Answer :
Answer:
[tex]t=\frac{310.9-300}{\frac{31.09}{\sqrt{10}}}=1.109[/tex]
The degrees of freedom are given by:
[tex]df=n-1=10-1=9[/tex]
And the p value would be given by:
[tex]p_v =P(t_{9}>1.109)=0.148[/tex]
Since the p value is higher than the significance level of 0.05 we don't have enough evidence to conclude that the true mean is significantly higher than $300 because we FAIL to reject the null hypothesis.
Step-by-step explanation:
Information given
314 305 344 283 285 310 383 285 300 300
We can calculate the sample mean and deviation with the following formula:
[tex]\bar X =\frac{\sum_{i=1}^n X_i}{n}[/tex]
[tex]s =\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)^2}{n-1}}[/tex]
[tex]\bar X=310.9[/tex] represent the sample mean
[tex]s=31.09[/tex] represent the standard deviation for the sample
[tex]n=10[/tex] sample size
[tex]\mu_o =300[/tex] represent the value to test
[tex]\alpha=0.05[/tex] represent the significance level
t would represent the statistic
[tex]p_v[/tex] represent the p value
Hypothesis to test
We want to verify if the true mean is greater than 300, the system of hypothesis would be:
Null hypothesis:[tex]\mu \leq 300[/tex]
Alternative hypothesis:[tex]\mu > 300[/tex]
The statistic 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{310.9-300}{\frac{31.09}{\sqrt{10}}}=1.109[/tex]
The degrees of freedom are given by:
[tex]df=n-1=10-1=9[/tex]
And the p value would be given by:
[tex]p_v =P(t_{9}>1.109)=0.148[/tex]
Since the p value is higher than the significance level of 0.05 we don't have enough evidence to conclude that the true mean is significantly higher than $300 because we FAIL to reject the null hypothesis.