Testing the hypothesis, we have that:
a)
The null hypothesis is: 
The alternative hypothesis is: 
b) The p-value of the test is of 0.1212.
c) Since the <u>p-value of the test is of 0.1212 > 0.05</u>, we cannot conclude that the mean daily number of texts for 25–34 year olds differs from the population daily mean number of texts for 18–24 year olds.
d) Since <u>|z| = 1.55 < 1.96</u>, we cannot conclude that the mean daily number of texts for 25–34 year olds differs from the population daily mean number of texts for 18–24 year olds.
Item a:
At the null hypothesis, we <u>test if the mean is the same</u>, that is, of 128 texts every day, hence:

At the alternative hypothesis, we <u>test if the mean is different</u>, that is, different of 128 texts every day, hence:

Item b:
We have the <u>standard deviation for the population</u>, thus, the z-distribution is used. The test statistic is given by:
The parameters are:
is the sample mean.
is the value tested at the null hypothesis.
is the standard deviation of the population.
n is the sample size.
For this problem, the values of the parameters are: 
Hence, the value of the test statistic is:



Since we have a two-tailed test, as we are testing if the mean is different of a value, the p-value is P(|z| < 1.55), which is 2 multiplied by the p-value of z = -1.55.
Looking at the z-table, z = -1.55 has a p-value of 0.0606
2(0.0606) = 0.1212
The p-value of the test is of 0.1212.
Item c:
Since the <u>p-value of the test is of 0.1212 > 0.05</u>, we cannot conclude that the mean daily number of texts for 25–34 year olds differs from the population daily mean number of texts for 18–24 year olds.
Item d:
Using a z-distribution calculator, the critical value for a <u>two-tailed test</u> with <u>95% confidence level</u> is |z| = 1.96.
Since <u>|z| = 1.55 < 1.96</u>, we cannot conclude that the mean daily number of texts for 25–34 year olds differs from the population daily mean number of texts for 18–24 year olds.
A similar problem is given at brainly.com/question/25369247