Question

As a part of her studies, Jolyn gathered data on the length of time between dentist visits for a sample of 23 individuals. She works through the testing procedure: H0:μ=187; Ha:μ>187 α=0.05 The test statistic is t0=x¯−μ0sn√=2.266. The critical value is t0.05=1.717. At the 5% significance level, does the data provide sufficient evidence to conclude that the mean length of time between dentist visits is more than 187 days? Select the correct answer below:
a. We should reject the null hypothesis because t0 b. We should not reject the null hypothesis because t0 c. We should reject the null hypothesis because t0>tα. So, at the 5% significance level, the data provide sufficient evidence to conclude that the average length of time between dentist visits is greater than 187 days.
d. We should not reject the null hypothesis because t0>tα. So, at the 5% significance level, the data do not provide sufficient evidence to conclude that the average length of time between dentist visits is greater than 187 days.

Answer 1

Answer:

c. We should reject the null hypothesis because t0>tα. So, at the 5% significance level, the data provide sufficient evidence to conclude that the average length of time between dentist visits is greater than 187 days.

Step-by-step explanation:

1) Data given and notation  

$\bar X$ represent the mean average

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

$n=23$ sample size  

$\mu_o =187$ 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)  

2) State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is hgiher than 187, the system of hypothesis would be:  

Null hypothesis:$\mu \leq 187$  

Alternative hypothesis:$\mu > 187$  

Since we don't know the population deviation, 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".  

3) Calculate the statistic  

On this case the value for the satistic calculated was 2.266

4)P-value   and critical value

First we need to calculate the degrees os freedom given by:

$df=n-1=23-1=22$

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

$p_v =P(t_{22}>2.266)=0.0168$  

The critical value owuld be a quantile of the t distribution with 22 degrees of freedom that accumulates 0.05 of the area on the right tail. And is given by:

$t_{crit}=1.717$

5) 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 reject the null hypothesis, and the best conclusion would be:

c. We should reject the null hypothesis because t0>tα. So, at the 5% significance level, the data provide sufficient evidence to conclude that the average length of time between dentist visits is greater than 187 days.