Question

3. The average age of online consumers a few years ago was 23.3 years with a standard deviation of 5.3 years. It is believed the average has changed since older individuals gained more confidence with internet use. Recently a random sample of 20 online shoppers were selected. The sample had an average age of 26.2. Determine whether or not the average age of online consumers has changed over the last few years, use α =0.01. What is the p-value of the result?

Answer 1

Answer:

We cant say, with α =0.01, that the average age changed. The p-value is 0.014

Step-by-step explanation:

As a consecuence of the Central Limit Theorem, the mean sample has a distribution approximately normal, with unknown mean and standard deviation $\sigma = 5.3/\sqrt{20} = 1.1851$ (the standard deviation of one single sample divided by the  sqaure root of the sample lenght).

The null hypothesis H₀ is that the average age is still 23.3 (the mean is 23.3). The alternative hypothesis is that the mean is different. We want to see if we can refute H₀ with significance level α =0.01.

Lets call X the mean of a random sample of 20 online shoppers. As we discuse above, X is approximately normal with unknown mean and standard deviation equal to 1.1851 . If we take the hypothesis H₀ as True, then the mean will be 23.3, and if we standarize X, we have that the distribution

$W = \frac{X - 23.3}{1.1851}$

Will have a distribution approximately normal, with mean 0 and standard deviation 1.

The values of the cummulative ditribution of the normal function can be found in the attached file. Since we want a significance level of 0.01, then we need a value Z such that

P(-Z < W < Z) = 0.99

For the symmetry of the standard Normal distribution, we have that Φ(Z) = 1-Φ(-Z), where Φ is the cummulative distribution function of the standard Normal random variable. Therefore, we want Z such that

Φ(Z) = 0.995

If we look at the table, we will found that Z = 2.57, thus,

$0.99 = P(-2.57 < W < 2.57) = P(-2.57 < \frac{X-23.3}{1.1851} < 2.57)\\ = P(-2.57*1.1851 + 23.3 < X < 2.57*1.1851 + 23.3) = P(20.254 < X < 26.345)$

Thus, we will refute the hypothesis if the observed value of X lies outside the interval [20.254, 26.345].

Since the observed value is 26.2, then we dont refute H₀, So we dont accept that the number average age of online consumers has changed.

For us to refute the hypothesis we need Z such that, for the observed value, |W| > Z. Replacing X by 26.2, we have that [tex] W = \frac{26.2-23.3}{1.1851} = 2.4470. We can observe that Φ(2.447) = 0.993, substracting that amount from 1 and multiplying by 2 (because it can take low values too), we obtain that the p-value is 0.014.  

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