Respuesta :
Answer:
[tex]t=\frac{-34.90-0}{\frac{95.66}{\sqrt{10}}}=-1.154[/tex]
The degrees of freedom are:
[tex] df = n-1= 10-1=9[/tex]
The p value would be given by:
[tex]p_v =P(t_{9}<-1.154)=0.139[/tex]
The p value is a large value and if we use a significance level of 0.05 or 0.1 we see that we can FAIL to reject the null hypothesis. And then we don't have enough evidence to conclude that her clients would save money on average by switching to the online company
Step-by-step explanation:
Information given
[tex]\bar X=-34.90[/tex] represent the mean for the difference
[tex]s=95.66[/tex] represent the sample deviation for the difference
[tex]n=10[/tex] sample size
[tex]\mu_o =0[/tex] represent the value that we want to test
[tex]\alpha[/tex] represent the significance level
t would represent the statistic
[tex]p_v[/tex] represent the p value
System of hypothesis
We want to determine if that her clients would save money on average by switching to the online company, the system of hypothesis would be:
Null hypothesis:[tex]\mu \geq 0[/tex]
Alternative hypothesis:[tex]\mu < 0[/tex]
The statistic for this case is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
Replacing the info given we got:
[tex]t=\frac{-34.90-0}{\frac{95.66}{\sqrt{10}}}=-1.154[/tex]
The degrees of freedom are:
[tex] df = n-1= 10-1=9[/tex]
The p value would be given by:
[tex]p_v =P(t_{9}<-1.154)=0.139[/tex]
The p value is a large value and if we use a significance level of 0.05 or 0.1 we see that we can FAIL to reject the null hypothesis. And then we don't have enough evidence to conclude that her clients would save money on average by switching to the online company
Answer:
Fail to reject the null because the evidence is not statistically significant that her clients would save money on average
Step-by-step explanation:
The null hypothesis is that car insurance from an online company is not cheaper than from local price.
using a left sided t test- t = (mean difference between online price (sample) and the local price) / (standard deviation/√n-1)
mean difference = -34.90, SD = 95.66, n = 10
t test = -34.90 / (95.66/√(10-1))
= -34.90 / (95.66/√9)
= -34.90 / (95.66/3)
= -34.90 / 31.88667
= -1.094.
df = n-1 = 9
Looking up the table for under 9 df you are about 85% at the 0.05 level of significance using the left sided table away (-1.833).
P value (using a t-test calculator) = .151184 which is higher than .05
So, this evidence is not statistically significant that her clients would save money on average .
