Mathematics, 23.03.2020 17:52 loveeegabbie7071

Use the t-distribution and the sample results to complete the test of the hypotheses. Use a 5% significance level. Assume the results come from a random sample, and if the sample size is small, assume the underlying distribution is relatively normal.

Test H0: μ= 4 vs Ha: μ≠4 using the sample results x= 4.8, s= 2.3, with n= 15.

(a) Give the test statistic and the p-value.

Round your answer for the test statistic to two decimal places and your answer for the p-value to three decimal places.

test statistic = _

p-value = _

(b) What is the conclusion?

Reject or H0.

or

Do not reject H0.

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Answer from: kaylarojascliff

a) $t=\frac{4.8-4}{\frac{2.3}{\sqrt{15}}}=1.347$

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

df=n-1=15-1=14

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

$p_v =2*P(t_{(14)}1.347)=0.200$

b) 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 DON'T have enough evidence to reject the null hypothesis, so we can't conclude that the true mean is different from 4 at 1% of signficance.

Do not reject H0.

Step-by-step explanation:

Data given and notation

$\bar X=4.8$ represent the sample mean

s=2.3 represent the sample standard deviation

n=15 sample size

$\mu_o =68$ 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 4, the system of hypothesis would be:

Null hypothesis: $\mu = 4$

Alternative hypothesis: $\mu \neq 4$

If we analyze the size for the sample is < 30 and 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".

Calculate the statistic

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

$t=\frac{4.8-4}{\frac{2.3}{\sqrt{15}}}=1.347$

P-value

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

df=n-1=15-1=14

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

$p_v =2*P(t_{(14)}1.347)=0.200$

Part b: Conclusion

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 DON'T have enough evidence to reject the null hypothesis, so we can't conclude that the true mean is different from 4 at 1% of signficance.

Do not reject H0.

Answer from: Quest

The answer is 16.64 feetyour welcome

Answer from: Quest

pretty sure 750 each month!

step-by-step explanation:

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