textRequest reports that adults 18–24 years old send and receive 128 texts every day. Suppose we take a sample of 25–34 year olds to see if their mean number of daily texts differs from the mean for 18–24 year olds reported by TextRequest.a. State the null and alternative hypotheses we should use to test whether the popu-lation mean daily number of texts for 25–34 year olds differs from the population daily mean number of texts for 18–24 year olds.b. Suppose a sample of thirty 25–34 year olds showed a sample mean of 118.6 texts per day. Assume a population standard deviation of 33.17 texts per day and com-pute the p-value.c. With a = .05 as the level of significance, what is your conclusion? d. Repeat the preceding hypothesis test using the critical value approach.d. Repeat the preceding hypothesis test using the critical value approach.

c) Since the p-value of the test is of 0.1212 > 0.05, 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 |z| = 1.55 < 1.96, 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 test if the mean is the same, that is, of 128 texts every day, hence:

[tex]H_0: \mu = 128[/tex]

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

[tex]H_1: \mu \neq 128[/tex]

Item b:

We have the standard deviation for the population, thus, the z-distribution is used. The test statistic is given by:

[tex]z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

The parameters are:

[tex]\overline{x}[/tex] is the sample mean.

[tex]\mu[/tex] is the value tested at the null hypothesis.

[tex]\sigma[/tex] is the standard deviation of the population.

n is the sample size.

For this problem, the values of the parameters are: [tex]\overline{x} = 118.6, \mu = 128, \sigma = 33.17, n = 30[/tex]

Hence, the value of the test statistic is:

[tex]z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

[tex]z = \frac{118.6 - 128}{\frac{33.17}{\sqrt{30}}}[/tex]

[tex]z = -1.55[/tex]

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 p-value of the test is of 0.1212 > 0.05, 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 two-tailed test with 95% confidence level is |z| = 1.96.

Since |z| = 1.55 < 1.96, 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 https://brainly.com/question/25369247