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olya-2409 [2.1K]
3 years ago
5

Please help....thanks :)

Chemistry
1 answer:
Naya [18.7K]3 years ago
6 0

Answer: someone help me with my work its so  hard

Explanation:

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Answer:

b) \bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}

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: \mu =7

Alternative hypothesis \mu \neq 7

The statistic is to check this hypothesi is given by:

t = \frac{\bar X -\mu}{\frac{s}{\sqrt{n}}}

We know the following info from the problem:

\bar X = 6.4 , s=0.5, n =30

Replacing we got:

t = \frac{6.4-7}{\frac{0.5}{\sqrt{30}}}= -6.573

And the p value would be:

p_v= 2*P(Z

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:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}

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.

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