothesis. There is no evidence to conclude that the population mean is greater than 29. The population mean is less than or equal to 29.
e) P-value is 0.0853
On the basis of p value we again accept the null hypothesis.