Question

A sample of 35 observations is selected from a normal population. The sample mean is 53, and the population standard deviation is 5. Conduct the following test of hypothesis using the 0.01 significance level.

H0: μ = 54

H1: μ ≠ 54

Is this a one- or two-tailed test?

What is the decision rule?

Answer 1

Answer:

Null hypothesis:$\mu = 54$  

Alternative hypothesis:$\mu \neq 54$  

Two tailed test

So we reject the null hypothesis is $z< -2.58$ or if $z>2.58$

Our calculated value is not on the rejection zone so then we don't have enough evidence to reject the null hypothesis at 0.01 of significance for this case.

Step-by-step explanation:

Data given and notation  

$\bar X=53$ represent the sample mean

$\sigma=5$ represent the population standard deviation

$n=35$ sample size  

represent the value that we want to test

$\alpha=0.01$ represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

$p_v$ represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is equal to 54 or no, the system of hypothesis would be:  

Null hypothesis:$\mu = 54$  

Alternative hypothesis:$\mu \neq 54$  

If we analyze the size for the sample is > 30 but and we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$  (1)  

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

$z=\frac{53-54}{\frac{5}{\sqrt{35}}}=-1.183$  

Decision rule

Since we are conducting a bilateral test we need to find a value on the normal standarrd distribution that accumulates 0.005 of the area on each tail. And we can use the following excel code: "=NORM.INV(0.005;0;1)"

So we reject the null hypothesis is $z< -2.58$ or if $z>2.58$

Our calculated value is not on the rejection zone so then we don't have enough evidence to reject the null hypothesis at 0.01 of significance for this case.