Question

5.11 Play the piano. Georgianna claims that in a small city renowned for its music school, the average child takes at least 5 years of piano lessons. We have a random sample of 20 children from the city, with a mean of 4.6 years of piano lessons and a standard deviation of 2.2 years. (a) Evaluate Georgianna’s claim using a hypothesis test. (b) Construct a 95% confidence interval for the number of years students in this city take piano lessons, and interpret it in context of the data. (c) Do your results from the hypothesis test and the confidence interval agree? Explain your reasoning.

Answer 1

Answer:

a) $t=\frac{4.6-5}{\frac{2.2}{\sqrt{20}}}=-0.813$    

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

$df=n-1=20-1=19$  

$p_v =P(t_{(19)}$  

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't conclude that the true mean is lower than 5 years at 5% of signficance.  

b) $4.6-2.09\frac{2.2}{\sqrt{20}}=3.572$    

$4.6+2.09\frac{2.2}{\sqrt{20}}=5.628$    

So on this case the 95% confidence interval would be given by (3.572;5.628)

c) Yes from the hypothesis test we fail to reject the null hypothesis that the true mean is equal or higher to 5  and from the confidence interval the limits contains the value 5, so then the two procedures are showing the same result FAIL to reject the null hypothesis.

Step-by-step explanation:

Data given and notation  

$\bar X=4.6$ represent the sample mean

$s=2.2$ represent the sample standard deviation

$n=20$ sample size  

$\mu_o =5$ 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 at least 5 years, the system of hypothesis would be:  

Null hypothesis:$\mu \geq 5$  

Alternative hypothesis:$\mu < 5$  

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:  

$t=\frac{4.6-5}{\frac{2.2}{\sqrt{20}}}=-0.813$    

P-value

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

$df=n-1=20-1=19$  

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

$p_v =P(t_{(19)}$  

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 have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is lower than 5 years at 5% of signficance.  

Part b

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$   (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=20-1=19$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,19)".And we see that $t_{\alpha/2}=2.09$

Now we have everything in order to replace into formula (1):

$4.6-2.09\frac{2.2}{\sqrt{20}}=3.572$    

$4.6+2.09\frac{2.2}{\sqrt{20}}=5.628$    

So on this case the 95% confidence interval would be given by (3.572;5.628)

And as we can see the confidence interval contains the value 5.

Part c

Yes from the hypothesis test we fail to reject the null hypothesis that the true mean is equal or higher to 5  and from the confidence interval the limits contains the value 5, so then the two procedures are showing the same result FAIL to reject the null hypothesis.