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3 years ago

How is a debit card like a check?

Mathematics

2 answers:

madam [21]3 years ago

5 0

Erm I would say it comes from the same money pot (account) but not 100% sure what your tryin to ask

Alexxandr [17]3 years ago

5 0

I'm not sure but a debit card can be prepaid and only have a certain amount of money on it sometimes and that's kind of like how a check is only a certain amount of money (+ you don't physically have the money with you)

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Testing the hypothesis, we have that:

The null hypothesis is: $H_0: \mu = 128$

The alternative hypothesis is: $H_1: \mu \neq 128$

b) The p-value of the test is of 0.1212.

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:

$H_0: \mu = 128$

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

$H_1: \mu \neq 128$

Item b:

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

$z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

The parameters are:

$\overline{x}$ is the sample mean.

$\mu$ is the value tested at the null hypothesis.

$\sigma$ is the standard deviation of the population.

n is the sample size.

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

Hence, the value of the test statistic is:

$z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{118.6 - 128}{\frac{33.17}{\sqrt{30}}}$

$z = -1.55$

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 brainly.com/question/25369247

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