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Suppose the standard deviation of a normal population is known to be 3, and H0 asserts that the mean is equal to 12. A random sa

blagie [28]

Answer:

$z=\frac{12.95-12}{\frac{3}{\sqrt{36}}}=1.9$

$p_v =P(z>1.9)=0.0287$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 12 at 5% of signficance.

Step-by-step explanation:

Data given and notation

$\bar X=12.95$ represent the sample mean

$\sigma=3$ represent the population standard deviation for the sample

$n=36$ sample size

$\mu_o =12$ represent the value that we want to test

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

z would represent the statistic (variable of interest)

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

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is equal to 12, the system of hypothesis would be:

Null hypothesis: $\mu \leq 12$

Alternative hypothesis: $\mu > 12$

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-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:

$z=\frac{12.95-12}{\frac{3}{\sqrt{36}}}=1.9$

P-value

Since is a right tailed test the p value would be:

$p_v =P(z>1.9)=0.0287$

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 12 at 5% of signficance.

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3 years ago

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suppose that you have 3000$ to invest. which investment yields the greater return over a 10 year period: 8.04% compounded daily

Ksenya-84 [330]

Answer: Option A: 8.04% compounded daily

Step-by-step explanation:

$A = P\bigg(1+\dfrac{r}{n}\bigg)^{nt}\qquad where\\\\\bullet A = \text{accrued amount (principal plus interest earned)}\\\bullet P = \text{principal (amount invested)}\\\bullet r = \text{rate (in decimal form)}\\\bullet n=\text{number of times compounded in one year}\\\bullet t=\text{time (number of years)}\\\\\\Option\ A:\\A = \text{unknown}\\P=3000\\r=8.04\%\rightarrow 0.0804\\n=\text{daily}\rightarrow 365\\t=10\\\\A=3000\bigg(1+\dfrac{0.0804}{365}\bigg)^{365\times 10}\\\\.\ =\$ 6,702.79$

$Option\ B:\\A = \text{unknown}\\P=3000\\r=8.1\%\rightarrow 0.081\\n=\text{quarterly}\rightarrow 4\\t=10\\\\A=3000\bigg(1+\dfrac{0.081}{4}\bigg)^{4\times 10}\\\\.\ =\$ 6,689.37$