I need help with #2!!!

In 2008, the average household debt service ratio for homeowners was 13.2. The household debt service ratio is the ratio of debt

ankoles [38]

Answer:

$t=\frac{13.88-13.2}{\frac{3.14}{\sqrt{44}}}=1.436$

$df=n-1=44-1=43$

$p_v =P(t_{(43)}>1.436)=0.079$

We see that the p value i higher than the significance level so then we FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly higher than 13.2 *the value of 2008 ).

Step-by-step explanation:

Information given

$\bar X=13.88$ represent the sample mean

$s=3.14$ represent the sample standard deviation

$n=44$ sample size

$\mu_o =13.2$ represent the value that we want to test

$\alpha=0.05$ represent the significance level

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test

Hypothesis to test

We want to conduct a hypothesis in order to check if the true mean has increased from 2008 , and the system of hypothesi are:

Null hypothesis: $\mu \leq 13.2$

Alternative hypothesis: $\mu > 13.2$

The statistic for this case is:

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

Calculating the statistic

Replacing the info given we got:

$t=\frac{13.88-13.2}{\frac{3.14}{\sqrt{44}}}=1.436$

P-value

The degrees of freedom are:

$df=n-1=44-1=43$

Since is a right tailed test the p value is:

$p_v =P(t_{(43)}>1.436)=0.079$

Decision

We see that the p value i higher than the significance level so then we FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly higher than 13.2 *the value of 2008 ).

8 0

4 years ago

F(x) = 4x+8<br> g(x) = x+3,<br> Find f(x)*g(x)

klemol [59]

Answer:

Step-by-step explanation:

5 0

4 years ago

Read 2 more answers

Of 1000 randomly selected cases of lung cancer, 823 resulted in death within 10 years.

Arada [10]

Answer:

a) $0.823 - 1.96\sqrt{\frac{0.823(1-0.823)}{1000}}=0.799$

$0.823 + 1.96\sqrt{\frac{0.823(1-0.823)}{1000}}=0.847$

The 95% confidence interval would be given by (0.799;0.847)

b) $n=\frac{0.823(1-0.823)}{(\frac{0.03}{1.96})^2}=621.79$

And rounded up we have that n=622

c) $n=\frac{0.5(1-0.5)}{(\frac{0.03}{1.96})^2}=1067.11$

And rounded up we have that n=1068

Step-by-step explanation:

Part a

$\hat p=\frac{823}{1000}=0.823$

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by: