Question

The producers of a new toothpaste claim that it prevents more cavities than other brands of toothpaste. A random sample of 60 people used the new toothpaste for 6 months. The mean number of cavities at their next checkup is 1.5. In the general population, the mean number of cavities at a 6-month checkup is 1.73 (σ = 1.12). a. Is this a one- or two-tailed test? b. What are H0 and Ha for this study? c. Compute zobt. d. What is zcv? e. Should H0 be rejected? What should the researcher conclude? f. Determine the 95% confidence interval for the population mean, based on the sample mean.

Answer 1

Answer:

a) Is a one tailed left test.

b) Null hypothesis: There is no difference between new and other toothpastes in the mean number of cavities of the general population $\mu \geq 1.73$

Alternative hypothesis : New toothpaste reduces cavities compared to other brands $\mu$

c) $z=\frac{1.5-1.73}{\frac{1.12}{\sqrt{60}}}=-1.59$  

d)  $z_{crit}=-1.64$

e) $p_v =P(z$  

If we compare the p value and the significance level for example $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the mean number of cavities at a 6-month checkup is not significantly lower than 1.73 at 5% of significance.  

f) The 95% confidence interval would be given by (1.217;1.783)

Step-by-step explanation:

1) Data given and notation  

$\bar X=1.5$ represent the mean annual premium value for the sample  

$\sigma=1.12$ represent the population standard deviation  

$n=60$ sample size  

$\mu_o =1.73$ represent the value that we want to test  

$\alpha$ 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.  

a) Is this a one- or two-tailed test?

Is a one tailed left test.

b) What are H0 and Ha for this study?

Null hypothesis: There is no difference between new and other toothpastes in the mean number of cavities of the general population $\mu \geq 1.73$

Alternative hypothesis : New toothpaste reduces cavities compared to other brands $\mu$

c) Compute zobt

The statistic for this case 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:  

$z=\frac{1.5-1.73}{\frac{1.12}{\sqrt{60}}}=-1.59$  

d) What is zcv?

In order to find the critical value we need to find a value on the normal standard distribution such that:

$P(Z$ where 0.05 represent the significance level selected.

And this value is $z_{crit}=-1.64$

e) Should H0 be rejected? What should the researcher conclude?

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

$p_v =P(z$  

Conclusion  

If we compare the p value and the significance level for example $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the mean number of cavities at a 6-month checkup is not significantly lower than 1.73 at 5% of significance.  

f) Determine the 95% confidence interval for the population mean, based on the sample mean

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

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

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 tabel to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $z_{\alpha/2}=1.96$

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

$1.5-1.96\frac{1.12}{\sqrt{60}}=1.217$    

$1.5+1.96\frac{1.12}{\sqrt{60}}=1.783$

So on this case the 95% confidence interval would be given by (1.217;1.783)