Two times a number added to another number is 25. Three times the first number minus the other number is 20
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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 (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
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,
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.