Respuesta :
Answer:
b) [tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]
The confidence interval for this case is given (6.21, 6.59)
So we can conclude at 95% of confidence that the true mean for the PH concentration is between 6.21 and 6.59 moles per liter
c) Since the confidence interval not contains the value 7 we reject the hypothesis that the true mean is equal to 7. And the same result was obtained with the t test for the true mean.
Explanation:
We assume that part a is test the claim. And we can conduct the following hypothesis test:
Null hypothesis: [tex]\mu =7[/tex]
Alternative hypothesis [tex]\mu \neq 7[/tex]
The statistic is to check this hypothesi is given by:
[tex] t = \frac{\bar X -\mu}{\frac{s}{\sqrt{n}}}[/tex]
We know the following info from the problem:
[tex]\bar X = 6.4 , s=0.5, n =30[/tex]
Replacing we got:
[tex]t = \frac{6.4-7}{\frac{0.5}{\sqrt{30}}}= -6.573[/tex]
And the p value would be:
[tex]p_v= 2*P(Z<-6.573) = 4.93x10^{-11}[/tex]
Since the p value is very low compared to the significance assumed of 0.05 we have enough evidence to reject the null hypothesis that the true mean is equal to 7 moles/liter
Part b
The confidence interval is given by:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]
The confidence interval for this case is given (6.21, 6.59)
So we can conclude at 95% of confidence that the true mean for the PH concentration is between 6.21 and 6.59 moles per liter
Part c
Since the confidence interval not contains the value 7 we reject the hypothesis that the true mean is equal to 7. And the same result was obtained with the t test for the true mean.
The confidence interval of the sample has been 6.21, 6.59.
The statistical analysis of the experiment has been helpful in the determination of the confidence in the result.
The part A of the question has been assumed to be the null hypothesis for the concentration equal to 7 moles per liter.
Confidence Interval of the Sample:
The hypothesis has been rejected and the mean for the sample value has been 6.4 moles/liter.
The confidence interval of 95% has been the representation of the percent of samples having true mean value 7.
The interval for the true mean value has been given as 6.21-6.59. The confidence level, CI has been interpreted by the z value.
[tex]CI=\dfrac{1.96\;\times\;\sigma}{\sqrt{n} }[/tex]
The population size has been n = 30
The standard deviation for the sample, [tex]\sigma=0.5[/tex]
Substituting the values:
[tex]CI=\dfrac{1.96\;\times\;0.5}{\sqrt{30 } }\\CI=0.17[/tex]
The margin intervals for the sample have been found to be 0.17.
The mean of the sample has been, 6.4 moles per liters.
The range of 95% confidence interval is given as:
[tex]CI= (6.4-0.17)-(6.4+0.17)\\CI=6.21-6.59[/tex]
The confidence interval of the sample has been 6.21, 6.59.
The confidence interval has been consistent with the result, as there has been no value from the null hypothesis.
For more information about confidence interval, refer to the link:
https://brainly.com/question/2396419
