Question

A few years ago the average age of online consumers was 23.3 years. A random sample of 50 individuals who made an online purchase in 2018 found a mean age of 24.5 years with a standard deviation of 4.1 years.

Is there sufficient evidence to show that the mean age of online consumers has decreased? Identify the appropriate null and alternative hypotheses.

a. H0: μ = 23.3 Ha: μ > 23.3

b. H0: μ = 23.3 Ha: μ ≠ 23.3

c. H0: μ = 23.3 Ha: μ < 23.3

d. H0: μ = 24.5 Ha: μ < 24.5

e. H0: μ = 24.5 Ha: μ > 24.5

Answer 1

c. H0: μ = 23.3 Ha: μ < 23.3

Explanation:

The null hypothesis is related to the expected value.

A few years ago the average age of online consumers was 23.3 years.

This means that the null hypoteshis, that is, H0, is μ = 23.3.

The alternate hypothesis is related to the question made.

Is there sufficient evidence to show that the mean age of online consumers has decreased?

The question is if the age of 23.3 has decreases. This means that the alternate hypothesis, that is, Ha, is μ < 23.3.

So the correct answer is:

c. H0: μ = 23.3 Ha: μ < 23.3

Answer 2

We want to verify if the mean age of online consumers has decreased (23.3), the system of hypothesis would be:

Null hypothesis:
`\mu \leq 23.3`

Alternative hypothesis:
`\mu > 23.3`

And the correct option would be:

a. H0: μ = 23.3 Ha: μ > 23.3

`t=(24.5-23.3)/((4.1)/(√(50)))=2.070`

The degrees of freedom are given by:

`df=n-1=50-1=49`

And the p value would be:

`p_v =P(t_((49))>2.070)=0.0219`

Explanation:

Information provided

`\bar X=24.5` represent the sample mean for the age of online purchases

`s=4.1` represent the sample standard deviation

`n=50` sample size

`\mu_o =23.3` represent the value to verify

t would represent the statistic

`p_v` represent the p value

System of hypothesis

We want to verify if the mean age of online consumers has decreased (23.3), the system of hypothesis would be:

Null hypothesis:
`\mu \leq 23.3`

Alternative hypothesis:
`\mu > 23.3`

And the correct option would be:

a. H0: μ = 23.3 Ha: μ > 23.3

The statistic is given by:

`t=(\bar X-\mu_o)/((s)/(√(n)))` (1)

Replacing the info we got:

`t=(24.5-23.3)/((4.1)/(√(50)))=2.070`

The degrees of freedom are given by:

`df=n-1=50-1=49`

And the p value would be:

`p_v =P(t_((49))>2.070)=0.0219`