Mathematics, 11.01.2020 02:31 almaloney01

The station manager thinks the station doesn't play 10 songs per hour, but he doesn't know whether the station plays more or less than 10 songs per hour.

a what are the null and alternative hypotheses for this test?

b what test will you use to evaluate the alternative hypothesis? what are the conditions of this test and are they met in this situation?

c calculate your test statistic and p-value. show your work.

d what's your critical t value and what's your conclusion at \alpha = .05? compare the critical t value and the p-value to those from the test from question 1 and explain the difference.

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Answer from: groverparham3

a) Null hypothesis: $\mu = 10$

Alternative hypothesis: $\mu \neq 10$

b) If we analyze the size for the sample is < 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

We assume that the distribution is approximately normal.

c) $t=\frac{10.333-10}{\frac{1.877}{\sqrt{15}}}=0.687$

P-value

The first step is calculate the degrees of freedom, on this case:

df=n-1=15-1=14

Since is a two side test the p value would be:

$p_v =2*P(t_{(14)}0.687)=0.503$

d) For this case since we are conducting a two tailed test we need to critical values. We need to find a t value that accumulates 0.025 of the area on each tail of the t distribution with 14 degrees of freedom. and for this case the critical values are:

$t_{\alpha/2}= \pm 2.145$

Since our calculated values is less than the critical we have enough evidence to FAIL to reject the null hypothesis.

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v\alpha$ so we can conclude that we have enough evidence to FAIL reject the null hypothesis, so we can conclude that the true mean is not significantly different from 10 songs per hour.

With both methods critical value and p value we got the same result, FAIL to reject the null hypothesis.

Step-by-step explanation:

Assuming this previous info: "Management at a radio station wants to know the number of songs it plays per hour. A random sample over the past month yields the following counts for songs per hour: 9, 13, 12, 8, 7, 10, 11, 10, 8, 12, 13, 9, 10, 12, and 11. "

Data given and notation

We can caclulate the sample mean with the following formula:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

$\bar X=10.333$ represent the mean from the data

And we can calculate the sample deviation with the following formula:

$s = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

s=1.877 represent the sample standard deviation for the sample

n=15 sample size

$\mu_o =10$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

p_v represent the p value for the test (variable of interest)

Part a: State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is different from 10, the system of hypothesis would be:

Null hypothesis: $\mu = 10$

Alternative hypothesis: $\mu \neq 10$

Part b

If we analyze the size for the sample is < 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

We assume that the distribution is approximately normal.

Part c

We can replace in formula (1) the info given like this:

$t=\frac{10.333-10}{\frac{1.877}{\sqrt{15}}}=0.687$

P-value

The first step is calculate the degrees of freedom, on this case:

df=n-1=15-1=14

Since is a two side test the p value would be:

$p_v =2*P(t_{(14)}0.687)=0.503$

Part d

For this case since we are conducting a two tailed test we need to critical values. We need to find a t value that accumulates 0.025 of the area on each tail of the t distribution with 14 degrees of freedom. and for this case the critical values are:

$t_{\alpha/2}= \pm 2.145$

Since our calculated values is less than the critical we have enough evidence to FAIL to reject the null hypothesis.

With both methods critical value and p value we got the same result, FAIL to reject the null hypothesis.

Answer from: Quest

Try a or d, hope that

Answer from: Quest

don't know what that says sorry i can't you

step-by-step explanation:

idk what it says

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