Help plez it is due tommorow

Mathematics

1 answer:

Sphinxa [80]3 years ago

6 0

9:=4/5 10:=7/9do you want me to do the strategies for you. what grade are you in

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Suppose that a recent article stated that the mean time spent in jail by a first-time convicted burglar is 2.5 years. A study wa

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Using the z-distribution, it is found that since the test statistic is greater than the critical value, it can be concluded that the mean length of jail time has increased.

At the null hypothesis, it is tested if the mean length of jail time is still of 2.5 years, that is:

$H_0: \mu = 2.5$

At the alternative hypothesis, it is tested if it has increased, that is:

$H_1: \mu > 2.5$

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 sample.
n is the sample size.

For this problem, the values of the parameters are: $\overline{x} = 3, \mu = 2.5, \sigma = 1.5, n = 26$

Hence, the value of the test statistic is:

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

$z = \frac{3 - 2.5}{\frac{1.5}{\sqrt{26}}}$

$z = 1.7$

The critical value for a right-tailed test, as we are testing if the mean is greater than a value, with a significance level of 0.05, is of $z^{\ast} = 1.645$