Question

The Late Show with David Leterman is seen by a relatively large percentage of household members who record the show for viewing at a more convenient time. The show’s marketing claims that the mean income of households with DVR’s is greater than $40,000. Test that claim using a 0.5% significance level.
A random sample of 1700 households with DVR’s produces a sample mean of $41,182 and a sample standard deviation of $19,990.
Perform a test of hypothesis to check the validity of the marketing managers claim by completing the following steps:

a) State the hypothesis
b) Find the standardized test statistic
c) Find the P-value
d) At a 0.01 state your decision

Answer 1

Answer:

a) Null hypothesis:$\mu \leq 40000$  

Alternative hypothesis:$\mu > 40000$  

b) $t=\frac{41182-40000}{\frac{19990}{\sqrt{1700}}}=2.438$    

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

$df=n-1=1700-1=1699$  

Since the sample size is large enough we cna use the z distribution as an approximation for the statsitic on this case.

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

$p_v =P(t_{(1699)}>2.438)=0.0074$  

d) If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis at 1% of signficance.  So we can conclude that the true mean is higher than 40000 at the significance level assumed.

Step-by-step explanation:

Data given and notation  

$\bar X=41182$ represent the sample mean

$s=19990$ represent the sample standard deviation

$n=1700$ sample size  

$\mu_o =40000$ represent the value that we want to test

$\alpha=0.01$ 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 true mean is greater than 40000, the system of hypothesis would be:  

Null hypothesis:$\mu \leq 40000$  

Alternative hypothesis:$\mu > 40000$  

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".  

Part b: Calculate the statistic

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

$t=\frac{41182-40000}{\frac{19990}{\sqrt{1700}}}=2.438$    

Part c: P-value

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

$df=n-1=1700-1=1699$  

Since the sample size is large enough we cna use the z distribution as an approximation for the statsitic on this case.

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

$p_v =P(t_{(1699)}>2.438)=0.0074$  

Part d: Conclusion  

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis at 1% of signficance.  So we can conclude that the true mean is higher than 40000 at the significance level assumed.